@@allozovsky it's not inconsistent. There's one in Q I and another in Q III separated by p u/2 rotation Much like using an inverse trig function to return an answer in a limited domain where the actual solution may be outside of that domain. Recall that -1 is a real number, I is imaginary, and thr square roots are complex. Complex numbers don't behave quite the same as real and purely imaginary numbers.
@@onradioactivewaves If √𝒊 = ±(1 + 𝒊)/√2 gives two complex square roots, then √(−1) also should return two complex square roots, that is √(−1) = ±𝒊, isn't it? Otherwise it is inconsistent. That's pretty strange that Steve often evaluates in his videos multivalued complex functions alright, but at the same time uses only single 𝒊 for the square root of −1.
I happened to watch this on my break from studying before my leaving cert maths exam. Square root of a complex number was on the exam and I got the right answer using this method. What a crazy lucky coincidence.
I don't know why people keep complaining about this guy's solution to the problem, and why they offer geometric proofs instead. I really like this guy's answer because it uses only the simplest arithmetic/algebra and the simplest definition of a complex number: a + bi. Also, he did it in an incredibly concise manner. (My only complaint was at the end where he could have used a² - b² = (a+b)(a-b) = 0 --> a = ±b. But, typing this out, I suddenly realise how clever he is that even *factorization* doesn't need to be used in his answer.) This answer is teachable on someone's very first lesson on complex numbers; even the average 15-year-old will comprehend it very well, and he's earned my amazement.
Billy Ma-gusta thank you Billy. I think people just got too excited. It's like once they see derivatives, they want to show algebra students to take the derivative to find the vertex of a parabola.
Billy Ma-gusta You want concise? Polar form: R*exp(iθ) = sqrt(i) Square: R^2*exp(2iθ) = i = exp(i(π/2 + 2πn)), n integer Identify: R=1, θ=π/4+nπ Done. Or if you prefer rectangular form the unique representations are: ±(1+i)/sqrt(2) Tl;dr: Rectangular is great for addition and subtraction, polar for multiplication, powers and roots. Right tool for the job.
My point: Your working is as opaque as it gets. For all the average high schooler or TH-camr can see, you're just writing a whole bunch of Greek because nothing is explained. The polar form of complex numbers isn't explained at all. The polar form of √i specifically also isn't explained, neither is i. Converting between the polar form and the rectangular form also isn't explained. You cannot call this concise when you leave so much unexplained; someone who doesn't know the required background knowledge would call this gibberish. (Like how I call bullshit when I am told "using string theory, we can show that 11 physical dimensions exist". Just an example, don't digress pls.) And the greater point: the polar form does not NEED to be explained in order to prove this result, as bprp has shown. Simpler is better in maths, and a short proof is not necessarily a simple proof.
I would happily agree that I assume too much if I for some video on an integral complain about someone not using residuals and Cauchy's integral formula to solve an integral on the real line since it would be trivial that way. But assuming knowledge of the polar form of a complex number, which is as fundamental as the rectangular form, is not a stretch imo. As for the specific polar forms. I never use the polar form of sqrt(i), only the polar form of i, to keep things simpler. I could happily have shown a justification of the polar form of i, but it's a bit clunky without being able to show any figures. Basically you have an angle of π/2 to i and any 2π increment of that still leaves you at i. If I had shown this on a blackboard the justification is explained in a few seconds. The conversion to rectangular form is really not part of my solution, I'm happy to stay in polar form. My point is that this result is trivial in the polar world. It also gives some really nice insights about how the root behaves in general, which is totally lost in rectangular form. Using polar form the result also trivially generalizes to any real power of i. Just like addition/subtraction is trivial in rectangular but a mess in polar form. Sure you could technically use the strategy as shown in this video to describe the 5th or 50th root of i as well, but it would be very cumbersome and very brute force.
you actually don't need factorization to get that a = b. 1. a² - b² = 0 -> a² = b², therefore a and b have the same magnitude. 2. 2ab = 1, ab = 1/2, therefore a and b have the same sign (if the product of 2 real numbers is positive, then 2 real numbers have the same sign) 3. a = b, because a and b have the same magnitude (1) and the same sign (2) (definition of equality for real numbers) 4.a*a = 1/2 = a², from (3) and (2) 5. a = ± sqrt(1/2), sqrt both sides of (4) 6. b = ± sqrt(1/2), from (3) I actually prefer this method of solving the system of equations over the video because it takes less steps and is more intuitive. the videos method involves a lot of seemingly arbitrary moving of symbols around while each of my 6 steps have much clearer purpose.
There is a simpler way without the polar form. We know a² - b² = 0 and 2ab = 1, therefore a² = b² a = b or a = - b. But because of the second equation we can cancel out a = - b as a possibility. That means a = b and if we plug this in the second equation we get: 2a² = 1 a = +/ - (1/sqrt(2)). Thats it. :D
I appreciate the depth of this explanation, rather than memorizing forms, for the sake of speed and ease, you showed me understanding. I appreciate knowing why, over mechanical speed.
@@CorvusSapien It's not a solution. You write just answer. Initial post suppose to write i, but not sqrt(i) in exponential form. And than use powering properties. One could write i = exp(i*π/2 + 2πk) i**0.5 = exp(i*π/4 + πk) But why we didn't lose something else in this solution? Using powering properties is totally wrong way of thinking in this case.
@@peted2783 this is not a question. The question is about using powering properties. My opinion is that your's approach is wrong. On this case it gives correct answer, but in general case not. One must use this formula while square rooting: en.m.wikipedia.org/wiki/De_Moivre%27s_formula But you shouldn't use this formula: (e^z)^0.5 ≠ e^(0.5z) to find all roots. Initial post is about powering. And my comment is about it. I know how to square rooting in complex field 😂
that only gives the positive sides of things because you're doing it in polar form which makes the angle be divided by two, and it was positive (90 cuz e^(pi*i) is on the Y axis) meaning it has an angle of 45 now, which is only the positive quarter, meaning you have to draw a 45 degree line, or Y = X line, and take the answers that sit on the unit circle which are two 45 degrees, and the +180 degrees from that, 215. meaning you'll get the one in the negative quarter
@@francescolorenzelli8912 didnt think id find a kingdom hearts reference here ..... i still dont get the joke tho edit: Ahh i see the vision is blury...
Only thing confusing was saying 1/(2*(1/√2)) = 1/√2 When I looked I instead got √2/2 but if you multiply by √2/√2 you get 2/2√2 which gives you 1/√2 so you were right but that part was the only thing I found to be unclear
@@TheWannaramble 2/sqrt(2) should be multiplied with sqrt(2)/sqrt(2) (which changes nothing since it's just 1). 2*sqrt(2)/sqrt(2)*sqrt(2) We know that square root of n multiplied with itself gives us n so: 2*sqrt(2)/2 Both sides divided by 2 Sqrt(2)
My professor taught me something valuable when writing my master’s: Never start an argument/discussion/presentation with ‘as we all know’; you never know who doesn’t know, and thus risk pushing away potentially interested readers.
@@callumross6290dont know about mathematicians but in engineering we never *know* i = sqrt(-1), we always *define* that i = sqrt(-1) since it's just a definition and needs to be said for the particular problem i guess
You can also think about it in polar form. i is on the unit circle, so it's roots are also on the unit circle. The argument for the principal value must be π/4 (½ the argument of i). So if you have your unit circle memorized the principal root is clearly sqrt(2)/2 + i*sqrt(2)/2. The other root is opposite the principal root at -sqrt(2)/2 -i*sqrt(2)/2
That's what I was think. You are rotating half way towards the imaginary number line from the real number line. That would be pi/4 rotation. Then figure out your polar coordinates and trig.
But wrong! The answer is wrong! There are two square roots of i. A value x is a square root of i if x^2 = i. So what you find are the square roots of i. But one of those square roots is ⎷i. Just like 2 and -2 are the square roots of 4, but ⎷4 = 2. and not also -2 So only the principle root is ⎷i. the symbol ⎷ is used for the principle root. So his answer is wrong. For ⎷i see also 5.1 on en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number
@@ZotyLisu Indeed: there are two square roots(as I mentioned in my reaction). But only one of them can be written with the ⎷ symbol. Example: 4 has two square roots: 2 and -2. But ⎷4 is just one of those values: 2 (the non negative) and is called the principle square root. So when he gives two values for ⎷i he's wrong. In the case of complex numbers we have a similar situation. There are again two square roots, but only one of them can be written with the symbol ⎷ See the paragraph Algebraic Formula. on en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number For the principle root of x+iy. You will see that that principle square root (written as ⎷(x+iy). stands for just 1 value. And in the video he gives two values for ⎷i. and that's not correct. May be he does not know the differents between the square roots of a value x and the value ⎷x. The problem he solved is not "find ⎷i." but "find the square roots of i."
