I wanted to make 2 comments about this video. 1. I intentionally let A and B be complex numbers to show that when everything is simplified it leads to the same solutions, I thought that was such a cool conclusion. A lot of people are upset I did this, I should have expressed more clearly what I was doing. 2. The final solution should have only been the principle root. I was so excited about A and B being complex detail, that I forgot to restrict the final solution.
Just one little flaw here: √i is per definitionem ONLY the principal root (1 + i) / √2. If you'd ask for x² = i, you'd get both solutions ± (1 + i) / √2 = ± (1 + i) * √2/2.
or you could just say it's a 45° and 225° rotation on a unit circle in the complex plane. This is because 45+45=90, and 225+225=450=360+90. Then just do sin and cos of those and you'll get the values which is why you also have sqrt(2)/2, which is sin and cos (45°), as well as -sqrt(2)/2, which is sin and cos (225°)
@@TonyFisher-lo8hhtbf this video is padded out by at least 4 minutes, right from his expansion of a+bi squared he took a detour onto a much longer calculation journey. Frankly if you get a^2-b^2=0 it’s obvious that a=b (or a=-b) and you can sub it in to 1=2ab to get 1=2a^2 solve for a and as a=b you also get b.
There is one detail that should be pointed out in your calculations. The value a is only allowed to be real because it is the real part of the complex number a + bi. Therefore, when you get to the equation a^2 = -1/2, this equation should be rejected because it is not possible that a real number squared equals a negative number. In fact, this is why you get two extra redundant answers. The answer is not wrong, but the math used to get to the answer can lead to some misunderstanding about the differences in the complex and real numbers. Note: I didn´t read all the comments before writing mine, but there were other comments about this exact same issue)
I wasn't going to comment on this issue at all, but then it took up so much time in the back half of the video. One thing to add: It's not incorrect to allow a and b to be complex numbers, but you're guaranteed to get all the solutions with less work if you force them to be real. This is because a+bi can already represent every possible complex number when a and b are real. Allowing complex a and b will only get you numbers a+bi that would have been represented regardless.
@@Mothuzad By definition, a and b are real. To make a and b complex is not helping in that situation and make the problem more 'complex'... When we do math, we have to stick with a definition. This is why sometimes we have to be careful with recreational maths; sometimes the person using the maths will abuse the definition and might gives the illusion that it can be done in other context too...
I’m not so sure about this comment. You’re relying on the assumption that a must be a real number but I’m fairly certain that’s not true. If a were a complex number and b was anything at all then the term a+bi would still be a complex number, which was our only real “requirement” or assumption. We didn’t try to claim that a or b were real numbers, only that a+bi is a complex number
@@Orrinn123You're right in one thing being "not sure". If you look up the definition of complex number, it is defined as "a,b are real numbers". You cant define complex number with the the definition of complex number. And yes, in this video, he did it wrong in some way and still got right solution.
@@deltalima6703 General consensus. From WIkipedia: "Principal square root: Each positive real number has two square roots, one positive and the other negative. The square root symbol refers to the principal square root, which is the positive one. The two square roots of a negative number are both imaginary numbers, and the square root symbol refers to the principal square root, the one with a positive imaginary part."
But for that you may prove Euler formula (which is one of the most beautiful things I’ve learned at uni) but this require knowledge of Taylor series and proving it (with all the relevant calculus), it’s very tedious. We can either skip these steps and say “just use Euler” without showing why we can use Euler, but it feels less like mathematical method and more like magic, or choose a different path from the ground up. What I like in the video is that it isn’t require any background or mathematical assumptions besides basic algebra.
@@DTier_fish It's not hard. Graph complex value z as (a + bi) on the complex plane. The distance to the origin is sqrt(a^2 + b^2) and that's r, the modulus. The angle from the positive real axis is theta, the argument, and that's atan(b/a). Once you know the modulus and the argument, you can say that z = re^(i*theta). And then you can use all the log stuff with e on complex numbers. So if z = 1 + sqrt(3)i, the modulus (r) is 2 (1^2 + sqrt(3)^2), and the argument (theta) is pi/3 (atan sqrt(3)/1). So z = 2e^(i * pi/3). Have fun!
Draw a unit circle centered on the origin of the complex plane. i is at the top of the circle at its intersection with the imaginary i axis. i is 90° or π/2 radians around the circle from the positive real axis. Also i=e^{iπ/2)=cos(π/2)+i*sin(π/2). √i is π/4 around the unit circle or √i=e^{iπ/4)=cos(π/4)+i*sin(π/4)=√2/2+i*√2/2=(√2/2)(1+i).
@@simplylmao the fact that you are dealing with imaginary numbers and solution, if someone who never ever met something like this before it can be overwhelming for them
@@therealnonoo as long as you know i = root of -1, everything else is really just squaring and multiplying really. Ofc i see how it can be challenging for someone new to maths
shouldnt sqrt(x) give only the principal value here... which should never be ngetaive? so iin this case its only the positive option... for example...you say sqrt(36)=6 and not ±6... correct me if im wrong plz
I really like your style of working through things. The visual equations are easy to follow and show all the steps, while not ending up in pages and pages of equations at the end, and I love that you are very careful not to skip over even the simple steps, which makes it really easy to see that there's really no magic in there. The solution at the end could actually just be written as ±(√2/2 + (√2/2)i). This is consistent with the fact that we already know that when taking a square root of something, the result should always come out to be "plus or minus (something)". (And then, as you said in the comment, since the radical symbol means specifically the principal root, you should technically just take the positive one as the answer to this question.)
Oh my gosh, he did so many unnecessary or even contradictory processes. It can be done in 30 seconds: a² = b², therefore a = ±b. Since 2ab = 1, then a = b. Substituting, we have 2a² = 1 and then, a = ± sqrt(2)/2. Since a = b, sqrt(i) = sqrt(2)/2 + i.sqrt(2)/2 or -sqrt(2)/2 -i.sqrt(2)/2
z = i ← this is a complex number The modulus of (i) is 1, so the modulus of √i is: 1^(1/2) = 1 The argument of (i) is (π/2), so the argument of √z is [(π/2)/2], i.e. (π/4) Thus we can deduce that the first root of (i) is: z1 = 1.[cos(π/4) + i.sin(π/4)] → and to get the second root, we add an angle of (2π)/2, i.e. (π) z2 = 1.[cos{(π/4) + π} + i.sin{(π/4) + π}] It gives z1 = 1.[cos(π/4) + i.sin(π/4)] z1 = cos(π/4) + i.sin(π/4) z1 = [(√2)/2] + i.[(√2)/2] z1 = [(√2)/2].(1 + i) z2 = 1.[cos{(π/4) + π} + i.sin{(π/4) + π}] z2 = cos{(π/4) + π} + i.sin{(π/4) + π} z2 = - cos(π/4) - i.sin(π/4) z2 = - [(√2)/2] - i.[(√2)/2] z2 = - [(√2)/2].(1 + i) Summarize √i = ± [(√2)/2].(1 + i) √i = ± (√2).(1 + i)/2 √i = ± (√2 + i√2)/2
The way I got around understanding this, was to first to define i in exponential form and raising it to exponent 1/2. Then just redefine it to polar form.
