complex numbers

Note that $i=\sqrt{-1}$ is an imagine number. Therefore,

$iz^4+1=0$

$iz^4=-1$

$iz^4=i^2$

$z^4=i$

$\Rightarrow \boxed{z = \cos{\frac{\pi}{8}}+i\sin{\frac{\pi}{8}}}$

To show that the above solution is correct, square $z$ twice as follows:

$z^2 = \cos^2{\frac{\pi}{8}} +2i\sin{\frac{\pi}{8}}\cos{\frac{\pi}{8}}-\sin^2{\frac{\pi}{8}}$

$\quad = \cos{\frac{\pi}{4}} +i\sin{\frac{\pi}{4}}$

Squaring $z$ we double the angle ( $\frac{\pi}{8}\times 2 = \frac{\pi}{4}$ ), similarly when we square $z^2$ , we get:

$z^4 = \cos{\frac{\pi}{2}} +i\sin{\frac{\pi}{2}}=0+i=\boxed{i}$