Great, but the radical symbol is generally taken to mean only the "principal root" of the argument, not all roots. so, for example, x = √i has only one solution. whereas X² = i has two solutions.
With this method, you're missing most of the roots. You've got the seed of a good idea though, certainly better than the algebra weeds the video wades into...but then again, the video's point is to futz about, see a pattern, and find the better way (which is often how real progress is made). This turns out to be your way, minus your mistake that eliminates so many valid solutions. The imaginary unit i is not just exp(i pi/2). It is also exp(i(pi/2 ± 2pi)) and exp(i(pi/2 ± 4pi)) etc...i.e. exp(i(pi/2 + k*pi)) for integer k. Divide (pi/2 + k*pi) by n, and tidy up. That's how one actually takes nth roots...this general method can be made to work for all complex numbers in polar form if you're careful.
What I was missing in the video is the following simple formula for the n nth roots of i, which follows directly from how complex multiplication works: pi (1 + 4k) / 2n, with 0
If you write z^n =i =exp(i π/2) ,then you get the answer z[k]= exp[i π k /(2 n)] ,k= 1,2, ,n. These are the n solutions. Since exp[i π k /(2 n)]= cos[i π k /(2 n)]+i sin[i π k/(2 n)] ,we see that these solutions form a regular n-gon on the unit circle.
This video seems to pretend that the radical sign does not entail the principal root. But it absolutely does. If you want to consider all roots (so a multivalue function) you should use another notation than the radical sign. This is like saying ✓5 = ±✓5 which is obviously wrong simply because the radical sign ALWAYS entails principal root (for any complex number).
@@dibeos honestly, a video on curved asymptotes and the multiple ways a graph behaves as it reaches infinity would be interesting although that's just my two cents cuz there isn't much to explore in this topic
Ascii text makes this so hard. Wikipedia has pictures. 1 = e^(i2pi) = e^(ik2pi) (k = 0 or positive integer) 1 = e^(i(k2pi)/N)^N, 1 is the identity operator of multiplication Nth roots of (ae^ib) (N a positive integer) =(ae^ib)^(1/N) (multiply times one as above) =((ae^ib)*e^((ik2pi)/N)^N)^1/N =((ae^(ib))^1/N*e^((ik2pi)/N) for k = 0 to N-1 (integers) Note that =(ae^(ib)^1/N =a^(1/N)*e^i(b/N) or the phase is 1/Nth of b. Note that for e^(i2pi)/N the phase is 2pi divided by N for all Nth roots, this phase is multiplied by k For square roots N = 2, phase of 2pi/2 = pi which leads to1 or -1 Square roots of 4 are +/- 2. For cube roots N = 3, 2pi/3 is the phase separation of all three roots. For fourth roots N = 4, 2pi/4 = pi/2 which leads to +1 or -1 and +i or -i. The fourth roots of 16 are +/- 2 and +/- 2i.
This comment is coming from a person who very much loves your videos. I LOVE the topics you cover and really appreciate all thencare you put in presenting them. I often find myself struggling with how quickly everything is presnted, especially when it comes to written text in the screen. I find that i will get into the video and realize that i missed something or am not fully grasping the logic. Maybe its just me, but I think i have learned that i should buckle up and get ready to rewatch if inam going to make it through one of your videos.
@@graf_paper thank you for the feedback! We are really trying to slow the videos down. But I guess we need to slow them down even more. Thanks for letting us know!! We will fix it 😎
@@mahmoudhabib95 A set A is dense in another set B if every point in B is either in A or arbitrarily close to a point in A. This means for any point in B and any given distance, you can find a point in A within that distance. Formally, the closure of A is equal to B.
Great, but the radical symbol is generally taken to mean only the "principal root" of the argument, not all roots.
so, for example, x = √i has only one solution. whereas X² = i has two solutions.
That’s what I came here to say.
wouldn't it be easier to convert to polar coordinates?
@@petefritz5679 probably yes, but I really wanted to calculate it this way. I don’t know, it feels more tangible
They're teaching, not solving for an IMO. This made sense for a broader audience.
What is IMO stand for in this case?
That's really good for problem solving. But I think the goal of the video is investigation from first principle and stuff, plus for curiousity's sake
@@schizoframia4874 International Math Olympics
5th root is pi/10 and every 72 degrees thereafter, 6th root is pi/12 and every 60 degrees thereafter, etc.
@@jesusthroughmary yeah, it does make sense
Fantastic episode. Very clear and concise. Keep it up!
Just use the exponential form:
i= exp(i pi/2),the nth root means bla^1/n so the nth root of i is exp(i pi/2n). Fi ished. Needed not even one minute
With this method, you're missing most of the roots. You've got the seed of a good idea though, certainly better than the algebra weeds the video wades into...but then again, the video's point is to futz about, see a pattern, and find the better way (which is often how real progress is made). This turns out to be your way, minus your mistake that eliminates so many valid solutions. The imaginary unit i is not just exp(i pi/2). It is also exp(i(pi/2 ± 2pi)) and exp(i(pi/2 ± 4pi)) etc...i.e. exp(i(pi/2 + k*pi)) for integer k. Divide (pi/2 + k*pi) by n, and tidy up. That's how one actually takes nth roots...this general method can be made to work for all complex numbers in polar form if you're careful.
negative N is easy, it's i^(-n) = (i^(-1))^n = (1/i)^n = (-i)^n = (e^(i*3pi/2))^n and here you are
@@محمدبورايب yeah, you are right 😎
What I was missing in the video is the following simple formula for the n nth roots of i, which follows directly from how complex multiplication works:
pi (1 + 4k) / 2n, with 0
If you write z^n =i =exp(i π/2) ,then you get the answer z[k]= exp[i π k /(2 n)] ,k= 1,2, ,n. These are the n solutions.