I do not agree that this should result in two answers. When we write "x^2 = 4", then we do expect two answers, +2 and -2. But when we write "sqrt(4)", we expect just the positive square root; or in a complex context, the primary branch. Likewise, when we write "x^2 = i", we expect two answers, but "sqrt(i)" should have just one. It is commonly accepted that sqrt(2)(1+i)/2 is the primary branch we expect to keep.
what determines the primary branch? acceptance in the community? or is there a mathematical reason (ie sqrt(x) for x>0 = something >0) ? do you take the one with the positive real?
When dealing with complex numbers things gets a little crazy. When we have a square root, sure, since the second root will always be the first root multiplied by -1. But when we have cube roots, forth roots and so on, it is not so simple anymore. In the real world, makes sense a radical giving only one solution, but in complex world that is not very useful.
@@ricoseb There is a mathematical reason. Is the root with the smallest positive angle in the polar form representation of the number. By the way, that reason have a curious implication: the principal cubic root of a negative real number is never a real number, because the real root would make a angle of pi and will aways exist a complex root with a smaller angle (pi/3). Because of that, Wolfram Alpha cannot plot certain graphs correctly when the function involves cubic roots, since it consider only the main branch of the radical.
There are two schools of thought here, and both are valid. One carries the spirit of the square root bijection from real arithmetic and as you suggest considers the value with the smallest non-negative argument the principal and refers to that as the proper square root. The other, which is more prevalent, is that once we enter the complex world we dispense with the "one square root" idea and we say that all complex values have two square roots, three cube roots, etc. We still can identify the principal as above, but all values of the root are considered. Personally, I prefer the latter point of view, but I see logic in yours as well.
From a mathematical point of view, the guy from the video should be clear on how is the symbol sqrt(a + bi) is defined. In mathematics, it is fairly standard to use sqrt(a + bi) for the principal branch of the complex square root function and (a + bi)^{1/2} for the set-valued square root function (there will be two values as the output for the two roots). With that being said, we can infer directly from the video how he interpreted the symbol sqrt(i): he interpreted it as (i)^{1/2}. Hope this helps.
But didn’t you define that “a” would be the real part and “b” the imaginary part? I assumed the first two answers would be excluded because of that. Although when you rearranged the answers were the same, the first two were “b+ai” instead of “a+bi.”
@@deltalima6703I disagree, when the complex answer is defined as a+bi, a and b must be real. When you draw the complex plane, you don't expect each axis to be itself complex.
Given the same fraction in both, I would factor it out and write as (-√2 / 2)(1 + i) and (√2 / 2)(1 + i) making the solution literally ±(√2 / 2)(1 + i). But since √ generally means the principal root, meaning no negative values, the only actual solution should be (√2 / 2)(1 + i). The other one is an extraneous solution introduced by the squaring at the very beginning.
All you have to do is convert _i_ to its polar form. Then square-root the magnitude (which does nothing in this case) and divide the phase by 2, which gives you a complex number with a magnitude of 1 and a phase of π/4. Converting back to rectangular form gives sqrt(2)/2 + sqrt(2)i/2. To get the second solution, express i's phase as 5π/2. Then applying the square root operation will give you a complex number with a magnitude of 1 and a phase of 5π/4, which turns out to be the negative of the first solution, -sqrt(2)/2 - sqrt(2)i/2.
I went down into a "plotting complex numbers and its square roots in the complex plane" rabbit hole after watching this! Hoping you'll also make a short video on this pleeease
My favourite way to do this is use polar form of i = e^i(pi/2 + k2pi), for any integer k. Then sqrt(i) = i^(1/2) = e^i(pi/2 + k2pi)*(1/2) = e^i(pi/4 + kpi). This gives the same 2 solutions in the video plus infinite others. Though they’re all equivalent to one of the two in the video. Conceptually how I think about it is multiplying a complex number z by i is the same as rotating it by 90 degrees. Square root of x asks, “what number multiplied twice is the same as multiplying by x?” Combined: “what rotation can we apply twice to get a 90 degree rotation?” The obvious answer is 45 degrees, but also 225 degrees (225*2 - 360 = 90). Another way to find these is 2x = 90, 450, 810, … => x = 45, 225, 405, …
You did it the hard way. There are two vectors. Both have a length of 1. Rotate the I vector half way to a angle of zero so the fist solution is cos(45 deg)+isin(45deg) . the other solution is to rotate I half way to zero deg counter clockwise.
Not really. He just did it the purely algebraic way (which I’m fairly certain is the method this channel almost exclusively uses for problems of this sort)
2 remarks. sqrt(z) "represents" two values, because "sqrt" is not a function. By choice, decision has been taken to say that sqrt(z) is positive if sqrt(z) € |R. But if z is imaginary, for example z=i, sqrt(i) represents two values which are opposed.
This community college math instructor is pleased to see so many alternative procedures to find the answer to sqrt(i) in the comments. Many of them come from principles and procedures that come from a moderately advanced study of complex numbers, but I don't think that's what the author is trying to do. It's pretty clear to me that he is showing a "first week" lesson to students who are just beginning to explore the novel thinking that accompanies an introduction to complex numbers. Kudos for an excellent use of technology to teach a detailed lesson in basic complex number procedures.
I mean geometrically the answers also make perfect sense, as in both have a lenght of 1 and the first one is225° when you square it it means 2*225° =90° and 45° also squared 45*2=90°, which means both have a Rotation of 90° and length 1 which is just i or e^iπ/2
The square root of i satisfies x^4 = -1, so factor x^4 + 1 = (x^2 - sqrt(2) x + 1)(x^2 + sqrt(2) x + 1) then solve each quadratic term for x, to get the square roots of i and -i.
Using the polar form of complex numbers this becomes a little easier. First of all, realize that multiplying 2 complex numbers in polar form can be done by multiplying their magnitudes and adding their angles. Knowing this, you can find the square root of a complex number by taking the square root of its magnitude and halving the angle. So, in this case, notice that i has a magnitude of 1 and an angle of 90 degrees (1/4 of the way around the unit circle) on the complex plane. To get the square root, we cut the angle in half (45 degrees or 1/8 of the way around the unit circle) and keep the magnitude (the square root of 1 is 1). This is a point on the unit circle 1 unit away from the origin at 45 degrees. Using the Pythagorean theorem, it's easy to see that the coordinates of this point would be sqrt(2)/2 in both the horizontal and vertical directions. That would be sqrt(2)/2 + sqrt(2)/2 i, which is one of Andy's results in this video. What about the other result? Notice that in polar form, the angle of i can also be thought of as 360 + 90 degrees (1 1/4 times around the origin), or 450 degrees. Half of that is 225 degrees (5/8 of the way around the unit circle). That's exactly opposite 45 degrees. So, it's just the negative of the first result, namely -sqrt(2)/2 - sqrt(2)/2 i, which is Andy's other result.
can you solve this question please: Two circles, each with a radius of 2 cm, are tangent to each other. A line is drawn that is tangent to both circles. A third circle is then drawn so that it is tangent to both of the original circles and the line. Find the radius of the smaller circle.