@@henkhu100 yeah he could've wrote w0 and w1 or whatever, but I disagree that that's an only definition, in my course the value of an expression with ⎷ symbol was defined as a set of all answers - I'm still a bit confused on how that works tbh
@@ZotyLisu If ⎷ is a set then what is the meaning of for instance (⎷5)/(⎷6) ? A quotient as a result of deviding one set by another? ⎷5 is not a set, it is just a value. Example: solve x^2=7. In your opinion the solution is x=⎷7 because ⎷7 stands for all answers . But the solution is x=⎷7. or x=-⎷7. The definition of the ⎷ symbol in your course was not correct. I am sure you know the solution of a standerd quadratic equation: x=(-b+⎷(b^2-4ac))/2a. or x=(-b-⎷(b^2-4ac))/2a Your definition of the ⎷ symbol would only give the first value because the ⎷ symbol already stands for the + end the - in the formula.
Make sure you don't focus too much on improving your accent first of all. I could comprehend it just fine. Chinese pronunciation works just fine on English language as long as you have practiced. More good videos please.
you are studying math you dont even know how to speak english to understand this. math is universa, i didnt even listen the audio to understand everything i jus tskipped it
My calc class is taught by a Romanian woman, half the class is Chinese. Communication is not a problem. Math is the universal language, numbers unite us all.
You went a really complicated way of solving these equations. In my head I did it like so a^2 - b^2 = 0; a^2 = b^2; a = +-b 2ab = 1; we know a and b should be of the same sign so we'll say a = b and get 2a^2 = 1 a^2 = 1/2 a = sqrt(1/2)
Thats why so many people don't get into maths. Maths can be wathever you want, depending on how you enter in. Many examples show very simple solution or very complex ones for the same question. Which one do you prefer ?
All the people complaining he didn't use polar co-ordinates are completely missing the point. If you haven't already studied exactly why e^ix = i sin(x) + cos(x) then that would make this video completely pointless, the kind of people who want to know the answer to this problem most likely haven't come across that level of mathematics yet
Harry Stuart, well assuming you have a calculator for inverse Tan or a book full with tables of precalculated values for not carefully selected examples. And dividing is multiplying with the power of -1 defined as (a-bi)/(a^2+b^2) but you are free to learn such tables like some people like studying phone book numbers 😋
I agree that it doesn't look nice when you first come across it, but polar representation is one of the reasons complex numbers are so useful in the sciences
Yeah, bro. Everyone that understand English would understand this man. Some people are talking about his accent in poor because they don’t understand English. As a matter of fact, they don’t even know the origin of English, to go further, English comprises of different European languages.
Everybody is like 'there' s an easy way:' and then has full paragraphs of calculations. Just think in polar coordinates and rotations and the answer is obvious
Right? It should just be "what rotation composed with itself brings you to where i sits (90 degrees)". Boom 45 degrees. Boom, express as cos(45d)+i sin(45d)
3:20 from these equations you could have just done: a^2-b^2=0 => |a|-|b|=0 => |a|=|b| Then, knowing that you can go to the next equation: 2ab=1 => ab=1/2 => sign(a)=sign(b) (Either both positive or both negative, because their multiplication results in a positive number) Then, because their signs are equal *AND* their absolute values are equal, you can assume their both equal. So now you have 2 solutions that differ by sign: a=b=+-sqrt(1/2)
IAAGO ARIEL SCHWOELK LOBO No one rationalizes denominators in 2017 - that's what people did back in the day when there were no calculators and you'd prefer dividing the memorized decimal form of the square root by the rationalized denominator.
Another thing is that rationalizing the denominator often hides the geometric connections between quantities. It's a bit harder to see that `√2/2` is the inverse of `√2`, but it is obvious when you didn't rationalize it: `1/√2` (one over something is the inverse). It's even more hard to see it with some more complex expressions with radicals. That's why I usually leave it unrationalized, as an inverse, unless I really have to rationalize it.
Sci Twi Look; I've got mixed feelings for the conventions of rationalizing denominators or not. You are completely right about maintaining the instant recognition for inverses, but then you'd be compromising the recognition of like terms. For example, 3/sqrt5 does not look like it could be added to 2sqrt5/2, but after rationalizing, you can see clearly that 3sqrt5/5 _can_. Now, on the issue of having the same answers as the teachers do (and trying to overlook the insult to many great math teachers I've met that I'm sure you were not trying to offend), you can't really be opposed to unification of measures or answers -at least to some extent you have to accept it. Of course it makes your life easier to save extra moments on a test or whatever, but using a more real life example with more important implications, the SIU (International System of Units)'s purpose is to ease scientists' endeavors at "sciencing", if you will, by having set standards as to what units are official, what they measure, and how much of that something they measure. This, of course, may mean little to a mathematician's job, but if you can apply this same smooth interchange of information through the answers and numbers you represent, what you try to state will be better conveyed and understood by the audience to whom you present the information to. Anyway, I know I can't force someone to think the way I do, and you have to use the methods that you know are better for your learning (very similar to the π/τ argument), but thanks for reading to the end.
Or you could just use polar form and the tiniest bit of trig. Complex multiplication is a rotation and a stretch, because the magnitude of i is 1, this case is just a rotation. i Is a 90° rotation counter clockwise from 1 so you need to find a number that you rotate twice to get 90°. That's 45° So sqrt(i)=cos(45)+isin(45) which is what you got
Well you missed out on the other solution there, which is rotation by 225º twice, i.e. √i = cos(225)+isin(225) = - [1/√2 + i/√2] (which is also in the video). All solutions to √i are on the form √i = cos(π*n + π/4) + isin(π*n + π/4), where n = 0, 1, 2 etc. Albeit they only result in two unique solutions: ± [1/√2 + i/√2].
Another way to see it is that multiplying by i makes the complex number rotate around origo by 90° (pi/2). Multiplying by i^(1/2) instead rotates 45° (pi/4). So, for example, 1 × i^(3/2) = -1/sqrt(2) + i/sqrt(2) since that is where a rotation of 135° from 0° takes us.
I think this method can only get one answer of the square root of i, but I'm not sure. If there is a way to get the other answer, then please reply. I want to know.
@@opposite342 There is multiple ways to express an angle as a multiple of pi. The angle pi/2 is the same angle as (5*pi)/2 (by just adding 2*pi = 4*pi/2). If you now say sqrt(i) = sqrt(e^(i*5*pi/2)) = e^(i*5*pi/4) = cos(5*pi/4) + sin(5*pi/4) * i = - 1/sqrt(2) - 1/sqrt(2) i
I like the more unit circle method, where squaring imaginary numbers doubles the angle to the positive real line, so square root must half. i is 90°, thus it's square root must be 45°, the coordinates of which are ((√2)/2,(√2)/2), or (1/√2)+(1/√2)i
This showed up on my recommended, and I could feel myself getting smarter throughout the video because of your amazing teaching style. You have earned yourself a new subscriber, so thank you
5:06 I noticed that not only are +1 and -1 solutions to a^4 = 1, but i and -i are too. More generally for a^n = 1, all nth roots of unity are solutions. Not that ignoring these solutions matters in this case, as they essentially just switch a and i*b but give the same final answers.
I dont think we need to consider a or b being i or -i since theyre respectively the real and imaginary parts of the solution were looking for, therefore a and b are real numbers
I just realized, this answer has a magnitude of 1 on the complex number plane. If you just looked at a and b while disregarding the i, you could say that with Pythagorean’s Theorem and with a and b as the x and y coordinates, the hypotenuse is 1. If you draw a unit circle with radius 1 on the complex plane (which touches points 1, -1, i, -i), then you can draw angles based on the points plotted on this unit circle. The angle of this answer with the real number 1 is 45 degrees. The angle between i and 1 is 90 degrees. The angle between -1 and 1 is 180 degrees. The angle between 1 and 1 is 360 degrees. This answer squared is i, i squared is -1, -1 squared is 1.
