When students learn more about complex numbers, I prefer writing x^3=27*e^(2*pi*i*k), so x=3*e^(2*pi*i*k/3), where k is any integer. k=0,1,2, give three distinct solutions.
That's my method. We could also just say it's the multivalued cube root. And if they're comfortable with rings of integers modulo n, tell them that for z^(1/n) just use k running through the elements of Zn.
The most general solution set is all 3a where a³ = 1 in whatever number system you're working with. In the real numbers, the values of a are {1}. In the complex numbers, the values of a are {1, 1/2 + i(√3)/2, 1/2 - i(√3)/2}. In the hyperbolic numbers, the values of a are {1} (because the hyperbolic nth roots of unity are {1} for odd n and {1, j, -1, -j} for even n). In the dual numbers, the values of a are {1} (because the dual nth roots of unity are {1} for odd n and {1, -1} for even n).
x^3 = 27 Just looking, we can tell x = 3 is a solution. By definition, this means that the polynomial x^3 - 27 divides evenly by x - 3. 3 | 1 0 0 -27 (0x^2 and 0x) | 3 9 27 ___________ 1 3 9 0 So x^3 - 27 = 0 factors to (x - 3)(x^2 + 3x + 9) = 0 We just need to solve the quadratic x^2 + 3x + 9. x = [-3 +- sqrt(9 - 36)]/2 x = (-3 +- 3sqrt(3)i)/2 And just like that, we have our solutions. x = 3, (-3 + 3sqrt(3)i)/2, and (-3 - 3sqrt(3)i)/2.
Probably an AS level further maths question, based on my memory of the late 1990s. There are three solutions in the complex plane, each with magnitude 3, and formula of form x=|x|e^(i2πn/a) where for the cubic form a=3 and n is any real integer. Note that n=0, n=3 and n=-3 etc. give the same solution.
the easiest way to do this is probably by using trigonometric representation for complex numbers and de Moivre formula. We get IzI^3 (cos3a + isin3a)=27 Im(27) = 0, therefore sin3a=0, so cos3a = +/- 1. Since IzI>0 => cos3a=1 and IzI^3=27 => IzI=3. Solving for "a" we get: 3a=2kπ => a = 2/3 kπ. Therefore z = 3(cos(2/3 kπ) + isin(2/3 kπ)). We expect 3 solutions (fundamental theorem of algebruh) so substituting k=0,1,2 gives all three solutions. It would probably be an overkill for this problem, because the first solution is somewhat trivial and using Bezout's theorem we can easily write z^3-27 it as a product of first and second order polynomial (the second one is easy to solve). However, for higher powers of "z" it is way faster and easier to program into an algorythm.
@@pelasgeuspelasgeus4634That’s obviously not true. sin(z) is perfectly defined for any complex number z. sin(i) = i/2 * (e - e^-1), Don’t you know this ?
@@Robert_Emu_Lee Of course I do know it. Another absurd meaningless equation that was calculated by erroneously using trig functions that are meant for real numbers and not for imaginary arbitrary inventions. But, in fantasyland it's perfectly fine.
@@pelasgeuspelasgeus4634 This formula i.e. z=IzI(cos(a) + isin(a)) is perfectly correct in this scenario and does not require any kind of imaginary input. Imaginary numbers may be represented on 2D Argand plane. X axis of this plane represents Re(z) and Y axis represents Im(z). For your information, both Re(z) and Im(z) are REAL numbers by definition. Therefore Argand plane represents complex numbers without actually using complex numbers. Every point on that plane (x,y) represents a distinct complex number which is x+yi. Let's say we pick a point (a,b) on that plane. This point represents a number z=a+bi. The shortest distance between (a,b) and (0,0) is IzI. Let θ be the angle meassured between IzI and x-axis anti-clockwise. You can write a=IzIcos(θ) and b=IzIsin(θ). This point represents a number IzI(cos(θ)+isin(θ)). Both coordinates on the plane are real, θ is real. It just represents a number that is not real. You don't even need to plug complex numbers into trig functions in this formula - all the "complexity" comes after you calculate a and b and plug them into a+bi to obtain the number you are looking for. btw. as @Robert_Emu_Lee said, sin(z) IS defined for complex "z".
That's like comman sense bro... A 3 degree polynomial obviously have 3 solutions But while studying at junior level we ignore the complex part and just take the real solution!