You're having a hard time because he ignored the geometrical interpretation. They're on the unit circle, at 45° and -135°. Just think of it in polar form, z = r cis θ, squared that's just z² = r² cis 2θ. We know that i is on the complex unit circle, at 90°. So r² = 1 and 2θ = 90°, meaning that the solutions are also on the unit circle, and rotating through θ twice gets you to 90°. i.e. 45° or -135°. That's all there is to it. He did all that rigor and nicely animated algebra for something you can do in your head. That is, how do I turn twice to get to 90°? NB: I figured degrees would be more intuitive than radians for this comment
Very simple doing it geometrically in 2D vector space .. multiplication multiplies length of a vector and adds angles. Square root squares the length and divides the angle by two. So the first solution is obvious .. length 1, angle 45 degrees. The other is easy to miss, but once you realize there are many infinite many ways how to express an angle by adding 360 degrees, it's also simple.
"When dealing with imaginary numbers you can't combine them with real numbers". That's the whole point of imaginary numbers. When get squared they literally become reals.
Even easier than Euler's formula (although equivalent, really): just use the geometry of complex numbers on the unit circle. i = 90 degrees, so root i = 45 degrees. Done.
it's good to see how the math works out like this. I find it very useful to use the the unit circle to understand complex numbers. what points when multiplied together rotate around to 0+1i? And the answer doesn't even need any calculations. sqrt2/2 + sqrt2/2 i, and -sqrt2/2 - sqrt2/2i. you can also use polar coordinates since they are easier to multiply and divide than cartesian coordinates. I = 1 @90degrees. So the answer is 1 @45 degrees and 1 @ 225 degrees.
Or just write i in exponential (polar form) as exp[(\pi/2 + n 2\pi) i], n = 0, 1, 2,... Raise that to the power of 1/2 to find the square root, and crank the handle on n=0 and n=1 (n=2,3,etc just repeat the first two soluions). For n=0, S_0 = exp[(\pi/4) i] For n=1, S_1 = exp[(5\pi/4) i] If you need the solutiosn in rectangular form use euler's to get +/- 1/sqrt(2) (1+i)
nice…also as i is a “rotator” (so to speak) the answer results in the intuitive 45 degree angle-vs i bring essentially a 90 degree angle (coordinate wise) I’m not very articulate this early…but I’m sure you catch my drift
Just wondering, as √4 = 2 and √4 ≠ -2 because √ is always positive, isn't it the same thing for this ? I mean, I think (-√2/2 - √2/2i) is as wrong as -2 for √4
Maybe. But I think that the rule of the radical having only one solution is not very useful with complex numbers. Specially with radical indexes bigger than 2 (cubic roots, forth roots, etc).
@@aldineisampaio I'll repost what I wrote earlier: We need a way to say √2 and have it be different from -√2. We need a way to say √i and have it be different from -√i. We all agree that √2 + √3 + √5 + √7 is a single value, not 16 different values. We don't want it to suddenly become 16 different values when it's √-2 + √-3 + √-5 + √-7. We don't want √a + √b +√c + √d to be either 1, 2, 4, 8, or 16 values depending how many of a, b, c, and d are real. So we have conventions. √ denotes the principle root. It's always defined, even with complex numbers. There are different conventions, and you are free to choose whichever branch cut you like, as long as you are consistent, but in no case does √a is denote two or more values.
just express it in polar and it's easy i = e^(i*pi/2 + 2n*pi) so srt(i) = e^(i*pi/4 + n*pi) = cos(pi/4 + n*pi) + i sin(pi/4 + n*pi) = 1/srt(2) + 1/srt(2) * i, -1/srt(2) - 1/srt(2) so srt(2)/2 + srt(2)/2, -srt(2)/2 -srt(2)/2
By the way i is equal to e^((pi/2)*i) and squareroot of that is e^((pi/4)*i) and sqaureroot of that is e^((pi/8)*i) and so on. Assuming you only care about the principle root
You’re missing that i = e^i(pi/2 + k2pi) for any integer k. So, sqrt(i) = e^i(pi/4 + kpi), which gives the 2 solutions from the video (as well as infinite others).
Simple, use argument π÷2 and module is 1, exp(iπ÷4), develop with trigonomic expression and developp again √2/2 +√2/2 i Works also with arg of the solution π/4, -3π/4
Taking the square of sqrt(i) introduces an extra incorrect answer of sqrt(i). sqrt(i) is a number and not a function with a variable, it cannot be equal to two different numbers. Think about if we want to calculate sqrt(1). If we take the square of it, we will end up with 1 and -1 are two "answers", but -1 != sqrt(1)
I think its easier both conceptually and algebraically to express the imaginary unit i in polar form as 1cis(pi/2) and then apply DeMoivre's Theorem to get +-1cis(pi/4) giving +-([sqrt2 + sqrt2*i]/2)
manyvpeople including me don’t know what that is if he just pulled some theorm out of nowhere without deriving it it would feel like it came from nothing the way the video is organized allows people to fully understand what is happening
@@baconheadhair6938 I understand what you mean but equally you should have learnt imaginary numbers in such a way that you also learnt the polar and euler forms to express complex numbers. Because the usual representation a + bi is really bad and doesnt segue well into other ideas whereas the polar and euler forms can help you grasp a better understanding of stuff like euler's identity and trig expressions. Furthermore, DeMoivre's Theorem is the most convenient way to find roots and also square complex numbers and is only applicable in polar / euler form, sooo those forms are almost always better i believe. I think the video shouldve actually like, explained what the imaginary unit is, what complex numbers are, why you express them as a + bi, what the polar / euler forms are and why theyre easier to use here than using the rectangular form. And then solved the question through the application of DeMoivre's Theorem in polar coordinates because there are many many questions similar to this in calculus courses, such as "find the roots of unity of z^n = 1" that are ONLY solvable through DeMoivre's Theorem and polar/euler form. This video wouldve helped more with the conceptual understanding of complex numbers and how to solve questions surrounding them if it had gone more in depth imo
@@mancerguy1841 no i haven’t even learned complex numbers in school and im sure theres many others who haven’t either. if he just grazed over polar form and whatever the euler form is, it wouldn’t be as valuable to newcomers in this subject i think you’re right though that it would be useful if he went in depth about each of these forms, but to do it effectively in a way that doesnt require extensive background knowledge would be lengthy . also even if that theorm is useful, it would still feel like magic if he didn’t derive it, even if he explained what it is
Here's the fun part. Thinking the square root of i as a complex number like a+bi isn't actually an assumption, it is 100% truth. Square root of i must be a complex number because complex numbers are the largest set of numbers, any real or imaginary number is also a complex number. For example, 2 is a complex number because it can be written as 2+0i, 3i is a complex number because it can be written as 0+3i. Therefore, square root of any real or imaginary number must be something like a+bi.
