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

1. Loser66

@ganeshie8

2. Loser66

$$z = |z|e^{i\theta}$$ $$z^4 = |z|^4 e^{4i\theta} = -4 = 4e^{i\pi }$$ 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?

3. freckles

$-1=\cos(\pi+2 k \pi)+i \sin(\pi+2 k \pi) \\ -4=4e^{ \pi+2 k \pi} \\ z^4=4 e^{\pi+2k \pi} \\ \text{ now we want 4 roots } \\ \text{ so } k=0,1,2,3$

4. Loser66

yes, 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

5. Loser66

@Crazy_questions senior in University.

6. anonymous

7. Loser66

I have another question, @freckles

8. freckles

$(1+i)^4=(1+i)^2(1+i)^2=(1+2i+i^2)(1+2i+i^2) \\ (1+i)^4=(2i)(2i)=4i^2=-4$

9. Loser66

$$|\dfrac{(2 + 3i)^4}{3-i)^2}|= \dfrac{|(2+3i)^4|}{|(3-i)^2|} =\dfrac{|2+3i|^4}{|3-i|^2}$$

10. Loser66

Can I do that? since $$|2+ 3i| = \sqrt{4+(3)^2} = \sqrt{13}$$ and $$|3-i|= \sqrt{3^2 + (-1)^2} = \sqrt {10}$$

11. Loser66

If I can do that, then I can save my time

12. Loser66

you 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%283-i%29^2|

13. freckles

yes 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

14. freckles

well I don't have the wolfram plus or whatver it is call to see what they have

15. freckles

or did

16. Loser66

17. freckles

they expanded first right?

18. Loser66

yea

19. Loser66

I tested many times. They did what I did.

20. Loser66

I have class in 15 minutes, still have many questions need help. Would you mind to help me out while I am not here ?

21. freckles

$|(re^{i \theta})^n|=|r^n| |e^{i \theta n}| \\ \ \text{ assume } r>0 \\ \text{ then } |r^n|=r^n \\ \\ |(r e^{i \theta})^n|=r^n |e^{i \theta n}|=r^n| \cos(\theta n)+ i \sin(\theta n)| \\ =r^n \sqrt{\cos^2(\theta n )+ \sin^2(\theta n)} =r^n \\ |r e ^{i \theta}|^n=|r|^n |e^{ i \theta}|^n =r^n |e^{i \theta}|^n \text{ still assuming } r>0 \\ |r e ^{ i \theta}|^n=r^n (\sqrt{\cos^2(\theta)+\sin^2(\theta)})^n=r^n(1)^n=r^n$

22. freckles

$|a+bi|^n=|(a+bi)^n| \text{ seems to hold }$

23. Loser66

I have to go right now. Thanks @freckles