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Loser66
 one year ago
Find all values of z such that z^4 = 4
Please, help
Loser66
 one year ago
Find all values of z such that z^4 = 4 Please, help

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(z = ze^{i\theta}\) \(z^4 = z^4 e^{4i\theta} = 4 = 4e^{i\pi }\) Hence \(z^4 = 4 \rightarrow z=\sqrt2\) \(4i\theta = i(\pi+2k\pi) : k\in \mathbb Z\) then \(\theta = \pi/4 + k\pi/2\), Now I stuck. what is k?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[1=\cos(\pi+2 k \pi)+i \sin(\pi+2 k \pi) \\ 4=4e^{ \pi+2 k \pi} \\ z^4=4 e^{\pi+2k \pi} \\ \text{ now we want 4 roots } \\ \text{ so } k=0,1,2,3\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0yes, I am crazy.This is what I did if k =0, then \(z = \sqrt2 e^{\pi/4}= \sqrt2(cos (\pi/4) + isin(\pi/4))= 1+i\) but then z^4 =(1+i)^4 from this I take 1+i to get \sqrt 2, then ^4 and get 4, not 4. That throws me off. Thanks @freckles

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@Crazy_questions senior in University.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I have another question, @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(1+i)^4=(1+i)^2(1+i)^2=(1+2i+i^2)(1+2i+i^2) \\ (1+i)^4=(2i)(2i)=4i^2=4\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(\dfrac{(2 + 3i)^4}{3i)^2}= \dfrac{(2+3i)^4}{(3i)^2} =\dfrac{2+3i^4}{3i^2}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Can I do that? since \(2+ 3i = \sqrt{4+(3)^2} = \sqrt{13} \) and \(3i= \sqrt{3^2 + (1)^2} = \sqrt {10}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0If I can do that, then I can save my time

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0you see what the wolfram does. It takes a very long process to get my answer http://www.wolframalpha.com/input/?i=evalulate+ %282%2B3i%29^4%2F%283i%29^2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yes I believe that is right I was going to try to prove: \[(a+bi)^n=a+bi^n\] but the way I was going about it was taking too long

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well I don't have the wolfram plus or whatver it is call to see what they have

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0You can access to it free up to 3 problems

freckles
 one year ago
Best ResponseYou've already chosen the best response.2they expanded first right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I tested many times. They did what I did.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I have class in 15 minutes, still have many questions need help. Would you mind to help me out while I am not here ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[(re^{i \theta})^n=r^n e^{i \theta n} \\ \ \text{ assume } r>0 \\ \text{ then } r^n=r^n \\ \\ (r e^{i \theta})^n=r^n e^{i \theta n}=r^n \cos(\theta n)+ i \sin(\theta n) \\ =r^n \sqrt{\cos^2(\theta n )+ \sin^2(\theta n)} =r^n \\ r e ^{i \theta}^n=r^n e^{ i \theta}^n =r^n e^{i \theta}^n \text{ still assuming } r>0 \\ r e ^{ i \theta}^n=r^n (\sqrt{\cos^2(\theta)+\sin^2(\theta)})^n=r^n(1)^n=r^n\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[a+bi^n=(a+bi)^n \text{ seems to hold }\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I have to go right now. Thanks @freckles
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