True, you can also solve multiplications using the complex plane: any two nubers mulitpilied will have an angle equal to the angle of the first and the angle of the second number summed up, and will have a distance from 0 equal to the distance of the first number multiplied by the distance of the second number I’m french sorry if this isn’t very clear
He didn't even need to do all that things for finding A, it just required the simple observation that a = b or -a = -b This would mean 2a² = 1 and a = ± 1/√2
The Best way for me is to think of complex numbers as vectors on a complex plane. And if you raise an imaginary nomber to some power you make a rotation. i^2 = i^(1+1) = -1 hense you make a 90 degree rotation anti-clockwise and went from the imaginary axis to the real one. i^0.5 = i^(1-0.5) means that you make a 45 degree rotation clockwise. Now you only need to decompose your vector. The real part is cos(45 deg), and the imaginary one is i*sin(45 deg).
i don't care about the haters and their circle formulas, using various methos is really useful and gives more options, i know that formula but using this for fun is really nice
There is a method for determining square roots which uses a system of 3 equations on module, real part and imaginary part. It wasnt fully used here, and wasn't relevant in this case. This guy sucks at math and saying so doesn't make anyone a hater, just a skeptical person who knows a tiny bit about math
So you just proceeded to state that there is such a method and didn't explain the method or at least name it, then you said he sucks at math even though he got the right answer in simple steps. Nice.
Actually it can be done way simpler, if we know on the complex plane multiplication means that the absolute value of complex numbers multiply and the angle to the Real axis add up, So i is 90° with an absolute value of 1 sqrt(i) is 45 ° with absolute value 1, use some cos and sin to get the actual values if necessary
Luka Popovic out of all of those infinitely many solutions only two lay between 0° and 360° all other solutions are only a different way to reach those two points.
coming from the other standard form of a complex number: z = a + i * b = r*e^(i * phi + 2*pi*k), with r = sqrt(a^2 + b^2), phi = atan2(b, a). For the complex number "z = i", r=1, phi=(pi/2), so i=e^(i*(pi/2+2*pi*k)), i^1/2 = the result above.
honestly though, it just truly added to the enjoyment of the video. He shows so much excitement when explaining and it altogether created a really good video
English is also not my motherlanguage and i understood you justy fine, no complains, ignore these morons :D, they should be thankful you didnt speak your language and just put subtitles
Gmod2012lo1 english is not my mother language either, so it got me thinking: what if non-native english speakers understand other accents better when compared to native speakers?
He was speaking bamboo English. Its when I get up in the morning and the memory is loading to the ram but it takes a short time so I'm incoherent in the mean time.
What is your mother tongue? I assume it is west Germanic because you made the typo "justy" which reminds me of German richtig. German? Dutch? Frisian? Afrikaans?
Hi there! If you like this video, you will also like: (a+bi)^(c+di) th-cam.com/video/Z4PsLt9_ky8/w-d-xo.html Be sure to subscribe for more interesting math videos! bprp
Another way1: İ=e^(i*pi/2) Sqrt(i)=e^(i*pi/4)= Cos (pi/4)+i*sin (pi/4)= (İ+1)/sqrt(2) Another way 2: Sqrt (i)=x X^4=-1 (X^2+sqrt (2)x+1)* (X^2-sqrt (2)x+1) Solutions (İ-1)/sqrt(2) (İ+1)/sqrt(2) (1-i)/sqrt(2) -(1+i)/sqrt(2) And later we found Sqrt (i)=(i+1)/sqrt(2)
and since in complex roots , the roots would divide a unit circle into n sectors each with 2pi/n angled sectors another solution will be after the rotation of pi radian , which implies the other solution as well.
Uh, easier to do the other substution no ? a²-b² = 0 gives a = +- b Plugged into 2ab =1 we get 2a² = +- 1 and so a² = 1/2 since a is real which also means a = b So a = root(1/2) and b = root(1/2) Or both negative.
Stil the long route. It's easier to write i=e^(0,5*pi*i+k*2pi*i). Then the square root just halves the angle. So Sqrt(i)=e^(0,25*pi*i+k*pi*i). Basic angles from sine and cosine can translate back to the answer he got.
6:31 isnt an easier way of doing that rearranging equation a² - b² = 0 to a² = b² in which case a = b then substitute a or b in the second equation so 2a² = 1 rearrange so a = +-√2/2
@@ampleman602 a = -b or b = -a is the same. And he's right, it can only be a = b because of the second condition 2ab = 1, that only works if a = b. If a = -b you would get a negative output.
This can also be done using re^iθ polar coordinates, where i = 1*e^[i(pi/2)], taking the square root yields sqrt(1)*e^[i(pi/2)/2], which is equal to 1*e^[i(pi/4)], in this case would be a vector 45° from the horizontal, with a length of 1, with the [ Re, Im ] coordinates being [ 1/sqrt(2), 1/sqrt(2) ], or in an equation, sqrt( i ) = 1/sqrt(2) + i / sqrt(2)
That was interesting to follow, I've forgotten so much math including basic algebra, this was very helpful and you did a great job of explaining it all.
In polar form, the answer is very obvious. Half of a 90° rotation to the left is a 45° rotation to left or a 135° rotation to the right, which is the exact same result you got.
Paul Paulson oh that makes sense. If the square root of negative one is represented as 180 degrees, then the square root is equal to a one half power, meaning the power controls the rotation (in a way). 1 is 0 degrees so yeah.
Cody Fan Yes, because with complex numbers, what is called "multiplication" is actually just the application of a rotation and a scaling. To square a complex number is just to apply the rotation and scaling twice. i is a rotation by 90°, while 2 i is the same rotation but also scales by a factor of 2. If you have e.g. 3 + 2i, you would have to convert it first into rotation and scaling. But just imagine it as the point (3,2) in the complex plane, draw a line to (0,0) and the length of that line is the scaling factor and its angle to the horizontal axis is the rotation. Actually, to me, this is what complex numbers are really about and what their true meaning is. The thing with the imaginary square root of negative one is just some kind of mathematical trick (which is still usefull because it's often easier to calculate with it).
We can use complex analysis to solve it . Simply by knowing that Log(z) = ln(z) + i Arg(z) with arg in ]-pi , pi[ . So z^1/2 = exp(1/2 * Log(z)) You’ll end up with exp(I pi/4) . Thanks
the word simply is doing a lot of work in that sentence. i had to google what complex analysis meant, and it sounds like something you need linear algebra for, which isn’t taught in high school. he managed to teach this in a way a 15 year old could understand
Using a geometric interpretation, since i is unit length, its square root would be a unit vector (real, img) with half the angle. If i is at 90 degrees, sqrt(i) is at 45 degrees: 0.707, 0.707.
You spoke very well. Anyone following the logic, algebraic manipulation and substitution do not require any elaborate explanation. Yes, very elegant and cool! Thank you.
You already know from a²-b²= 0 that the squares must be the same, a = ±b. From 2ab = 1 you know that they both need to be the same sign, so you know a = b. Then by solving 2a² = 1 you get a,b = ±1/√2, so √i = ±(1 + i)/√2
My thought exactly. Makes me wonder rif he was just trying to prolong it for some reason. I like this channel, but this felt like it was intentionally taking the long way around when on the 2nd board.
You have very interesting questions you solve. The bits where you rationalize to create a very simplified fraction is genius. Did you practice stuff like this a lot to have such a deep intuition? You make it look so easy and natural. Thanks for these vids!
When a^4 = 1/4, there will be four values of a and not just two values like you explained. These values would be 1/√2, -1/√2, i/√2, -i/√2. Likewise, there will be four values of b as well.
I understood it (and similar problems through school) by utilising index laws. These consist of laws like a^b ⋅ a^c = a^(b+c), however the law that applies here is a^b / a^c = a^(b-c). So, in principle, 2 = 2^1 and √2 = 2^(1/2). Therefore 2^1 / 2^(1/2) = 2^(1 - 1/2), which equates to 2^(1/2), or more simply, √2. That's my interpretation and understanding of why that works.
in a^2-b^2=1 you can simply add b^2 to both sides to get a^2=b^2 and then cut the roots then use this to get a=b then in 2ab=1 divide both sides by 2 to get ab=1/2 and then use a=b to get a^2=1/2 and then root both sides to get the answer to a and therefore b
sqrt(a+bi)=?
Answer here: th-cam.com/video/CeVdh5LH908/w-d-xo.html
👍
White Chalk Red Chalk, nice 😊
Steve:
0:00 As we all know, √(-1) = 𝒊
Also Steve:
8:50 √𝒊 = ±(1 + 𝒊)/√2 - *two answers*
That's inconsistent!
@@allozovsky it's not inconsistent. There's one in Q I and another in Q III separated by p u/2 rotation Much like using an inverse trig function to return an answer in a limited domain where the actual solution may be outside of that domain. Recall that -1 is a real number, I is imaginary, and thr square roots are complex. Complex numbers don't behave quite the same as real and purely imaginary numbers.
@@onradioactivewaves If √𝒊 = ±(1 + 𝒊)/√2 gives two complex square roots, then √(−1) also should return two complex square roots, that is √(−1) = ±𝒊, isn't it? Otherwise it is inconsistent. That's pretty strange that Steve often evaluates in his videos multivalued complex functions alright, but at the same time uses only single 𝒊 for the square root of −1.
