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# Halp! I am stuck...

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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

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z = -1 - i or z = -1 + i

z^4 = -4

Solve:

z = -1 - i or z = -1 + i

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
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Yes. Thanks!

Jun 3, 2021
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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
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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!

Jun 4, 2021