Hi, could you please make your merch in smaller sizes? I wanted to buy the fundemental theorem of engineering hoodie, but size small would reach mid way down my thighs
Could you at any point, if it´s even humanly possible to make a simplified proof of Andrew Wiles' proof of Fermat's last theorem? I heard it't 119 pages long. Seems a bit ridiculous to me but then again I have to clue about elliptic curves. 😅 It's a brilliant anecdote non the less
@@JeremyGluckStuff oh okay, I've never seen this one being used except in one Presh Talwalker video, so I thought this just wasn't an identity that people found important enough to remember
you go through it as if you understand, but let's take a second to check if you really do. so you're leveraging the fact that a^(m/n) has n solutions in Complex space if m/n is Rational. but what if m/n is not Rational? for x = 5^(22/7) there are 7 solutions. for x = 5^(355/113) there are 113 solutions. so seemingly as our approximation of pi improves, the number of solutions increases. but for x = 5^(3 +sqrt(2)/10) there is only one solution, and for x = 5^pi there is also only one solution. why is that? I think the reason is simply that you don't understand what you're doing, and that x = 5^pi actually does have infinitely many solutions that are all entirely inaccessible to you except for one.
look at the progression I set up: the limiting behavior of Rational approximations of pi as an exponent is that the number of solutions approaches infinity, but when we have pi itself as the exponent there is only one solution. 22/7 is a worse approximation of pi than 3 +sqrt(2)/10, but 5^(22/7) has more recognized solutions 3 +sqrt(2)/10 is a worse approximation of pi than 355/113, and 5^(355/113) has more recognized solutions 355/113 is a worse approximation of pi than pi, but 5^(355/113) has more recognized solutions there is no principled reason for this pattern. it occurs only because of ignorance of how to handle irrationals. that is not rigor. this progression does indicate that the behavior of the full system is chaotic, and that irrationals do not behave in accordance with the limiting behavior of Rationals, but what this really should be telling you is that you have a fundamentally wrong understanding of the Real field, since Dedekind Completeness predicts smoothness where there unambiguously is none.
Please, use your brain and understand that complex numbers mean nothing. If you don't believe me, try to buy i kilos of potatoes next time you go to grocery store...
Oh goody! Another hyperrealist crank! Let me shut you down and lead you on the right path, as a mathematician: Math isn't made to conform to the concrete real world. Mathematics is the science of abstraction. Though, complex numbers have found uncountably many applications. This is, in essence, a strawman.
Guys, use your brain and understand that negative numbers and 0 also mean nothing. Proof? Go to the grocery store and try to buy -2 or 0 kilos of potatoes! 🤦♂️
@@Gordy-io8sb Really? Math is about abstraction? Was that what Pythagoras or Euclid had in mind? No buddy. Math is about real tangible data. You may enjoy your abstract imaginary world but real soon everybody will see how naked your theory is. And BTW, because you mentioned the usual ridiculous argument about applications, complex numbers are indeed used but their use leads to gross errors. One simple question: if Cartesian system has 3 real axis where is the imaginary axis? Can you describe it?
@@pelasgeuspelasgeus4634 3D Cartesian space is inherently real. There is no imaginary axis. If 3-dimensional Cartesian space was complex, then we would have something like (a, bi, cj) or (a, b, ci) for coordinates. Which is not what we have. n-dimensional Cartesian space is made from stitching together n copies of the real numbers via Cartesian product. That makes no sense. And gross errors...how? I think there's very little error in our understanding of the quantum world, signals, and etc. from complex numbers. If there is, it's likely from algebraic errors which are not baked into the complex numbers in any magnitude.
![[Part 4] หนุ่มน้อยรักคุณย่ายิ่งกว่าใคร ไม่ว่าใครก็รังแกดูถูกท่านไม่ได้!! #โลกการละคร](http://i.ytimg.com/vi/sLsBxw99FR0/mqdefault.jpg)
Easy just use the cubic formula
This is the true trivial solution 😎
Nah 💀
No one remembers that formula 😭
Easy, just use polar coordinates. Actually. It makes everything so much easier. LEARN THEM!!
the quintic is more fun
When students learn more about complex numbers, I prefer writing x^3=27*e^(2*pi*i*k), so x=3*e^(2*pi*i*k/3), where k is any integer. k=0,1,2, give three distinct solutions.
That's my method. We could also just say it's the multivalued cube root. And if they're comfortable with rings of integers modulo n, tell them that for z^(1/n) just use k running through the elements of Zn.
The most general solution set is all 3a where a³ = 1 in whatever number system you're working with.
In the real numbers, the values of a are {1}. In the complex numbers, the values of a are {1, 1/2 + i(√3)/2, 1/2 - i(√3)/2}. In the hyperbolic numbers, the values of a are {1} (because the hyperbolic nth roots of unity are {1} for odd n and {1, j, -1, -j} for even n). In the dual numbers, the values of a are {1} (because the dual nth roots of unity are {1} for odd n and {1, -1} for even n).
