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Loser66

  • one year ago

Find all values of z such that z^4 = -4 Please, help

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  1. Loser66
    • one year ago
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    @ganeshie8

  2. Loser66
    • one year ago
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    \(z = |z|e^{i\theta}\) \(z^4 = |z|^4 e^{4i\theta} = -4 = 4e^{i\pi }\) 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
    • one year ago
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    \[-1=\cos(\pi+2 k \pi)+i \sin(\pi+2 k \pi) \\ -4=4e^{ \pi+2 k \pi} \\ z^4=4 e^{\pi+2k \pi} \\ \text{ now we want 4 roots } \\ \text{ so } k=0,1,2,3\]

  4. Loser66
    • one year ago
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    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
    • one year ago
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    @Crazy_questions senior in University.

  6. anonymous
    • one year ago
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    grade is it 10th?

  7. Loser66
    • one year ago
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    I have another question, @freckles

  8. freckles
    • one year ago
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    \[(1+i)^4=(1+i)^2(1+i)^2=(1+2i+i^2)(1+2i+i^2) \\ (1+i)^4=(2i)(2i)=4i^2=-4\]

  9. Loser66
    • one year ago
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    \(|\dfrac{(2 + 3i)^4}{3-i)^2}|= \dfrac{|(2+3i)^4|}{|(3-i)^2|} =\dfrac{|2+3i|^4}{|3-i|^2}\)

  10. Loser66
    • one year ago
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    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
    • one year ago
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    If I can do that, then I can save my time

  12. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    well I don't have the wolfram plus or whatver it is call to see what they have

  15. freckles
    • one year ago
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    or did

  16. Loser66
    • one year ago
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    You can access to it free up to 3 problems

  17. freckles
    • one year ago
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    they expanded first right?

  18. Loser66
    • one year ago
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    yea

  19. Loser66
    • one year ago
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    I tested many times. They did what I did.

  20. Loser66
    • one year ago
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    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
    • one year ago
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    \[|(re^{i \theta})^n|=|r^n| |e^{i \theta n}| \\ \ \text{ assume } r>0 \\ \text{ then } |r^n|=r^n \\ \\ |(r e^{i \theta})^n|=r^n |e^{i \theta n}|=r^n| \cos(\theta n)+ i \sin(\theta n)| \\ =r^n \sqrt{\cos^2(\theta n )+ \sin^2(\theta n)} =r^n \\ |r e ^{i \theta}|^n=|r|^n |e^{ i \theta}|^n =r^n |e^{i \theta}|^n \text{ still assuming } r>0 \\ |r e ^{ i \theta}|^n=r^n (\sqrt{\cos^2(\theta)+\sin^2(\theta)})^n=r^n(1)^n=r^n\]

  22. freckles
    • one year ago
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    \[|a+bi|^n=|(a+bi)^n| \text{ seems to hold }\]

  23. Loser66
    • one year ago
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    I have to go right now. Thanks @freckles

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