+13 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 \(a+bi\).
z = -1 - i or z = -1 + i
z^4 = -4
Solve:
z = -1 - i or z = -1 + i
Does the answer help you?
+13 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.
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.