There ya go psychologists; the root of imagination.
This is so underrated
This deserves more likes...
Bro what 30 likes 😂
The blackboard was designed by a guy named Hilbert.
Woo
Plot twist: He has endless layers of boards.
Hahhaa
Truuue
blackboardwhiteboard
😂😂😂
@BLVGaming / Y1000 Couldn't not agree
I happened to watch this on my break from studying before my leaving cert maths exam. Square root of a complex number was on the exam and I got the right answer using this method. What a crazy lucky coincidence.
Congrats man
Yeah I remember the exam. I used a different method tho.
@@artemis_furrson What was it? 🧐
@Mosinlogan Being interested in what you study is a blessing many students would strive to have.
@@Someone-wj1lf there's no way people are like that? Lol I didn't know that, that's strange
What I miss when I zone out for 30 seconds in class:
Underrated comment
ADHD moment. Felt it.
@@kdog3908what is ADHD!?
@@swetkataria5282 Attention Deficit Hyperactivity Disorder. More commonly called ADD in the US, I think.
accurate
Me: trying to go to sleep
TH-cam: BuT wHaTs ThE sQuArE rOoT oF i???
-1
@@laerciocivali no
lol
@@laerciocivali thats i squared
Same here XD
I, as a non native english speaker, watched your video at 2x speed. Got everything you said. Keep it up.
You should be an English teacher
Cough cough the pinned comment cough cough
me too, I guess as a non english native speaker you are used to a bigger variety of accents
OMG EXACTLY ME!!!!!!!!!!
Me too. I think improving the accent would be nice still, though.
I don't know why people keep complaining about this guy's solution to the problem, and why they offer geometric proofs instead. I really like this guy's answer because it uses only the simplest arithmetic/algebra and the simplest definition of a complex number: a + bi. Also, he did it in an incredibly concise manner. (My only complaint was at the end where he could have used a² - b² = (a+b)(a-b) = 0 --> a = ±b. But, typing this out, I suddenly realise how clever he is that even *factorization* doesn't need to be used in his answer.) This answer is teachable on someone's very first lesson on complex numbers; even the average 15-year-old will comprehend it very well, and he's earned my amazement.
Billy Ma-gusta thank you Billy. I think people just got too excited. It's like once they see derivatives, they want to show algebra students to take the derivative to find the vertex of a parabola.
Billy Ma-gusta You want concise?
Polar form: R*exp(iθ) = sqrt(i)
Square: R^2*exp(2iθ) = i = exp(i(π/2 + 2πn)), n integer
Identify: R=1, θ=π/4+nπ
Done.
Or if you prefer rectangular form the unique representations are: ±(1+i)/sqrt(2)
Tl;dr: Rectangular is great for addition and subtraction, polar for multiplication, powers and roots. Right tool for the job.
My point: Your working is as opaque as it gets. For all the average high schooler or TH-camr can see, you're just writing a whole bunch of Greek because nothing is explained. The polar form of complex numbers isn't explained at all. The polar form of √i specifically also isn't explained, neither is i. Converting between the polar form and the rectangular form also isn't explained. You cannot call this concise when you leave so much unexplained; someone who doesn't know the required background knowledge would call this gibberish. (Like how I call bullshit when I am told "using string theory, we can show that 11 physical dimensions exist". Just an example, don't digress pls.)
And the greater point: the polar form does not NEED to be explained in order to prove this result, as bprp has shown. Simpler is better in maths, and a short proof is not necessarily a simple proof.
I would happily agree that I assume too much if I for some video on an
integral complain about someone not using residuals and Cauchy's
integral formula to solve an integral on the real line since it would be
trivial that way. But assuming knowledge of the polar form of a complex number, which is as fundamental as the rectangular form, is not a stretch imo.
As for the specific polar forms. I never use the polar form of sqrt(i), only the polar form of i, to keep things simpler. I could happily have shown a justification of the polar form of i, but it's a bit clunky without being able to show any figures. Basically you have an angle of π/2 to i and any 2π increment of that still leaves you at i. If I had shown this on a blackboard the justification is explained in a few seconds.
The conversion to rectangular form is really not part of my solution, I'm happy to stay in polar form.
My point is that this result is trivial in the polar world. It also gives some really nice insights about how the root behaves in general, which is totally lost in rectangular form. Using polar form the result also trivially generalizes to any real power of i.
Just like addition/subtraction is trivial in rectangular but a mess in polar form. Sure you could technically use the strategy as shown in this video to describe the 5th or 50th root of i as well, but it would be very cumbersome and very brute force.
you actually don't need factorization to get that a = b.
1. a² - b² = 0 -> a² = b², therefore a and b have the same magnitude.
2. 2ab = 1, ab = 1/2, therefore a and b have the same sign (if the product of 2 real numbers is positive, then 2 real numbers have the same sign)
3. a = b, because a and b have the same magnitude (1) and the same sign (2) (definition of equality for real numbers)
4.a*a = 1/2 = a², from (3) and (2)
5. a = ± sqrt(1/2), sqrt both sides of (4)
6. b = ± sqrt(1/2), from (3)
I actually prefer this method of solving the system of equations over the video because it takes less steps and is more intuitive. the videos method involves a lot of seemingly arbitrary moving of symbols around while each of my 6 steps have much clearer purpose.
There is a simpler way without the polar form. We know a² - b² = 0 and 2ab = 1, therefore a² = b² a = b or a = - b. But because of the second equation we can cancel out a = - b as a possibility. That means a = b and if we plug this in the second equation we get: 2a² = 1 a = +/ - (1/sqrt(2)). Thats it. :D
yeah I instantly saw that, its cool how there are so many different ways to arrive at the same answer
No , just cause a² = b² , Doesn't mean a = b , cause (-1)² = 1² but obviously -1 ≠ 1
@@gamerdude7800 read my comment again
@@gamerdude7800 he mentioned a=±b
This is how I did it too, nice
I litterally read it as "Squ(i)rt"
I'll remember that forever
@@rafaelwoods4 same
Time to stop waxing your carrot
Me too
So did I
I appreciate the depth of this explanation, rather than memorizing forms, for the sake of speed and ease, you showed me understanding. I appreciate knowing why, over mechanical speed.
You'll get mechanical speed with practice. Idiots just don't know to practice.
@@woophereigo9755 Smart guy over here. Shut up.
@@manperson6234 Bunch of morons. Get good.
@@woophereigo9755 yikes bud 😬
@@woophereigo9755 One thing is be fast while doing your own calculations, another is following another one being fast doing his calculations
Okay, that's easy. The real question is:
How many blackboards does this guy have?
lol the same here ;
@The Daily Egg
yes
He must be teaching in many different classrooms or renting classrooms.
He's like people in the math problem
i
You can also convert to e^(i*π/2). Then sqrt(e^(i*π/2))=e^(i*π/4). Then convert back to get 1/sqrt(2)+i/sqrt(2)
And losing second solution )
You can get the second solution by generally writing sqrt(i)=exp(i*(pi/2+k*pi)) for k any integer
@@CorvusSapien It's not a solution. You write just answer. Initial post suppose to write i, but not sqrt(i) in exponential form. And than use powering properties.
One could write
i = exp(i*π/2 + 2πk)
i**0.5 = exp(i*π/4 + πk)
But why we didn't lose something else in this solution?
Using powering properties is totally wrong way of thinking in this case.
@@at_one doing square root gives 2 solutions
sqrt(i)= ±sqrt(e^πi/2)= ±e^πi/4 = ±(cos(π/4)+isin(π/4))
giving the 2 solutions: sqrt(i)=1/sqrt(2) + i/sqrt(2) or sqrt(i)=-1/sqrt(2) -i/sqrt(2)
@@peted2783 this is not a question. The question is about using powering properties.
My opinion is that your's approach is wrong. On this case it gives correct answer, but in general case not.
One must use this formula while square rooting:
en.m.wikipedia.org/wiki/De_Moivre%27s_formula
But you shouldn't use this formula:
(e^z)^0.5 ≠ e^(0.5z)
to find all roots.
Initial post is about powering. And my comment is about it. I know how to square rooting in complex field 😂
Wanna know what's behind my board? It's another board!
bruh this teacher is so prepared. he uses 2 different colors of chalk to distinguish between terms and grouping symbols. Good Job!
Wanna know what's behind the second board? A third board!
@@_carrbgamingjr yep
I'm board.
