What are the solutions to $z^4+1=0$? I can't seem to find the  solutions to  $z^4+1=0 $.
$z$ is in the complex plane. 
The solutions show four roots; however, how do I find them once $z^4 = -1$? 
 A: You can write $z^4=-1$ as $(z^2)^2=-1$. The two square roots of $-1$ are $i$ and $-i$, so we get the two equations $z^2=\pm i$. 
Since $i$ corresponds to $\pi/2$ on the unit circle, its square root will have to correspond to $\pi/4$ (or use De Moivre if you don't see this). So 
$$
z=\pm\frac{1+i}{\sqrt 2},\ \ z=\pm\frac{1-i}{\sqrt 2}
$$
are the roots. 
A: $$z^4=-1=e^{\pi i+2k\pi i}=e^{\pi i(1+2k)}\Longrightarrow z=e^{\frac{\pi i}{4}(1+2k)}\,\,,\,k=0,1,2,3$$
A: You can write $-i$  in polar form: $$-i = e^{i \cdot 3 \pi /2} $$
Then to find a fourth root...
A: Since we have $i^2=-1$ $$z^4+1=(z^2)^2-(i)^2$$
$a^2-b^2=(a-b)(a+b)$, so we can factor to have
$$z^4+1=(z^2)^2-(i)^2=(z^2-i)(z^2+i)$$
It's easy to solve from here on.
$$z^4+1=0 \implies \left \{ \begin{align}&z^2-i=0\implies z=\pm\sqrt i \\&z^2+i=0 \implies z=\pm\sqrt{-i}\end{align}\right.$$
Using the properties 


*

*$i=e^{i(\pi/ 2)}$ 

*$-i=e^{i(3\pi/ 2)}$

*$e^{i\theta}=\cos \theta + i\cdot \sin \theta$


you  can express the result in much more interesting forms.
A: Here is how i was taught to find roots. I'll try to give a fully worked out answer, with no shortcuts, for the first 2 roots; then you should be able to do the 2nd two roots on your own.
$1+z^4=0$ gives $z^4=-1$
We know that $-1=e^{i \pi}$ because Euler tells us $e^{i\pi}= \cos \pi + i \sin \pi =-1 +i(0)=-1$
So, in this problem, we can write: $z^4 =e^{i \pi}$
But $e^{i\pi}=e^{i(\pi + 2\pi n)}$ for n=0,1,2,3,... (think of a point on the unit circle and do n complete rotations, for each integer n you end up right back where you started). 
So let us say that $$z_n^4=-1=e^{i(\pi +2\pi n)}$$
Then we have, $$z_n =[e^{i(\pi + 2 \pi n)}]^{\frac{1}{4}}=e^{i(\frac{\pi}{4}+\frac{\pi}{2}n)}$$
Now we can find the roots for n=0,1,2,3; that is, the 4 desired roots. 
First, $$z_0 =e^{i(\frac{\pi}4 +\frac{\pi}2(0))}=e^{i\frac{\pi}{4}}=\cos \frac{\pi}4 +i\sin \frac{\pi}4=\frac{1+i}{\sqrt{2}}$$
Second, $$z_1 =e^{i(\frac{\pi}4 +\frac{\pi}2 (1))} =e^{i\frac{3\pi}4}=\cos \frac{3\pi}4 +i \sin \frac{3\pi}4=\frac{-1+i}{\sqrt{2}}$$
Using this method you should be able to work out the next two roots :)
Can you guess what $z_4$ would be? Hint: think of a unit circle; you've been working your way around it from $z_0$ to $z_3$...
A: $$z^4+1 = (z^2+1)^2-2z^2 = (z^2+1-z\sqrt2)(z^2+1+z\sqrt2)$$
