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Problem:

Find all complex numbers \(a\) such that \(z^4 = -4.\)

Note: All solutions should be expressed in the form \(a+bi \), where \(a\) and \(b\) are real numbers. 

 

What I have done so far: 

 

\(\begin{align*} z^4 &= -4 \\ (z^2)^2 &= -4 \\ z^2 &= \sqrt{-4} \\ &= \sqrt{-1}\cdot \sqrt{4} \\ &= \pm 2i \\ \end{align*} \)

 

Can someone guide me? I dont know how to express this in the form \(a+bi\)

 Jun 3, 2021
edited by JonJoseph  Jun 3, 2021
 #1
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z = -1 - i or z = -1 + i 

 

 

z^4 = -4

 

  Solve:

 

z = -1 - i or z = -1 + i 

 

Does the answer help you?

 Jun 3, 2021
 #5
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Could you plz tell me how you got that answer? I am not able to make out how you got it. I know that is the right answer though. 

JonJoseph  Jun 3, 2021
edited by JonJoseph  Jun 3, 2021
 #2
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Yes. Thanks!

 Jun 3, 2021
 #3
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https://youtu.be/8wVAbsv-VJw

Guest Jun 3, 2021
 #4
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????

.
 Jun 3, 2021
 #6
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Can you please tell me how u got that answer? I am being asked to show my working. I tried to backtrack to see how I can explain it, but I could not. 

 Jun 3, 2021
 #7
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We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

 

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

 

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

 Jun 3, 2021
 #8
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OK. Thanks!

 

laughlaughlaugh

 Jun 4, 2021

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