1/√2=√2/2 (and, more generally, 1/√x=√x/x) so the original was just the complex conjugate of the right answer? Or more likely, someone just mixing up the principle square root with the other one. Interestingly, that wrong answer was the square root of -i.
First. let √i = a + ib where *a and b are real.* Now we can use your algebraic method, although i = e^(iπ/2), so √i = e^(iπ/4) = cos(π/4) + i.sin(π/4) = (1+i)/√2) is quicker. So i = a^2 - b^2 + 2abi. Equate real parts giving a^2 - b^2 = 0. So a^2 = b^2, hence a = ± b. Equate imaginary parts giving i = 2abi, therefore 2ab = 1. If a = b, then 2a^2 = 1, which makes a = √(1/2) and b = √(1/2) by the convention that the principal square root of a real is positive. So √i = (1 + i)/√2 If a = -b, then -2a^2 = 1, leading to a being imaginary, which contradicts the initial condition. One solution and it doesn't take 7 minutes.
If only you have used Euler's formula you would've done it in 1 minute. Since √a = a^(1/2) And i=e^i(π/2 +2kπ) Then √i=e^i(π/2+2kπ)/2=e^i(π/4+kπ) while k is an integer Then we change the Euler's formula into an algebric one using the r(cos(theta)+isin(theta)) which will give us the same results you got. How simple and fast
isn't it faster to simply write i = e( i pi/2) ; if you want the sqrt, then i ^ (1/2) = e (i pi /2 + 2 k pi)^(1/2) = e ( i pi /4) for K=0 and e ( i 5 pi /4) for K = 1 If you really, really really really want to write Cos pi/4 and sin pi/4 as sqr 2/2 that works 2. e ( i pi/4) is just a lot easier to visualize for students on the unit circle I'd say
I would say you have found 2 solutions to the problem z^2=i, that is not finding the value of sqrt(i). If you are treating sqrt(z) as a function, it needs to have only 1 solution.
I didn't like how you set out to find "a" and "b" to satisfy "a + bi", and found values for "a and b" that were imaginary. I thought coefficients a and b for a complex number had to be real numbers
@@manibabai2115 Square root of a square root is a hypercube root. They're correct. √x = x^(1/2) √(√x) = x^((1/2)*(1/2)) = x^(1/4) = ∜x So if [ x = i ] for √(√x), we get: √(√i) = ∜i
((sqrt(2) + i(sqrt(2)/4) For folks who don't like radicals in their denominators. I think. I remember doing it in my head during a smoke break at work once. If you know how to work in the complex plane, and that the principle cube root of -8 is NOT -2, but sqrt(3)i, then this is just basic complex arithmetic. i^i? Now that's hard.
I wanted to make 2 comments about this video.
1. I intentionally let A and B be complex numbers to show that when everything is simplified it leads to the same solutions, I thought that was such a cool conclusion. A lot of people are upset I did this, I should have expressed more clearly what I was doing.
2. The final solution should have only been the principle root. I was so excited about A and B being complex detail, that I forgot to restrict the final solution.
what is a principle root?
Finding these roots by viewing i as a point in the plain is much simpler
@@ZincAddictbasically just the positive root
Andy at the wedding altar: “This looks important, let’s put a box around it”
What about the honeymoon?
Actually, this morbidly applies to a funeral even better.
@ I can just picture him saying this at a funeral 🤣
@@Qermaq- after they consummate he’ll give a very concerted “how exciting.”
@@jackmandu Hey, honey, this looks like a fun one.
The honeymoon: How exciting. A box. I bet this is important.
Who knew the 9th letter of the alphabet, so innocent looking, held so much power...
Eldritch Abomination.
I hate _i_
a look into a plane beyond mortal perception. imaginary, if you will.
well 9th letter of english alphabet or 9th letter of greek ‘iota ‘😢
*Iota
*Greek letter
Just one little flaw here: √i is per definitionem ONLY the principal root (1 + i) / √2.
If you'd ask for x² = i, you'd get both solutions ± (1 + i) / √2 = ± (1 + i) * √2/2.
or you could just say it's a 45° and 225° rotation on a unit circle in the complex plane.
This is because 45+45=90, and 225+225=450=360+90.
Then just do sin and cos of those and you'll get the values which is why you also have sqrt(2)/2, which is sin and cos (45°), as well as -sqrt(2)/2, which is sin and cos (225°)
hi
what day it it going
Bam! That's it.
Complex numbers are the bridge between algebra and geometry.
Interesting way to put it
I am so glad I learned how to use the polar form of writing imaginary numbers. Great Video!
exactly
Using the polar form drops the entire calculation to four or five lines compared to the video's interminable length.
@@TonyFisher-lo8hhtbf this video is padded out by at least 4 minutes, right from his expansion of a+bi squared he took a detour onto a much longer calculation journey. Frankly if you get a^2-b^2=0 it’s obvious that a=b (or a=-b) and you can sub it in to 1=2ab to get 1=2a^2 solve for a and as a=b you also get b.
There is one detail that should be pointed out in your calculations. The value a is only allowed to be real because it is the real part of the complex number a + bi. Therefore, when you get to the equation a^2 = -1/2, this equation should be rejected because it is not possible that a real number squared equals a negative number. In fact, this is why you get two extra redundant answers. The answer is not wrong, but the math used to get to the answer can lead to some misunderstanding about the differences in the complex and real numbers. Note: I didn´t read all the comments before writing mine, but there were other comments about this exact same issue)
i thought he forgot about it
I wasn't going to comment on this issue at all, but then it took up so much time in the back half of the video.
One thing to add: It's not incorrect to allow a and b to be complex numbers, but you're guaranteed to get all the solutions with less work if you force them to be real. This is because a+bi can already represent every possible complex number when a and b are real. Allowing complex a and b will only get you numbers a+bi that would have been represented regardless.
@@Mothuzad By definition, a and b are real. To make a and b complex is not helping in that situation and make the problem more 'complex'... When we do math, we have to stick with a definition. This is why sometimes we have to be careful with recreational maths; sometimes the person using the maths will abuse the definition and might gives the illusion that it can be done in other context too...
I’m not so sure about this comment. You’re relying on the assumption that a must be a real number but I’m fairly certain that’s not true.
If a were a complex number and b was anything at all then the term a+bi would still be a complex number, which was our only real “requirement” or assumption. We didn’t try to claim that a or b were real numbers, only that a+bi is a complex number
@@Orrinn123You're right in one thing being "not sure". If you look up the definition of complex number, it is defined as "a,b are real numbers". You cant define complex number with the the definition of complex number. And yes, in this video, he did it wrong in some way and still got right solution.
7:51 you can just put the ± symbol and parenthesis
Bro did it the unnecessarily hard way
My thoughts exactly.
Easier than explaining the complex logarithm.
I was gonna use the argand diagram way of reasoning
@@sandpaper7781what’s one of those?
I think there should be only one answer, since the radical symbol is defined to give the Principal Root and not all roots.
Defined by whom?