@@stevens5541 ok
There is an other easy way :
We have i=e^(i(π/2))
So √i =[e^(i(π/2))]^((1/2))
So √i=e^(i(π/4))=√2/2 +i √2/2
It means that √i = √2/2 +i √2/2
Brilliant 👍
Damn bro nice flex lol
Aka 1+i
i don't understand
@@ericw2391 ok l will check it.Thanks+
My man’s too drippy for us, wearing supreme and teaching maths
😆
When the number is imaginary!
Ù⁸⁸
He’s on another level
Root_4 (-1)
Another way can be e^(i*π/2)= i for r=1, square root both sides and it will be e^(i*π/4)=√i, which will give √i= (1+i)√2
that only gives the positive sides of things because you're doing it in polar form which makes the angle be divided by two, and it was positive (90 cuz e^(pi*i) is on the Y axis)
meaning it has an angle of 45 now, which is only the positive quarter, meaning you have to draw a 45 degree line, or Y = X line, and take the answers that sit on the unit circle
which are two
45 degrees, and the +180 degrees from that, 215. meaning you'll get the one in the negative quarter
Why does WolframAlpha only gives a positive answer?@@amorios1092
@@amorios1092also one of my scientific calculators I use on my phone gives it
6:05 when he got the answer he started moving closer to the speed of light
Lmao underrated
When you beat a boss in Kingdom Hearts
@@francescolorenzelli8912 didnt think id find a kingdom hearts reference here ..... i still dont get the joke tho
edit: Ahh i see the vision is blury...
lmaoo
I thought the same thing haha
whitechalkredchalk
Lol
Real
@@Saikiisdoneimaginary
**He lifts up the black board**
Me - what the hell is thisss?
*lifts second blackboard revealing a third*
“Oh my god!”
@@SuperUghe yeah I guess the future is here!
What? This is pretty common xD
@@SuperUghe It's blackboards all the way down.
No it is -green board-
Just a heads up, writing it in Euler form is way faster.
That is,
i = e^[iπ/2]
=> √i = e^[iπ/4]
= 1/√2 + i.1/√2.
Edit: forgot a - for the second root.
Ok fine but, Why is he holding a grenade in his hands
that's a microphone
@@kayjaad3349 $Thanks, I did not know that_
@@kayjaad3349 there's something called sarcasm sis......
@AFancySpoon you'll only get attention if you comment on the board that this guy is using XD
*Bitter truth tho*
His parents made him hold it, it'll go off the moment he makes a mistake.
Very understandable even as a non native speaker
thanks!!
I agree. Your work is very good. Dont worry about trolls who complain and then fight.
*egg*
@@YellowToad egg
Es verdad mientras voy en un bus lo miro, por el alto volumen de bus no puedo oír el vídeo, soy hispanohablante, aún así se entiende todo.
Only thing confusing was saying 1/(2*(1/√2)) = 1/√2
When I looked I instead got √2/2 but if you multiply by √2/√2 you get 2/2√2 which gives you 1/√2 so you were right but that part was the only thing I found to be unclear
glad you commented because that tripped me up as well
@@TheWannaramble 2/sqrt(2) should be multiplied with sqrt(2)/sqrt(2) (which changes nothing since it's just 1).
2*sqrt(2)/sqrt(2)*sqrt(2)
We know that square root of n multiplied with itself gives us n so:
2*sqrt(2)/2
Both sides divided by 2
Sqrt(2)
@@someoneunimportant3064 very clear, thanks
@@TheWannaramble you are very welcome, glad it helped
@@TheWannaramble that is what I thought
My professor taught me something valuable when writing my master’s: Never start an argument/discussion/presentation with ‘as we all know’; you never know who doesn’t know, and thus risk pushing away potentially interested readers.
I completely agree with your professor about that.
I think if youre a math channel that posts exclusively calculus content its okay to assume your viewers know the definition of i
@@callumross6290 What if you are interested in math but never learnt calculus, and someone recommended this channel as it’s really good content?
@@callumross6290dont know about mathematicians but in engineering we never *know* i = sqrt(-1), we always *define* that i = sqrt(-1) since it's just a definition and needs to be said for the particular problem i guess
I don't understand how this can be SO FUKING PERFECT
Bruno Amezcua, because mathematics is a series of quantitative tautologies, where each system builds off the previous system.
@@stumpfightskills571 Well said!!⚡🔥
He almost made a writing mistake at the end (= instead of or) but fixed it immediately...
You can also think about it in polar form. i is on the unit circle, so it's roots are also on the unit circle. The argument for the principal value must be π/4 (½ the argument of i). So if you have your unit circle memorized the principal root is clearly sqrt(2)/2 + i*sqrt(2)/2. The other root is opposite the principal root at -sqrt(2)/2 -i*sqrt(2)/2
Damn Its so much simpler
That's what I was think. You are rotating half way towards the imaginary number line from the real number line. That would be pi/4 rotation. Then figure out your polar coordinates and trig.
x = √i
x² = i
x⁴ = i² = -1
x⁴ + 1 = 0
x⁴ + 2x² + 1 = 0 + 2x²
(x²)² + 2x² + 1 = 2x²
(x² + 1)² = 2x²
x² + 1 = x √2
x² - x√2 = -1
x² - 2(x)(√2 / 2) = -1
x² - 2(x)(1/√2) + ½ = -1 + ½
x² - 2(x)(1/√2) + (1/√2)² = -½
(x - 1/√2)² = -½
x - 1/√2 = ±√(-½) = ± √(-1) / √2
x = 1/√2 ± i/√2
√i = 1/√2 (1 ± i)
I also solved the problem that way.
Yep, that's the way I learned it.
Surprisingly complex yet also surprisingly simple
But wrong!
The answer is wrong! There are two square roots of i. A value x is a square root of i if x^2 = i. So what you find are the square roots of i. But one of those square roots is ⎷i. Just like 2 and -2 are the square roots of 4, but ⎷4 = 2. and not also -2
So only the principle root is ⎷i. the symbol ⎷ is used for the principle root. So his answer is wrong.
For ⎷i see also 5.1 on en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number
@@henkhu100 what do you mean? in complex numbers n-th root always has n answers, and he showed both
@@ZotyLisu Indeed: there are two square roots(as I mentioned in my reaction). But only one of them can be written with the ⎷ symbol. Example: 4 has two square roots: 2 and -2. But ⎷4 is just one of those values: 2 (the non negative) and is called the principle square root. So when he gives two values for ⎷i he's wrong.
In the case of complex numbers we have a similar situation. There are again two square roots, but only one of them can be written with the symbol ⎷
See the paragraph Algebraic Formula. on
en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number
For the principle root of x+iy.
You will see that that principle square root (written as ⎷(x+iy). stands for just 1 value.
And in the video he gives two values for ⎷i. and that's not correct. May be he does not know the differents between the square roots of a value x and the value ⎷x. The problem he solved is not "find ⎷i." but "find the square roots of i."
@@henkhu100 yeah he could've wrote w0 and w1 or whatever, but I disagree that that's an only definition, in my course the value of an expression with ⎷ symbol was defined as a set of all answers - I'm still a bit confused on how that works tbh
@@ZotyLisu If ⎷ is a set then what is the meaning of for instance (⎷5)/(⎷6) ? A quotient as a result of deviding one set by another? ⎷5 is not a set, it is just a value.
Example: solve x^2=7. In your opinion the solution is x=⎷7 because ⎷7 stands for all answers . But the solution is
x=⎷7. or x=-⎷7.
The definition of the ⎷ symbol in your course was not correct.
I am sure you know the solution of a standerd quadratic equation: x=(-b+⎷(b^2-4ac))/2a. or x=(-b-⎷(b^2-4ac))/2a
Your definition of the ⎷ symbol would only give the first value because the ⎷ symbol already stands for the + end the - in the formula.
Make sure you don't focus too much on improving your accent first of all. I could comprehend it just fine. Chinese pronunciation works just fine on English language as long as you have practiced. More good videos please.
you are studying math you dont even know how to speak english to understand this. math is universa, i didnt even listen the audio to understand everything i jus tskipped it
@TheMasterfulcreator I see what you did there.
Nice I like what u did
Math is its own language
My calc class is taught by a Romanian woman, half the class is Chinese. Communication is not a problem. Math is the universal language, numbers unite us all.
Teacher : What is the value of infinity?
Me : The amount of boards that blackpenredpen has .
Teacher : 🙏
you know he can erase the chalk, right?
🤣🤣🤣
@Stev Da Great you're*
Teacher said namaste!!🙏
You went a really complicated way of solving these equations. In my head I did it like so
a^2 - b^2 = 0; a^2 = b^2; a = +-b
2ab = 1; we know a and b should be of the same sign so we'll say a = b and get
2a^2 = 1
a^2 = 1/2
a = sqrt(1/2)
yup.. same here
Exactly what I was thinking. But we would probably get less marks than him cause he did the longer method. 😂
Thats why so many people don't get into maths. Maths can be wathever you want, depending on how you enter in. Many examples show very simple solution or very complex ones for the same question. Which one do you prefer ?