Great. Now, do it with polar coordinates
x=3e^(i2πn/3)
x^3 = 27
Just looking, we can tell x = 3 is a solution. By definition, this means that the polynomial x^3 - 27 divides evenly by x - 3.
3 | 1 0 0 -27 (0x^2 and 0x)
| 3 9 27
___________
1 3 9 0
So x^3 - 27 = 0 factors to (x - 3)(x^2 + 3x + 9) = 0
We just need to solve the quadratic x^2 + 3x + 9.
x = [-3 +- sqrt(9 - 36)]/2
x = (-3 +- 3sqrt(3)i)/2
And just like that, we have our solutions. x = 3, (-3 + 3sqrt(3)i)/2, and (-3 - 3sqrt(3)i)/2.
This is the type of problem where working in polar coordinates makes the problem easier to work out
DTI intro is crazy work honestly
'cleanest' way when thinking about roots of unity: 3*(-sqrt(3)/2 +/- i/2)
C didn't trigger me... I was like "it is the constant term, makes sense"
I've always wondered about the digamma function at rational values.
Probably an AS level further maths question, based on my memory of the late 1990s. There are three solutions in the complex plane, each with magnitude 3, and formula of form x=|x|e^(i2πn/a) where for the cubic form a=3 and n is any real integer. Note that n=0, n=3 and n=-3 etc. give the same solution.
the easiest way to do this is probably by using trigonometric representation for complex numbers and de Moivre formula.
We get IzI^3 (cos3a + isin3a)=27
Im(27) = 0, therefore sin3a=0, so cos3a = +/- 1. Since IzI>0 => cos3a=1 and IzI^3=27 => IzI=3. Solving for "a" we get:
3a=2kπ => a = 2/3 kπ. Therefore z = 3(cos(2/3 kπ) + isin(2/3 kπ)). We expect 3 solutions (fundamental theorem of algebruh) so substituting k=0,1,2 gives all three solutions.
It would probably be an overkill for this problem, because the first solution is somewhat trivial and using Bezout's theorem we can easily write z^3-27 it as a product of first and second order polynomial (the second one is easy to solve). However, for higher powers of "z" it is way faster and easier to program into an algorythm.
Trigonometric formulas, like sin(x), are valid only for real x. Don't you know that?
Tell me, what is the value of sin(i)?
@@pelasgeuspelasgeus4634That’s obviously not true. sin(z) is perfectly defined for any complex number z. sin(i) = i/2 * (e - e^-1), Don’t you know this ?
@@Robert_Emu_Lee Of course I do know it. Another absurd meaningless equation that was calculated by erroneously using trig functions that are meant for real numbers and not for imaginary arbitrary inventions.
But, in fantasyland it's perfectly fine.
@@pelasgeuspelasgeus4634 This formula i.e. z=IzI(cos(a) + isin(a)) is perfectly correct in this scenario and does not require any kind of imaginary input.
Imaginary numbers may be represented on 2D Argand plane. X axis of this plane represents Re(z) and Y axis represents Im(z). For your information, both Re(z) and Im(z) are REAL numbers by definition. Therefore Argand plane represents complex numbers without actually using complex numbers. Every point on that plane (x,y) represents a distinct complex number which is x+yi.
Let's say we pick a point (a,b) on that plane. This point represents a number z=a+bi. The shortest distance between (a,b) and (0,0) is IzI. Let θ be the angle meassured between IzI and x-axis anti-clockwise. You can write a=IzIcos(θ) and b=IzIsin(θ). This point represents a number IzI(cos(θ)+isin(θ)). Both coordinates on the plane are real, θ is real. It just represents a number that is not real. You don't even need to plug complex numbers into trig functions in this formula - all the "complexity" comes after you calculate a and b and plug them into a+bi to obtain the number you are looking for.
btw. as @Robert_Emu_Lee said, sin(z) IS defined for complex "z".
@@wernerheisenberg1624 OK. Tell me, the multiplication of imaginary unit with a real number gives an imaginary number or a real one?
Huge thanks to Brian for sponsoring this video
Loved the video, I now have a headache.
I love your German accent it makes you sound like a mad scientist by default :P
Good video
if we know that x = 3 is a solution, why not just divide the cubic by x - 3 to get our quadratic? saves us a lot of time
The obvious solution is *x = 3,* but *x = 1.5√(3)i - 1.5* and *x = -1.5√(3)i - 1.5* are two other solutions.
you should have completed the square for completeness xx
That's like comman sense bro...
A 3 degree polynomial obviously have 3 solutions
But while studying at junior level we ignore the complex part and just take the real solution!
I like it *because* it is trivial
But what about the set of infinite quaternionic solutions?