@@deltalima6703 a root as a function must have only 1 output
The equation should be x^2 = i to have 2 different answers (which is x = +- root of i)
nah
@@deltalima6703 General consensus.
From WIkipedia: "Principal square root: Each positive real number has two square roots, one positive and the other negative. The square root symbol refers to the principal square root, which is the positive one. The two square roots of a negative number are both imaginary numbers, and the square root symbol refers to the principal square root, the one with a positive imaginary part."
@@deltalima6703 By near universal convention.
Dude, why going the hard way? Just use Euler’s formula: sqrt(i)=sqrt(e^(i*90))= e^(i*90/2)=e^(i*45)=cos(45)+i*sin(45) and the rest is history
Cant wait till i go to college and see this!
But you will "lose" a solution.
But for that you may prove Euler formula (which is one of the most beautiful things I’ve learned at uni) but this require knowledge of Taylor series and proving it (with all the relevant calculus), it’s very tedious.
We can either skip these steps and say “just use Euler” without showing why we can use Euler, but it feels less like mathematical method and more like magic, or choose a different path from the ground up.
What I like in the video is that it isn’t require any background or mathematical assumptions besides basic algebra.
@@OmerNachmaniZ I guess I didn’t view it from this side. Agreed!
@@DTier_fish It's not hard. Graph complex value z as (a + bi) on the complex plane. The distance to the origin is sqrt(a^2 + b^2) and that's r, the modulus. The angle from the positive real axis is theta, the argument, and that's atan(b/a). Once you know the modulus and the argument, you can say that z = re^(i*theta). And then you can use all the log stuff with e on complex numbers. So if z = 1 + sqrt(3)i, the modulus (r) is 2 (1^2 + sqrt(3)^2), and the argument (theta) is pi/3 (atan sqrt(3)/1). So z = 2e^(i * pi/3). Have fun!
Draw a unit circle centered on the origin of the complex plane. i is at the top of the circle at its intersection with the imaginary i axis. i is 90° or π/2 radians around the circle from the positive real axis. Also i=e^{iπ/2)=cos(π/2)+i*sin(π/2). √i is π/4 around the unit circle or √i=e^{iπ/4)=cos(π/4)+i*sin(π/4)=√2/2+i*√2/2=(√2/2)(1+i).
Yeah. Geometrically it’s easy to see the 1/√2.
More obvious to use polar form.
Beat me to it 😊. I, too, find the geometric interpretation much simpler. Does it constitute a proof? I think so… worth a follow-up, IMO.
@ if you use axioms of geometry or other proven theorems, yes. It’s a proof.
@@rbettsx I don't think a proof was asked for.
What a beautiful evening to absolutely fry my brain cells 😭😭😭😭
It was basic addition multiplication and some substitution, what exactly fried your head?
@@simplylmao the fact that you are dealing with imaginary numbers and solution, if someone who never ever met something like this before it can be overwhelming for them
@@therealnonooYes
@@therealnonoo as long as you know i = root of -1, everything else is really just squaring and multiplying really.
Ofc i see how it can be challenging for someone new to maths
it aint even bad at all, it is so simple.
you can also notate it as just ±(1+i)(√2/2)
ok einstein we get it, you're smart
@@huseynhuseynli2922 su
shouldnt sqrt(x) give only the principal value here... which should never be ngetaive? so iin this case its only the positive option... for example...you say sqrt(36)=6 and not ±6... correct me if im wrong plz
@@Zkflames ± is usually used with complex numbers to avoid confusion. its normal for most operations with them to have 2 answers
@@Zkflames if you solve for sqrt(x) you only get the principal value however solving for x^2 = 36 (for example) you would get x = ±6
So many boxes, so many colors.
I really like your style of working through things. The visual equations are easy to follow and show all the steps, while not ending up in pages and pages of equations at the end, and I love that you are very careful not to skip over even the simple steps, which makes it really easy to see that there's really no magic in there.
The solution at the end could actually just be written as ±(√2/2 + (√2/2)i). This is consistent with the fact that we already know that when taking a square root of something, the result should always come out to be "plus or minus (something)". (And then, as you said in the comment, since the radical symbol means specifically the principal root, you should technically just take the positive one as the answer to this question.)
Oh my gosh, he did so many unnecessary or even contradictory processes. It can be done in 30 seconds: a² = b², therefore a = ±b. Since 2ab = 1, then a = b. Substituting, we have 2a² = 1 and then, a = ± sqrt(2)/2. Since a = b, sqrt(i) = sqrt(2)/2 + i.sqrt(2)/2 or -sqrt(2)/2 -i.sqrt(2)/2
Verdade, muito mais simples!
Shush
There’s so much missing from that and it’s not an intuitive way to think about it
@@justyceleague698 What is missing?
z = i ← this is a complex number
The modulus of (i) is 1, so the modulus of √i is: 1^(1/2) = 1
The argument of (i) is (π/2), so the argument of √z is [(π/2)/2], i.e. (π/4)
Thus we can deduce that the first root of (i) is:
z1 = 1.[cos(π/4) + i.sin(π/4)] → and to get the second root, we add an angle of (2π)/2, i.e. (π)
z2 = 1.[cos{(π/4) + π} + i.sin{(π/4) + π}]
It gives
z1 = 1.[cos(π/4) + i.sin(π/4)]
z1 = cos(π/4) + i.sin(π/4)
z1 = [(√2)/2] + i.[(√2)/2]
z1 = [(√2)/2].(1 + i)
z2 = 1.[cos{(π/4) + π} + i.sin{(π/4) + π}]
z2 = cos{(π/4) + π} + i.sin{(π/4) + π}
z2 = - cos(π/4) - i.sin(π/4)
z2 = - [(√2)/2] - i.[(√2)/2]
z2 = - [(√2)/2].(1 + i)
Summarize
√i = ± [(√2)/2].(1 + i)
√i = ± (√2).(1 + i)/2
√i = ± (√2 + i√2)/2
Girolamo Cardano(discovered imaginary numbers ) watching through that right door and saying
" THIS guy is cooking "🤣🤣
The way I got around understanding this, was to first to define i in exponential form and raising it to exponent 1/2. Then just redefine it to polar form.
His hotness is what carried me through this video. I'm glad I got to the end, though, because this is really interesting!
I do not agree that this should result in two answers. When we write "x^2 = 4", then we do expect two answers, +2 and -2. But when we write "sqrt(4)", we expect just the positive square root; or in a complex context, the primary branch. Likewise, when we write "x^2 = i", we expect two answers, but "sqrt(i)" should have just one. It is commonly accepted that sqrt(2)(1+i)/2 is the primary branch we expect to keep.
what determines the primary branch? acceptance in the community? or is there a mathematical reason (ie sqrt(x) for x>0 = something >0) ? do you take the one with the positive real?
When dealing with complex numbers things gets a little crazy. When we have a square root, sure, since the second root will always be the first root multiplied by -1. But when we have cube roots, forth roots and so on, it is not so simple anymore. In the real world, makes sense a radical giving only one solution, but in complex world that is not very useful.
