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

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

I know this has been asked a million times but I'm wondering if someone can explain it clearly and easy to understand, sorry

Guest Apr 5, 2021

edited by
Guest
Apr 5, 2021

#1**0 **

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.

Guest Apr 25, 2021