Since exp[i π k /(2 n)]= cos[i π k /(2 n)]+i sin[i π k/(2 n)] ,we see that these solutions form a regular n-gon on the unit circle.
This video seems to pretend that the radical sign does not entail the principal root. But it absolutely does. If you want to consider all roots (so a multivalue function) you should use another notation than the radical sign.
This is like saying ✓5 = ±✓5 which is obviously wrong simply because the radical sign ALWAYS entails principal root (for any complex number).
Esse vídeo tá muito legal! Gostei que tenha calculado as raízes sem usar a fórmula do Euler
@@nabla_mat obrigado! Sim, fazendo assim a compreensão do problema é mais profunda na minha opinião, ainda que se torna mais complicado o problema
This video finally helped me to understand why roots of an imaginary number has more forms in polar coordinates
@@syphon5899 that’s awesome! We are glad to help you. Please, let us know what kind of videos you’d like us to make so that we can help further 😎
@@dibeos honestly, a video on curved asymptotes and the multiple ways a graph behaves as it reaches infinity would be interesting although that's just my two cents cuz there isn't much to explore in this topic
@@syphon5899 it’s ok, math is creative. We can make it interesting 😉
Euler’s formula or de moivre formula
Ascii text makes this so hard. Wikipedia has pictures.
1 = e^(i2pi) = e^(ik2pi) (k = 0 or positive integer)
1 = e^(i(k2pi)/N)^N, 1 is the identity operator of multiplication
Nth roots of (ae^ib) (N a positive integer)
=(ae^ib)^(1/N) (multiply times one as above)
=((ae^ib)*e^((ik2pi)/N)^N)^1/N
=((ae^(ib))^1/N*e^((ik2pi)/N) for k = 0 to N-1 (integers)
Note that =(ae^(ib)^1/N =a^(1/N)*e^i(b/N) or the phase is 1/Nth of b.
Note that for e^(i2pi)/N the phase is 2pi divided by N for all Nth roots, this phase is multiplied by k
For square roots N = 2, phase of 2pi/2 = pi which leads to1 or -1
Square roots of 4 are +/- 2.
For cube roots N = 3, 2pi/3 is the phase separation of all three roots.
For fourth roots N = 4, 2pi/4 = pi/2 which leads to +1 or -1 and +i or -i.
The fourth roots of 16 are +/- 2 and +/- 2i.
This comment is coming from a person who very much loves your videos. I LOVE the topics you cover and really appreciate all thencare you put in presenting them.
I often find myself struggling with how quickly everything is presnted, especially when it comes to written text in the screen. I find that i will get into the video and realize that i missed something or am not fully grasping the logic.
Maybe its just me, but I think i have learned that i should buckle up and get ready to rewatch if inam going to make it through one of your videos.
@@graf_paper thank you for the feedback! We are really trying to slow the videos down. But I guess we need to slow them down even more. Thanks for letting us know!! We will fix it 😎
I love the PDF attached by the way, working through the problem was a treat.
@@graf_paper awesome. Could you please tell me if you prefer its background black or white? Hahah just a random question
@@dibeos I am used the black background. Most math channels use a darker background so I would think your audience would be used to that?
loved it !!! :D
0:05 I thought the triple bar meant modular arithmetic
@@schizoframia4874 not only. It means equivalent, but of course a particular application of this symbol is in modular arithmetic.
@@dibeos so it is another way of writing =, or to say something is equivalent under mod arithmetic
@@schizoframia4874 exactly. I just tried to be fancy right there 😝
What does the phrase "dense in" mean?
@@mahmoudhabib95 A set A is dense in another set B if every point in B is either in A or arbitrarily close to a point in A. This means for any point in B and any given distance, you can find a point in A within that distance. Formally, the closure of A is equal to B.
@@dibeos thank you so much!
Very cool, got some practice brushing up on the complex numbers today! You guys definitely felt like doing some algebra for this video! haha
@@jammasound yeah hahaha it took a loooooong time, but it was definitely worth it 😎
Bruh some comments Should show more apparition
❤great video and editing as always thanks for the vid.
@@tobyendy9508 thank you so much for the encouragement!! Sofia and I put a lot of effort on each of these videos, it means a lot to us
What is apparition?
@@aplusbi I think he meant “appreciation” haha
Cool video, I amazed
💓
You took something easy and you made it complicated. That’s not the goal of maths nor is it the goal of teaching.
This is way too much work for something obvious.
@@stewartbrown7907 yeah, that’s why we did all the “heavy lifting” so that you guys can just appreciate the final result without the hard work 😌
@@dibeos that's not at all what I meant. i=e^{i*pi/4}, you can then just get all the roots you want just using exponents rules.