Question

HELP SOMEONE!!!!!! PLZZZZ!!!
Find all complex numbers $z$ such that $z^4 = -4.$

Answer 1

Since -4 = 4 exp(iπ), by DeMoivre's theorem and using the fact that cos and sin are both 2π-periodic,

`z^4 = 4 \exp\left(i\pi\right) \implies z = 4^(1/4) \exp\left(\frac14 \left(i\pi + 2i\pi k\right)\right)`

where k = 0, 1, 2, or 3. Then the 4th roots of -4 are

`k = 0 \implies z = 4^(1/4) \exp\left(\frac14 \left(i\pi+0\right)\right) = \sqrt2 \exp\left(i\frac\pi4\right)`

`k = 1 \implies z = 4^(1/4) \exp\left(\frac14 \left(i\pi+2i\pi\right)\right) = \sqrt2 \exp\left(i\frac{3\pi}4\right)`

`k = 2 \implies z = 4^(1/4) \exp\left(\frac14 \left(i\pi+4i\pi\right)\right) = \sqrt2 \exp\left(i\frac{5\pi}4\right)`

`k = 3 \implies z = 4^(1/4) \exp\left(\frac14 \left(i\pi+6i\pi\right)\right) = \sqrt2 \exp\left(i\frac{7\pi}4\right)`