Even simpler:
sqrt(i) = (e^(i pi/2))^(1/2) = e^(i pi/4) = cos(pi/4) + i sin(pi/4)
Even if signs are same It doesn't make a=b
I had an aneurysm trying to figure out how u were changing the chalk color for like a solid 10 sec until I realized it was just 2 pieces of chalk
Like two pieces of sqrt(i) ;=)
All the people complaining he didn't use polar co-ordinates are completely missing the point. If you haven't already studied exactly why e^ix = i sin(x) + cos(x) then that would make this video completely pointless, the kind of people who want to know the answer to this problem most likely haven't come across that level of mathematics yet
Harry Stuart thanks!
Harry Stuart, well assuming you have a calculator for inverse Tan or a book full with tables of precalculated values for not carefully selected examples. And dividing is multiplying with the power of -1 defined as (a-bi)/(a^2+b^2) but you are free to learn such tables like some people like studying phone book numbers 😋
I agree that it doesn't look nice when you first come across it, but polar representation is one of the reasons complex numbers are so useful in the sciences
densch123 you'll wind up being thankful for the exponential form when you hit differential equations :P
or theoretical physics... or having to find real and imaginary roots of numbers... and complex functions and variables... lol ;P
Really good explanation! Thanks! As an Englishman, I can completely understand your accent!
James Saker thank you James
Yeah, bro. Everyone that understand English would understand this man. Some people are talking about his accent in poor because they don’t understand English. As a matter of fact, they don’t even know the origin of English, to go further, English comprises of different European languages.
Everybody is like 'there' s an easy way:' and then has full paragraphs of calculations.
Just think in polar coordinates and rotations and the answer is obvious
Right? It should just be "what rotation composed with itself brings you to where i sits (90 degrees)". Boom 45 degrees. Boom, express as cos(45d)+i sin(45d)
Yep, I got it that way in seconds. When I was a teenager I was into the Mandelbrot/Julia sets, and the complex plane became my main jam.
@@AlanCanon2222 r/iamverysmart
MrFierMath what I did, like sqr of i is equal to sqrt -1, leave that sqrt and make -1 to polar, and then the moivre method and that is right?
I didn't have good teachers so this is the first time I understood this out
3:20 from these equations you could have just done:
a^2-b^2=0 => |a|-|b|=0 => |a|=|b|
Then, knowing that you can go to the next equation:
2ab=1 => ab=1/2 => sign(a)=sign(b)
(Either both positive or both negative, because their multiplication results in a positive number)
Then, because their signs are equal *AND* their absolute values are equal, you can assume their both equal.
So now you have 2 solutions that differ by sign:
a=b=+-sqrt(1/2)
8:54
10 people got *TRIGGERED* because he didn't rationalize the denominator
IAAGO ARIEL SCHWOELK LOBO lol!!!!
IAAGO ARIEL SCHWOELK LOBO No one rationalizes denominators in 2017 - that's what people did back in the day when there were no calculators and you'd prefer dividing the memorized decimal form of the square root by the rationalized denominator.
Another thing is that rationalizing the denominator often hides the geometric connections between quantities. It's a bit harder to see that `√2/2` is the inverse of `√2`, but it is obvious when you didn't rationalize it: `1/√2` (one over something is the inverse). It's even more hard to see it with some more complex expressions with radicals. That's why I usually leave it unrationalized, as an inverse, unless I really have to rationalize it.
Sci Twi I wonder what modern books have answer keys that use rationalised denominators..
Sci Twi Look; I've got mixed feelings for the conventions of rationalizing denominators or not. You are completely right about maintaining the instant recognition for inverses, but then you'd be compromising the recognition of like terms. For example, 3/sqrt5 does not look like it could be added to 2sqrt5/2, but after rationalizing, you can see clearly that 3sqrt5/5 _can_. Now, on the issue of having the same answers as the teachers do (and trying to overlook the insult to many great math teachers I've met that I'm sure you were not trying to offend), you can't really be opposed to unification of measures or answers -at least to some extent you have to accept it. Of course it makes your life easier to save extra moments on a test or whatever, but using a more real life example with more important implications, the SIU (International System of Units)'s purpose is to ease scientists' endeavors at "sciencing", if you will, by having set standards as to what units are official, what they measure, and how much of that something they measure. This, of course, may mean little to a mathematician's job, but if you can apply this same smooth interchange of information through the answers and numbers you represent, what you try to state will be better conveyed and understood by the audience to whom you present the information to. Anyway, I know I can't force someone to think the way I do, and you have to use the methods that you know are better for your learning (very similar to the π/τ argument), but thanks for reading to the end.
I’m not even doing this for school. I’m just interested.
same
Lucky
True
Same
Same, I'm really into this while i cant comprehend every step but i can feel his teaching energy pulling me into it.
Or you could just use polar form and the tiniest bit of trig.
Complex multiplication is a rotation and a stretch, because the magnitude of i is 1, this case is just a rotation.
i Is a 90° rotation counter clockwise from 1 so you need to find a number that you rotate twice to get 90°. That's 45° So sqrt(i)=cos(45)+isin(45) which is what you got
Nicely
Well you missed out on the other solution there, which is rotation by 225º twice, i.e. √i = cos(225)+isin(225) = - [1/√2 + i/√2] (which is also in the video).
All solutions to √i are on the form √i = cos(π*n + π/4) + isin(π*n + π/4), where n = 0, 1, 2 etc. Albeit they only result in two unique solutions: ± [1/√2 + i/√2].
Yeh, but it's for beginners so many people wouldn't know about those
@Rallynub You are right, I was typing a bit fast. Correcting my comment... :)
Wow just......wow
Another way to see it is that multiplying by i makes the complex number rotate around origo by 90° (pi/2). Multiplying by i^(1/2) instead rotates 45° (pi/4). So, for example, 1 × i^(3/2) = -1/sqrt(2) + i/sqrt(2) since that is where a rotation of 135° from 0° takes us.
Thank you for sharing this.
I may be the only one who liked you accent. And could you do a video on differential equations?
Pink Floyd is the Best Band of All Time. Hi there, thanks!! I do have diff eq videos here www.blackpenredpen.com/math/DiffEq.html
blackpenredpen Thanks!
The funny thing is that 1/sqr(2)=cos(45)=sin(45)
That's not really surprising, you get this immediately using Euler's identity. e^ix = cosx + isinx, so i = e^(i*pi/2). sqrt(i) = i^(1/2) = e^(i*pi/4)
@@kanekeylewer5704 Wow, what is that, I couldn't understand it much but I found it really interesting!
I think this method can only get one answer of the square root of i, but I'm not sure. If there is a way to get the other answer, then please reply. I want to know.
@@opposite342 There is multiple ways to express an angle as a multiple of pi. The angle pi/2 is the same angle as (5*pi)/2 (by just adding 2*pi = 4*pi/2). If you now say sqrt(i) = sqrt(e^(i*5*pi/2)) = e^(i*5*pi/4) = cos(5*pi/4) + sin(5*pi/4) * i = - 1/sqrt(2) - 1/sqrt(2) i
Opposite34 I know I’m a month late but as a person who doesn’t study maths at that level, you explained it really well and i could follow every step
I like the more unit circle method, where squaring imaginary numbers doubles the angle to the positive real line, so square root must half. i is 90°, thus it's square root must be 45°, the coordinates of which are ((√2)/2,(√2)/2), or (1/√2)+(1/√2)i
This showed up on my recommended, and I could feel myself getting smarter throughout the video because of your amazing teaching style. You have earned yourself a new subscriber, so thank you
5:06 I noticed that not only are +1 and -1 solutions to a^4 = 1, but i and -i are too. More generally for a^n = 1, all nth roots of unity are solutions. Not that ignoring these solutions matters in this case, as they essentially just switch a and i*b but give the same final answers.
I dont think we need to consider a or b being i or -i since theyre respectively the real and imaginary parts of the solution were looking for, therefore a and b are real numbers
Then shut up
a and b are real they can't be i or -i
@@sebm2334 exactly
I just realized, this answer has a magnitude of 1 on the complex number plane. If you just looked at a and b while disregarding the i, you could say that with Pythagorean’s Theorem and with a and b as the x and y coordinates, the hypotenuse is 1.