@@ricoseb There is a mathematical reason. Is the root with the smallest positive angle in the polar form representation of the number. By the way, that reason have a curious implication: the principal cubic root of a negative real number is never a real number, because the real root would make a angle of pi and will aways exist a complex root with a smaller angle (pi/3). Because of that, Wolfram Alpha cannot plot certain graphs correctly when the function involves cubic roots, since it consider only the main branch of the radical.
There are two schools of thought here, and both are valid. One carries the spirit of the square root bijection from real arithmetic and as you suggest considers the value with the smallest non-negative argument the principal and refers to that as the proper square root. The other, which is more prevalent, is that once we enter the complex world we dispense with the "one square root" idea and we say that all complex values have two square roots, three cube roots, etc. We still can identify the principal as above, but all values of the root are considered. Personally, I prefer the latter point of view, but I see logic in yours as well.
From a mathematical point of view, the guy from the video should be clear on how is the symbol sqrt(a + bi) is defined. In mathematics, it is fairly standard to use sqrt(a + bi) for the principal branch of the complex square root function and (a + bi)^{1/2} for the set-valued square root function (there will be two values as the output for the two roots). With that being said, we can infer directly from the video how he interpreted the symbol sqrt(i): he interpreted it as (i)^{1/2}. Hope this helps.
The hard part a lot of times is how do you know what “looks important” and what doesn’t
兩種更快的方法：
（1） 2:47 用 2ab=1, a=±b
（2）複數平面的兩數相乘等同於長度相乘和角度相加，就像 i^2=-1, i^(1/2) 就是位於45°、長度1的複數
（其實還有135°的，但因為要取正根所以不符合）
But didn’t you define that “a” would be the real part and “b” the imaginary part? I assumed the first two answers would be excluded because of that. Although when you rearranged the answers were the same, the first two were “b+ai” instead of “a+bi.”
Yes, he should have excluded those answers.
Not true actually. He did sub them in properly, a and b are complex also, not real as you assumed.
@@deltalima6703I disagree, when the complex answer is defined as a+bi, a and b must be real. When you draw the complex plane, you don't expect each axis to be itself complex.
The definition of a complex number is z=a+bi where a,b belong to R. That would help simplify the equation a bit, since a^2>0 given a is a real number.
Given the same fraction in both, I would factor it out and write as (-√2 / 2)(1 + i) and (√2 / 2)(1 + i) making the solution literally ±(√2 / 2)(1 + i).
But since √ generally means the principal root, meaning no negative values, the only actual solution should be (√2 / 2)(1 + i). The other one is an extraneous solution introduced by the squaring at the very beginning.
I will take your word for it.
All you have to do is convert _i_ to its polar form. Then square-root the magnitude (which does nothing in this case) and divide the phase by 2, which gives you a complex number with a magnitude of 1 and a phase of π/4. Converting back to rectangular form gives sqrt(2)/2 + sqrt(2)i/2.
To get the second solution, express i's phase as 5π/2. Then applying the square root operation will give you a complex number with a magnitude of 1 and a phase of 5π/4, which turns out to be the negative of the first solution, -sqrt(2)/2 - sqrt(2)i/2.
I went down into a "plotting complex numbers and its square roots in the complex plane" rabbit hole after watching this! Hoping you'll also make a short video on this pleeease
My favourite way to do this is use polar form of i = e^i(pi/2 + k2pi), for any integer k. Then sqrt(i) = i^(1/2) = e^i(pi/2 + k2pi)*(1/2) = e^i(pi/4 + kpi). This gives the same 2 solutions in the video plus infinite others. Though they’re all equivalent to one of the two in the video.
Conceptually how I think about it is multiplying a complex number z by i is the same as rotating it by 90 degrees. Square root of x asks, “what number multiplied twice is the same as multiplying by x?” Combined: “what rotation can we apply twice to get a 90 degree rotation?” The obvious answer is 45 degrees, but also 225 degrees (225*2 - 360 = 90). Another way to find these is 2x = 90, 450, 810, … => x = 45, 225, 405, …
You did it the hard way. There are two vectors. Both have a length of 1. Rotate the I vector half way to a angle of zero so the fist solution is cos(45 deg)+isin(45deg) . the other solution is to rotate I half way to zero deg counter clockwise.
Not really. He just did it the purely algebraic way (which I’m fairly certain is the method this channel almost exclusively uses for problems of this sort)
I just used the graph to do 45° on 1, to get a triangle where cos(45)=(√2)/2 and sin(45)=(√2)/2 for both a and b. Solved using the graph!
2 remarks.
sqrt(z) "represents" two values, because "sqrt" is not a function.
By choice, decision has been taken to say that sqrt(z) is positive if sqrt(z) € |R.
But if z is imaginary, for example z=i, sqrt(i) represents
two values which are opposed.
This community college math instructor is pleased to see so many alternative procedures to find the answer to sqrt(i) in the comments. Many of them come from principles and procedures that come from a moderately advanced study of complex numbers, but I don't think that's what the author is trying to do. It's pretty clear to me that he is showing a "first week" lesson to students who are just beginning to explore the novel thinking that accompanies an introduction to complex numbers. Kudos for an excellent use of technology to teach a detailed lesson in basic complex number procedures.
I mean geometrically the answers also make perfect sense, as in both have a lenght of 1 and the first one is225° when you square it it means 2*225° =90° and 45° also squared 45*2=90°, which means both have a Rotation of 90° and length 1 which is just i or e^iπ/2
The square root of i satisfies x^4 = -1, so factor x^4 + 1 = (x^2 - sqrt(2) x + 1)(x^2 + sqrt(2) x + 1) then solve each quadratic term for x, to get the square roots of i and -i.
Using the polar form of complex numbers this becomes a little easier. First of all, realize that multiplying 2 complex numbers in polar form can be done by multiplying their magnitudes and adding their angles. Knowing this, you can find the square root of a complex number by taking the square root of its magnitude and halving the angle.
So, in this case, notice that i has a magnitude of 1 and an angle of 90 degrees (1/4 of the way around the unit circle) on the complex plane. To get the square root, we cut the angle in half (45 degrees or 1/8 of the way around the unit circle) and keep the magnitude (the square root of 1 is 1). This is a point on the unit circle 1 unit away from the origin at 45 degrees. Using the Pythagorean theorem, it's easy to see that the coordinates of this point would be sqrt(2)/2 in both the horizontal and vertical directions. That would be sqrt(2)/2 + sqrt(2)/2 i, which is one of Andy's results in this video.
What about the other result? Notice that in polar form, the angle of i can also be thought of as 360 + 90 degrees (1 1/4 times around the origin), or 450 degrees. Half of that is 225 degrees (5/8 of the way around the unit circle). That's exactly opposite 45 degrees. So, it's just the negative of the first result, namely -sqrt(2)/2 - sqrt(2)/2 i, which is Andy's other result.
this may be a long explanation, but this was really cool to watch :>
can you solve this question please: Two circles, each with a radius of 2 cm, are tangent to each other. A line is drawn that is tangent to both circles. A third circle is then drawn so that it is tangent to both of the original circles and the line. Find the radius of the smaller circle.
basically britheguy but alot of headaches and brain damage
This problem becomes trivial if you use phasors
Functions shouldn't have only 1 x value?
