Since -4 = 4 exp(iπ), by DeMoivre's theorem and using the fact that cos and sin are both 2π-periodic,
[tex]z^4 = 4 \exp\left(i\pi\right) \implies z = 4^{1/4} \exp\left(\dfrac14 \left(i\pi + 2i\pi k\right)\right)[/tex]
where k = 0, 1, 2, or 3. Then the 4th roots of -4 are
[tex]k = 0 \implies z = 4^{1/4} \exp\left(\dfrac14 \left(i\pi+0\right)\right) = \sqrt2 \exp\left(i\dfrac\pi4\right)[/tex]
[tex]k = 1 \implies z = 4^{1/4} \exp\left(\dfrac14 \left(i\pi+2i\pi\right)\right) = \sqrt2 \exp\left(i\dfrac{3\pi}4\right)[/tex]
[tex]k = 2 \implies z = 4^{1/4} \exp\left(\dfrac14 \left(i\pi+4i\pi\right)\right) = \sqrt2 \exp\left(i\dfrac{5\pi}4\right)[/tex]
[tex]k = 3 \implies z = 4^{1/4} \exp\left(\dfrac14 \left(i\pi+6i\pi\right)\right) = \sqrt2 \exp\left(i\dfrac{7\pi}4\right)[/tex]