Hello can make a video on the G function and the H function? Pleaseeeeeee😢
The only real life solution is 3. The rest is science fiction for dorks.
Hi, could you please make your merch in smaller sizes? I wanted to buy the fundemental theorem of engineering hoodie, but size small would reach mid way down my thighs
Oh Yh, please do that, I had the same issue!
I mean, it goes up to 3XL, is xs too much to ask for?
Short people problem
I wanted to buy it for my daughter as a birthday present, but too large
Watching this with no AdBlock.
Watching this with no pants
@@rey5392 true chad.
Could you at any point, if it´s even humanly possible to make a simplified proof of Andrew Wiles' proof of Fermat's last theorem? I heard it't 119 pages long. Seems a bit ridiculous to me but then again I have to clue about elliptic curves. 😅
It's a brilliant anecdote non the less
Me in the exam after not seeing the obvious solution
No way there is a Roblox meme 😂
smart shirt. you must come from smart people.
Wow lmao this is the only legit problem where Indian math has helped me. We're forced to memorize this identity (x-y)(x^2 + xy + y^2) = x^3 - y^3 lol
They made us memorize that one in American school too.
Yeah dude they make us memorize this identity in LatAm, India isn't as special as you think it is lil bro
@@JeremyGluckStuff oh okay, I've never seen this one being used except in one Presh Talwalker video, so I thought this just wasn't an identity that people found important enough to remember
you go through it as if you understand, but let's take a second to check if you really do.
so you're leveraging the fact that a^(m/n) has n solutions in Complex space if m/n is Rational. but what if m/n is not Rational?
for x = 5^(22/7) there are 7 solutions. for x = 5^(355/113) there are 113 solutions. so seemingly as our approximation of pi improves, the number of solutions increases. but for x = 5^(3 +sqrt(2)/10) there is only one solution, and for x = 5^pi there is also only one solution. why is that?
I think the reason is simply that you don't understand what you're doing, and that x = 5^pi actually does have infinitely many solutions that are all entirely inaccessible to you except for one.
look at the progression I set up:
the limiting behavior of Rational approximations of pi as an exponent is that the number of solutions approaches infinity, but when we have pi itself as the exponent there is only one solution.
22/7 is a worse approximation of pi than 3 +sqrt(2)/10, but 5^(22/7) has more recognized solutions
3 +sqrt(2)/10 is a worse approximation of pi than 355/113, and 5^(355/113) has more recognized solutions
355/113 is a worse approximation of pi than pi, but 5^(355/113) has more recognized solutions
there is no principled reason for this pattern. it occurs only because of ignorance of how to handle irrationals. that is not rigor.
this progression does indicate that the behavior of the full system is chaotic, and that irrationals do not behave in accordance with the limiting behavior of Rationals, but what this really should be telling you is that you have a fundamentally wrong understanding of the Real field, since Dedekind Completeness predicts smoothness where there unambiguously is none.
Please, learn math properly before posting nonsense.
Please, use your brain and understand that complex numbers mean nothing. If you don't believe me, try to buy i kilos of potatoes next time you go to grocery store...
Oh goody! Another hyperrealist crank! Let me shut you down and lead you on the right path, as a mathematician:
Math isn't made to conform to the concrete real world. Mathematics is the science of abstraction. Though, complex numbers have found uncountably many applications. This is, in essence, a strawman.
Guys, use your brain and understand that negative numbers and 0 also mean nothing. Proof? Go to the grocery store and try to buy -2 or 0 kilos of potatoes! 🤦♂️
@@Gordy-io8sb Really? Math is about abstraction? Was that what Pythagoras or Euclid had in mind?
No buddy. Math is about real tangible data. You may enjoy your abstract imaginary world but real soon everybody will see how naked your theory is.
And BTW, because you mentioned the usual ridiculous argument about applications, complex numbers are indeed used but their use leads to gross errors.
One simple question: if Cartesian system has 3 real axis where is the imaginary axis? Can you describe it?
@@pelasgeuspelasgeus4634 3D Cartesian space is inherently real. There is no imaginary axis. If 3-dimensional Cartesian space was complex, then we would have something like (a, bi, cj) or (a, b, ci) for coordinates. Which is not what we have. n-dimensional Cartesian space is made from stitching together n copies of the real numbers via Cartesian product. That makes no sense.
And gross errors...how? I think there's very little error in our understanding of the quantum world, signals, and etc. from complex numbers. If there is, it's likely from algebraic errors which are not baked into the complex numbers in any magnitude.
@@pelasgeuspelasgeus4634 No. It's the study of abstract concepts and logic. Even geometry and trig are abstract in their own way.