You could also say plus or minus since the equations are similar
or you can use Euler formula: sqrt(i)=i^(1/2)=e^(π/2*1/2) or e^(5π/2*1/2) = e^(π/4) or e^(5π/4) =sqrt(2)/2+ sqrt(2)*i/2 or -sqrt(2)/2-sqrt(2)*i/2
I have never seen it explained so brilliantly. 🤗
Hi Andy, could you plot the solutions on the complex plane, please? I'm having a hard time picturing where they would be.
You're having a hard time because he ignored the geometrical interpretation. They're on the unit circle, at 45° and -135°.
Just think of it in polar form, z = r cis θ, squared that's just z² = r² cis 2θ.
We know that i is on the complex unit circle, at 90°. So r² = 1 and 2θ = 90°, meaning that the solutions are also on the unit circle, and rotating through θ twice gets you to 90°.
i.e. 45° or -135°.
That's all there is to it. He did all that rigor and nicely animated algebra for something you can do in your head. That is, how do I turn twice to get to 90°?
NB: I figured degrees would be more intuitive than radians for this comment
”I am going to solve this in 3, 2, 1.” ”How exiting!” Great rituals 😊. Thank you for exellent videos!
Very simple doing it geometrically in 2D vector space .. multiplication multiplies length of a vector and adds angles. Square root squares the length and divides the angle by two.
So the first solution is obvious .. length 1, angle 45 degrees. The other is easy to miss, but once you realize there are many infinite many ways how to express an angle by adding 360 degrees, it's also simple.
"When dealing with imaginary numbers you can't combine them with real numbers".
That's the whole point of imaginary numbers. When get squared they literally become reals.
Polar plots are the solution to a complex number function
aw man you almost got it, it's the principal root
Cool now find the square root of that
Even easier than Euler's formula (although equivalent, really): just use the geometry of complex numbers on the unit circle. i = 90 degrees, so root i = 45 degrees. Done.
it's good to see how the math works out like this. I find it very useful to use the the unit circle to understand complex numbers.
what points when multiplied together rotate around to 0+1i? And the answer doesn't even need any calculations. sqrt2/2 + sqrt2/2 i, and -sqrt2/2 - sqrt2/2i.
you can also use polar coordinates since they are easier to multiply and divide than cartesian coordinates. I = 1 @90degrees. So the answer is 1 @45 degrees and 1 @ 225 degrees.
sqrt(i) = (e^i*(pi/2 +2*pi*n)^1/2 =(e^i*pi/4)*(e^i*n*pi) =
n=0 => e^i*pi/4 = cos(pi/4) + i*sin(pi/4) = 1/sqrt(2) * (1 + i)
n=1 => e^i*(pi/4 + pi) = cons(5pi/4) + i*sin(5pi/4) = - 1/sqrt(2) * (1 + i)
n>1 repeat itself
Or just write i in exponential (polar form) as exp[(\pi/2 + n 2\pi) i], n = 0, 1, 2,... Raise that to the power of 1/2 to find the square root, and crank the handle on n=0 and n=1 (n=2,3,etc just repeat the first two soluions).
For n=0, S_0 = exp[(\pi/4) i]
For n=1, S_1 = exp[(5\pi/4) i]
If you need the solutiosn in rectangular form use euler's to get +/- 1/sqrt(2) (1+i)
so this video is basically talking about the root of imagination
why cant we say that root(i) = i^2* root(1) and root 1 is equal to 1 therefore root(i) is equal to i^2
Loved this video! What’s your favourite math concept to teach?
nice…also as i is a “rotator” (so to speak) the answer results in the intuitive 45 degree angle-vs i bring essentially a 90 degree angle (coordinate wise) I’m not very articulate this early…but I’m sure you catch my drift
Putting i on the left of a fraction is confusing and makes it look like a mixed fraction.
That was a fantastic equation
Just wondering, as √4 = 2 and √4 ≠ -2 because √ is always positive, isn't it the same thing for this ?
I mean, I think (-√2/2 - √2/2i) is as wrong as -2 for √4
Maybe. But I think that the rule of the radical having only one solution is not very useful with complex numbers. Specially with radical indexes bigger than 2 (cubic roots, forth roots, etc).
@@aldineisampaio I'll repost what I wrote earlier: We need a way to say √2 and have it be different from -√2. We need a way to say √i and have it be different from -√i. We all agree that √2 + √3 + √5 + √7 is a single value, not 16 different values. We don't want it to suddenly become 16 different values when it's √-2 + √-3 + √-5 + √-7. We don't want √a + √b +√c + √d to be either 1, 2, 4, 8, or 16 values depending how many of a, b, c, and d are real.
So we have conventions. √ denotes the principle root. It's always defined, even with complex numbers. There are different conventions, and you are free to choose whichever branch cut you like, as long as you are consistent, but in no case does √a is denote two or more values.
Really cool, what software was used to create the equations? I guess PowerPoint for the animations
just express it in polar and it's easy i = e^(i*pi/2 + 2n*pi) so srt(i) = e^(i*pi/4 + n*pi) = cos(pi/4 + n*pi) + i sin(pi/4 + n*pi) = 1/srt(2) + 1/srt(2) * i, -1/srt(2) - 1/srt(2)
so srt(2)/2 + srt(2)/2, -srt(2)/2 -srt(2)/2
2:29 that means abs(a)^2=abs(b)^2 wich also means abs(a)=abs(b)
By the way i is equal to e^((pi/2)*i) and squareroot of that is e^((pi/4)*i) and sqaureroot of that is e^((pi/8)*i) and so on. Assuming you only care about the principle root
(cos(2kπ+π/2)+sin(2kπ)+π/2)^1/2 then apply de-movire's theorem
This is easy. By Euler's formula, i = e^i(pi/2). So sqrt(i) = e^i(pi/4) = sqrt(2)/2 + i sqrt(2)/2
You’re missing that i = e^i(pi/2 + k2pi) for any integer k. So, sqrt(i) = e^i(pi/4 + kpi), which gives the 2 solutions from the video (as well as infinite others).
Simple, use argument π÷2 and module is 1, exp(iπ÷4), develop with trigonomic expression and developp again
√2/2 +√2/2 i
Works also with arg of the solution π/4, -3π/4
Taking the square of sqrt(i) introduces an extra incorrect answer of sqrt(i). sqrt(i) is a number and not a function with a variable, it cannot be equal to two different numbers.
Think about if we want to calculate sqrt(1). If we take the square of it, we will end up with 1 and -1 are two "answers", but -1 != sqrt(1)
I just don't understand how anyone can assume that square root of i will be a complex number in the form a+bi
Much easier way to solve. Multiply by 2/2 inside the square root:
√i=√(i * 2/2)=
√(2i)/√2=
√(1+i)²/√2=
(1+i)/√2=[(1+i)/√2]*(√2/√2)=
(√2/2)(1+i).
