+0

# HELP!!

+1
95
3

Find all complex numbers $z$ such that $z^4 = -4.$

Jun 12, 2020

#1
+17
0

Hi! I'm not sure about the answer, but I tried to do your problem.  Here's my solution:

First, take the square root of both sides to get +- x^2 = 2i. Then, you will get two cases:

Case #1; x^2 = 2i: We take the square root of both sides of the equation x^2 = 2i to get +- x = 1 + i. Thus, x can be either 1 + i or -1 - i in this case.

Case #2; -x^2 = 2i: We move the terms in the equation around to get x^2 = -2i. Then, taking the square root of both sides, we get

+-x = 1 - i. Therefore, this case yields the solutions x = 1 - i and x = -1 + i.

In the end, all of the complex numbers z such that z^4 = -4 are 1 + i, -1 -i, 1 - i, and -1 + i.

Jun 12, 2020
#2
+17
0

Ziggy  Jun 12, 2020
#3
+1130
0

so we get that z=$$-\sqrt2$$ since when we square it we get 2 and when we cube it we get $$-2\sqrt2$$ so we get...

but there are no values of even number roots of negative numbers so there is no value to this solution

Jun 12, 2020