If you draw a unit circle with radius 1 on the complex plane (which touches points 1, -1, i, -i), then you can draw angles based on the points plotted on this unit circle. The angle of this answer with the real number 1 is 45 degrees. The angle between i and 1 is 90 degrees. The angle between -1 and 1 is 180 degrees. The angle between 1 and 1 is 360 degrees. This answer squared is i, i squared is -1, -1 squared is 1.
True, you can also solve multiplications using the complex plane: any two nubers mulitpilied will have an angle equal to the angle of the first and the angle of the second number summed up, and will have a distance from 0 equal to the distance of the first number multiplied by the distance of the second number
I’m french sorry if this isn’t very clear
x = √i
x² = i
x⁴ = i² = -1
x⁴ + 1 = 0
x⁴ + 2x² + 1 = 0 + 2x²
(x²)² + 2x² + 1 = 2x²
(x² + 1)² = 2x²
x² + 1 = x √2
x² - x√2 = -1
x² - 2(x)(√2 / 2) = -1
x² - 2(x)(1/√2) + ½ = -1 + ½
x² - 2(x)(1/√2) + (1/√2)² = -½
(x - 1/√2)² = -½
x - 1/√2 = ±√(-½) = ± √(-1) / √2
x = 1/√2 ± i/√2
√i = 1/√2 (1 ± i)
since it is a power of i, it lands on the unit circle on the complex plane
@@alexcarpentier5698 That's really interesting, I never realized that before.
so -1, I, and sqrt(i) are all on the unit circle. I wonder if the fourth root of i is as well.
Another way to solve the fourth root of four being equal to four is this:
Sqrt(Sqrt(4))=x
Sqrt(4)=2, so
Sqrt(2)=x
He didn't even need to do all that things for finding A, it just required the simple observation that a = b or -a = -b
This would mean 2a² = 1 and a = ± 1/√2
The Best way for me is to think of complex numbers as vectors on a complex plane. And if you raise an imaginary nomber to some power you make a rotation.
i^2 = i^(1+1) = -1 hense you make a 90 degree rotation anti-clockwise and went from the imaginary axis to the real one.
i^0.5 = i^(1-0.5) means that you make a 45 degree rotation clockwise. Now you only need to decompose your vector.
The real part is cos(45 deg), and the imaginary one is i*sin(45 deg).
@@lc1777 Yes, substituting a=±b from the first equation saves about five minutes of algebra compared to substituting b=1/2a from the second.
@@lc1777 I had the same approach. Much easier.
@@beeble2003 yup
bro has infinite blackboards
🎅🎅🐡🥕
i don't care about the haters and their circle formulas, using various methos is really useful and gives more options, i know that formula but using this for fun is really nice
There is a method for determining square roots which uses a system of 3 equations on module, real part and imaginary part. It wasnt fully used here, and wasn't relevant in this case. This guy sucks at math and saying so doesn't make anyone a hater, just a skeptical person who knows a tiny bit about math
Clearly this guy sucks at math even though he got the right answer
So you just proceeded to state that there is such a method and didn't explain the method or at least name it, then you said he sucks at math even though he got the right answer in simple steps. Nice.
Actually it can be done way simpler, if we know on the complex plane multiplication means that the absolute value of complex numbers multiply and the angle to the Real axis add up,
So i is 90° with an absolute value of 1
sqrt(i) is 45 ° with absolute value 1, use some cos and sin to get the actual values if necessary
it can also be 225° since you're going for 2*α mod 360° = 90°
MrRoyalChicken Actually, there is whole spectrum of solutions. They are
e^(i*pi/4 + k*pi), where k is a whole number (positive or negative, or zero)
niklas schüller So there are infinitely many solutions ;)
Luka Popovic out of all of those infinitely many solutions only two lay between 0° and 360° all other solutions are only a different way to reach those two points.
coming from the other standard form of a complex number: z = a + i * b = r*e^(i * phi + 2*pi*k), with r = sqrt(a^2 + b^2), phi = atan2(b, a). For the complex number "z = i", r=1, phi=(pi/2), so i=e^(i*(pi/2+2*pi*k)), i^1/2 = the result above.
i may be complex, but 1 is still the loneliest number.
Just 2 punny!!!
This shit got me sad asf.
But can't debate bt being lonely he is also the glory coprime
I isnt complex
at least he's positive about it, poor -1, lonely and negative :(
Input
4^(1/4) = sqrt(2)
Result
True
Logarithmic form
log(4, 4)/4 = log(4, sqrt(2))=0.25
I like how you hold your microphone the whole time.
thanks!
And two pens in the other hand. That blew my mind.
honestly though, it just truly added to the enjoyment of the video. He shows so much excitement when explaining and it altogether created a really good video
English is also not my motherlanguage and i understood you justy fine, no complains, ignore these morons :D, they should be thankful you didnt speak your language and just put subtitles
Gmod2012lo1 english is not my mother language either, so it got me thinking: what if non-native english speakers understand other accents better when compared to native speakers?
@@JoaoVictor-gy3bk We really do. That's why I prefer being a non native English speaker.
@@JoaoVictor-gy3bk that doesn't make sense, nor is it true
He was speaking bamboo English. Its when I get up in the morning and the memory is loading to the ram but it takes a short time so I'm incoherent in the mean time.
What is your mother tongue? I assume it is west Germanic because you made the typo "justy" which reminds me of German richtig. German? Dutch? Frisian? Afrikaans?
does anyone else notice how oddly motion blurred he get when he's on the edge of the screen, and only the edge
the camera's not in focus?
yesss I've noticed
@@kzushii it does not look like because it just seems out of temporal sinc, not spatial distortion
This is genuinely the first time I've watched math videos for entertainment. Props to bprp
Hi there! If you like this video, you will also like: (a+bi)^(c+di) th-cam.com/video/Z4PsLt9_ky8/w-d-xo.html
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bprp
Where's the hashtag #YAY that we've been waiting for so long?
almost reached 0.5M views!
Yoav Carmel I know!!!! I am super excited!!
When it's power 4 we will find 4 solution i guess!!
Another way1:
İ=e^(i*pi/2)
Sqrt(i)=e^(i*pi/4)=
Cos (pi/4)+i*sin (pi/4)=
(İ+1)/sqrt(2)
Another way 2:
Sqrt (i)=x
X^4=-1
(X^2+sqrt (2)x+1)*
(X^2-sqrt (2)x+1)
Solutions
(İ-1)/sqrt(2)
(İ+1)/sqrt(2)
(1-i)/sqrt(2)
-(1+i)/sqrt(2)
And later we found
Sqrt (i)=(i+1)/sqrt(2)
What if we have infinite nested square roots of i
sqrt(i+sqrt(i+sqrt(i+...))): th-cam.com/video/4EZRXWW607c/w-d-xo.html
no
Can we also solve it by polar form ???
no
Juste use the i=e**(i*pi*0.5)
1.3 + 0.675i
I think 😅
In a trigonometric form, we have i = exp(i*pi/2). Therefore: sqrt(i) = exp(i*pi/4) = 1/sqrt{2} + i/sqrt{2}. I think, it is an easier way :)
Yeah, I wonder why he didn't just use polar notation/Euler's formula to begin with...
and since in complex roots , the roots would divide a unit circle into n sectors each with 2pi/n angled sectors another solution will be after the rotation of pi radian , which implies the other solution as well.
It could be done in seconds using euler and polar form of a complex number
√i=e^iπ/4=cos(π/4)+isin(π/4)
=1/√2+i/√2
Did anyone realise he is holding an Ood translator sphere?
Dem Rottensoul u realized too late. th-cam.com/video/STQPuHCiR8Y/w-d-xo.html
Uh, easier to do the other substution no ?
a²-b² = 0 gives a = +- b
Plugged into 2ab =1 we get
2a² = +- 1 and so a² = 1/2 since a is real which also means a = b
So a = root(1/2) and b = root(1/2)
Or both negative.
This is better. In Math Simpler is Better.
More direct = simpler = more elegant
No danger, it's simple algebra. This is the more elegant way to finish this line of reasoning.
I agree but is a matter of taste. You need to add the idea that they have the same sign based on the fact that 2ab is one.
Stil the long route. It's easier to write i=e^(0,5*pi*i+k*2pi*i). Then the square root just halves the angle. So Sqrt(i)=e^(0,25*pi*i+k*pi*i). Basic angles from sine and cosine can translate back to the answer he got.
i understood everything but i would never come with an idea like this
@@super_super_super485 😂😂
@@super_super_super485also 2/2...😂😂🤣and infinite can be made xd
Fax
6:31 isnt an easier way of doing that rearranging equation a² - b² = 0 to a² = b² in which case a = b
then substitute a or b in the second equation so 2a² = 1
rearrange so a = +-√2/2
same doubt
It can be a=-b
If a² = b²
a≠b
Never make that assumption
But it can be a = -b
Or
b = -a
@@ampleman602 a = -b or b = -a is the same. And he's right, it can only be a = b because of the second condition 2ab = 1, that only works if a = b. If a = -b you would get a negative output.
i love how happy he is when talking about it
i = e^(i*pi/2), sqrt(i) = e^(i*pi/4) = (1 + i) / sqrt(2). To get the other root, remember that i = e^(i*5*pi/4) and follow the same process
ikr!