I think its easier both conceptually and algebraically to express the imaginary unit i in polar form as 1cis(pi/2) and then apply DeMoivre's Theorem to get +-1cis(pi/4) giving +-([sqrt2 + sqrt2*i]/2)
no
manyvpeople including me don’t know what that is
if he just pulled some theorm out of nowhere without deriving it it would feel like it came from nothing
the way the video is organized allows people to fully understand what is happening
unobvious assumptions from unobvious theorems ruin math videos because they just make it worthless to the average viewer
@@baconheadhair6938 I understand what you mean but equally you should have learnt imaginary numbers in such a way that you also learnt the polar and euler forms to express complex numbers. Because the usual representation a + bi is really bad and doesnt segue well into other ideas whereas the polar and euler forms can help you grasp a better understanding of stuff like euler's identity and trig expressions.
Furthermore, DeMoivre's Theorem is the most convenient way to find roots and also square complex numbers and is only applicable in polar / euler form, sooo those forms are almost always better i believe.
I think the video shouldve actually like, explained what the imaginary unit is, what complex numbers are, why you express them as a + bi, what the polar / euler forms are and why theyre easier to use here than using the rectangular form. And then solved the question through the application of DeMoivre's Theorem in polar coordinates because there are many many questions similar to this in calculus courses, such as "find the roots of unity of z^n = 1" that are ONLY solvable through DeMoivre's Theorem and polar/euler form.
This video wouldve helped more with the conceptual understanding of complex numbers and how to solve questions surrounding them if it had gone more in depth imo
@@mancerguy1841 no i haven’t even learned complex numbers in school and im sure theres many others who haven’t either. if he just grazed over polar form and whatever the euler form is, it wouldn’t be as valuable to newcomers in this subject
i think you’re right though that it would be useful if he went in depth about each of these forms, but to do it effectively in a way that doesnt require extensive background knowledge would be lengthy . also even if that theorm is useful, it would still feel like magic if he didn’t derive it, even if he explained what it is
check: (√2/2 + i√2/2) (√2/2 + i√2/2) = 2/4 + i2/4 + i2/4 +i^2(2/4) = 1/2 +I1/2 + I/2 -1 = i
-(√2/2 + i√2/2) -(√2/2 + i√2/2) = 2/4 + i2/4 + i2/4 +i^2(2/4) = 1/2 +I1/2 + I/2 -1 = i
What tool do u use for your videos
you can figure out pretty early on that a=b
Since a and b (of a + bi) are real numbers, there is no reason to consider the complex forth roots of 1/4 when calculating a or b.
Here's the fun part. Thinking the square root of i as a complex number like a+bi isn't actually an assumption, it is 100% truth. Square root of i must be a complex number because complex numbers are the largest set of numbers, any real or imaginary number is also a complex number. For example, 2 is a complex number because it can be written as 2+0i, 3i is a complex number because it can be written as 0+3i. Therefore, square root of any real or imaginary number must be something like a+bi.
1/√2=√2/2 (and, more generally, 1/√x=√x/x) so the original was just the complex conjugate of the right answer? Or more likely, someone just mixing up the principle square root with the other one. Interestingly, that wrong answer was the square root of -i.
Could've posted this before my specialist maths exam
First. let √i = a + ib where *a and b are real.* Now we can use your algebraic method, although i = e^(iπ/2), so √i = e^(iπ/4) = cos(π/4) + i.sin(π/4) = (1+i)/√2) is quicker.
So i = a^2 - b^2 + 2abi. Equate real parts giving a^2 - b^2 = 0. So a^2 = b^2, hence a = ± b. Equate imaginary parts giving i = 2abi, therefore 2ab = 1.
If a = b, then 2a^2 = 1, which makes a = √(1/2) and b = √(1/2) by the convention that the principal square root of a real is positive. So √i = (1 + i)/√2
If a = -b, then -2a^2 = 1, leading to a being imaginary, which contradicts the initial condition. One solution and it doesn't take 7 minutes.
Could you not have used De Moivre’s Theorem
If only you have used Euler's formula you would've done it in 1 minute. Since √a = a^(1/2)
And i=e^i(π/2 +2kπ)
Then √i=e^i(π/2+2kπ)/2=e^i(π/4+kπ) while k is an integer
Then we change the Euler's formula into an algebric one using the r(cos(theta)+isin(theta)) which will give us the same results you got. How simple and fast
What I don’t get is that, I often gets used a the Y coordinate on a 2D plain.
So why wouldn’t just behave like 1 but in the y direction?
isn't it faster to simply write i = e( i pi/2) ; if you want the sqrt, then i ^ (1/2) = e (i pi /2 + 2 k pi)^(1/2) = e ( i pi /4) for K=0 and e ( i 5 pi /4) for K = 1
If you really, really really really want to write Cos pi/4 and sin pi/4 as sqr 2/2 that works 2. e ( i pi/4) is just a lot easier to visualize for students on the unit circle I'd say
quand vous dites que sqr(i)=a+bi, vous signifiez que a et b sont réels, alors le cas a^2=-1/2 n'est pas recevable...
You shall not consider a2=-1/2 in the solution
If √i = x
X^8 = i^4
X^8 = 1
Therefore x=1
√i = 1
Im pretty sure a few things are wrong . If there are any mistakes please tell me
x can also be i!!! i^8=1
Just use De Moivre's theorem to find the roots of unity.
I would say you have found 2 solutions to the problem z^2=i, that is not finding the value of sqrt(i).
If you are treating sqrt(z) as a function, it needs to have only 1 solution.
Very interesting, but what does it mean when it’s graphed?
In a complex plane, one number is at 45 degrees and the other at 225 degrees both of which when squared (angles add up) leads to 90 degree
Root of i is the eighth root of 1 which is a ring.
I didn't like how you set out to find "a" and "b" to satisfy "a + bi", and found values for "a and b" that were imaginary. I thought coefficients a and b for a complex number had to be real numbers
I wanted to show that even if we don't set that restriction from the beginning, the complex a and b yield the same results.
√i = √(√i) = 4th root of √-1
Okay, but you have to simplify that. I still haven't figured that out yet.
No they are not the same
@@manibabai2115 Square root of a square root is a hypercube root. They're correct.
√x = x^(1/2)
√(√x) = x^((1/2)*(1/2)) = x^(1/4) = ∜x
So if [ x = i ] for √(√x), we get: √(√i) = ∜i
No. i think you meant to write √i = √( √(-1) ) = 4th root of -1
What is the Square Root of -i? Is it even possible?
((sqrt(2) + i(sqrt(2)/4) For folks who don't like radicals in their denominators.
I think. I remember doing it in my head during a smoke break at work once. If you know how to work in the complex plane, and that the principle cube root of -8 is NOT -2, but sqrt(3)i, then this is just basic complex arithmetic.
i^i? Now that's hard.
Pro forgot De Moivre's formula😏
Just use Euler's formula and call it a day!