The area of the rectangle connecting i, -i, sqrt(i), and -sqrt(i) is sqrt(2)
Beautiful.
sqrt(-1) = i
(using indices)
-1^1/2 = i^1
sqrt(-1) =-1^1/2
3 dots*
-1^1/2 = i^1, multiply on both sides by a sqrt() or ^1/2
-1^1/4 = i^1/2
it's obvious when you write i as e^i*pi/2 then you consider the sqrt as power 1/2 then the sqrt is e^i*pi/4
The answer is squirt
The other solution doesn't fall out that way, but on the other hand there are infinitely many solutions in polar form.
This can also be done using re^iθ polar coordinates, where i = 1*e^[i(pi/2)], taking the square root yields sqrt(1)*e^[i(pi/2)/2], which is equal to 1*e^[i(pi/4)], in this case would be a vector 45° from the horizontal, with a length of 1, with the [ Re, Im ] coordinates being [ 1/sqrt(2), 1/sqrt(2) ], or in an equation, sqrt( i ) = 1/sqrt(2) + i / sqrt(2)
Smart guy: reads the title as "sqrt(1)
Me: reads the title as SKRRRR
squirt)
SQUIRT
Actually is "sqrt(-1)"
Tzar C. Incorrect. That would be sqrt(i^2) - not sqrt(i).
In my dirty mind is squirrrrtttttttt
That was interesting to follow, I've forgotten so much math including basic algebra, this was very helpful and you did a great job of explaining it all.
me: So, how much layers of blackboard do you have?
him: *YES*
6:50 from there on it’s unnecessary, because you already got the equation a^2=b^2
When you 4th root the 4, don't you get 4 roots which include 2 complex roots?
The other 2 roots are - +sqrt(2)i so the solutions just become mirrored.
a and b were defined as Real numbers
Euler's formula just makes everything easier
In polar form, the answer is very obvious. Half of a 90° rotation to the left is a 45° rotation to left or a 135° rotation to the right, which is the exact same result you got.
Paul Paulson oh that makes sense. If the square root of negative one is represented as 180 degrees, then the square root is equal to a one half power, meaning the power controls the rotation (in a way). 1 is 0 degrees so yeah.
The square root of negative one or i would then be 90 degrees
Cody Fan Yes, because with complex numbers, what is called "multiplication" is actually just the application of a rotation and a scaling. To square a complex number is just to apply the rotation and scaling twice. i is a rotation by 90°, while 2 i is the same rotation but also scales by a factor of 2. If you have e.g. 3 + 2i, you would have to convert it first into rotation and scaling. But just imagine it as the point (3,2) in the complex plane, draw a line to (0,0) and the length of that line is the scaling factor and its angle to the horizontal axis is the rotation.
Actually, to me, this is what complex numbers are really about and what their true meaning is. The thing with the imaginary square root of negative one is just some kind of mathematical trick (which is still usefull because it's often easier to calculate with it).
Paul Paulson It would be amazing if you could do a video where you illustrate this magnificent explanation.
Ayoub Merzak th-cam.com/video/mvmuCPvRoWQ/w-d-xo.html
You should watch this.
Thanks I thought we can't get the square root of ( i ) because my calculator said so
'F' for the calculator
so did i think, because years ago my middle high school math teacher said so. :(
Your calculator probably can’t return the rectangular coordinate form of a complex number (a + bi). It exists, but had an imaginary and real part.
Me : Calculate sqrt(i)
Calculator : Calculate 91×19
Me :
Artificial Intelligence : 1729
Switch to the complex mode and it would.
We can use complex analysis to solve it . Simply by knowing that Log(z) = ln(z) + i Arg(z) with arg in ]-pi , pi[ . So z^1/2 = exp(1/2 * Log(z))
You’ll end up with exp(I pi/4) .
Thanks
the word simply is doing a lot of work in that sentence. i had to google what complex analysis meant, and it sounds like something you need linear algebra for, which isn’t taught in high school. he managed to teach this in a way a 15 year old could understand
@@Hazel-xl8in I mean I’m 14 and I didn’t understand most of this… probably cuz I’m really bad at algebra…
@@deadpool9124 watch Algebra 1 full course video to get all the basics and fundamental understanding, probably ur school didn't teach you well.
هذا المثال في المنهج الدراسي في دولة العراق وهو من أبسط الامثلة في المرحلة الاخيرة (السادس الاعدادي) ❤
Very understandable, couldn't be more detailed.
And nice that you already did a video on the polar way ;).
Keep up the good work.
my nigga wearing a supreme shirt
real shit
sweater is fresh my g
Thanks!!!
Using a geometric interpretation, since i is unit length, its square root would be a unit vector (real, img) with half the angle. If i is at 90 degrees, sqrt(i) is at 45 degrees: 0.707, 0.707.
You spoke very well. Anyone following the logic, algebraic manipulation and substitution do not require any elaborate explanation. Yes, very elegant and cool! Thank you.
3:33 You could have saved time.
a^2-b^2=0 gives a=(-+)b. Then plug it into 2ab=1
Good idea
I think a more interesting approach is by putting I in the form of an immaginary exponential with the argument written with the +2i(pi)n
You already know from a²-b²= 0 that the squares must be the same, a = ±b. From 2ab = 1 you know that they both need to be the same sign, so you know a = b. Then by solving 2a² = 1 you get a,b = ±1/√2, so √i = ±(1 + i)/√2
2:47 is legendary
a²+b²=0 --> a=b
2a.a=1 --> 2a²=1 --> a²=1/2 --> a=+-1/sqrt2 seems another easier way to do that function system
My thought exactly. Makes me wonder rif he was just trying to prolong it for some reason. I like this channel, but this felt like it was intentionally taking the long way around when on the 2nd board.
De Moivre's therorem: Am I a joke to you?
Therorerorerorerorero
What is it??
It would have been easier if you just said
a² - b² = 0
a² = b²
a = +b or -b
Seems like only a = b, since in the other case we'll get -2b² = 1 (for a _real_ b). But yeah, that would've been much faster.
basically :
45 degrees + 45 degrees = 90 degrees
or -135 degree - 135 degree = -270 degree = 90 degree
use radians u peasant
@@BlowitchAlt lolol
your energy is so uplifting. We love you. Keep doing what you're doing, with that little smile of yours you'll get anywhere :)
He has no energy. SJW's love his gimmick of holding a weird mic.
You have very interesting questions you solve. The bits where you rationalize to create a very simplified fraction is genius. Did you practice stuff like this a lot to have such a deep intuition? You make it look so easy and natural. Thanks for these vids!
Agreed!
When a^4 = 1/4, there will be four values of a and not just two values like you explained. These values would be 1/√2, -1/√2, i/√2, -i/√2. Likewise, there will be four values of b as well.
True, but we're assuming a and b are real numbers
I admire this guy's ability so much. I lack mathematical knowledge, but it's so inspiring to see him perform this well with numbers. God bless you sir
why does 1/(2/√2) = 1/√2 totally lost at 7:36
He did not rationalize the fraction. If u divide 1/√2 both numerator and denominator by √2 you will get back 1/(2/√2). Both correct answers
WOW I just did what you said for the denominator (2/√2) x (√2/√2) = (2√2) / (√2^2) = (2√2) / 2 = √2. Thanks man
Look at en.wikipedia.org/wiki/Rationalisation_(mathematics)
i just answered it why are people still replying lmao
I understood it (and similar problems through school) by utilising index laws. These consist of laws like a^b ⋅ a^c = a^(b+c), however the law that applies here is a^b / a^c = a^(b-c). So, in principle, 2 = 2^1 and √2 = 2^(1/2). Therefore 2^1 / 2^(1/2) = 2^(1 - 1/2), which equates to 2^(1/2), or more simply, √2. That's my interpretation and understanding of why that works.
I love how he just holding the snowball in his bare hands like a dynamic mic
in a^2-b^2=1 you can simply add b^2 to both sides to get a^2=b^2 and then cut the roots then use this to get a=b then in 2ab=1 divide both sides by 2 to get ab=1/2 and then use a=b to get a^2=1/2 and then root both sides to get the answer to a and therefore b