What are the solutions to $z^4+1=0$? [duplicate]

Solution 1:

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 $z_3$...

Solution 2:

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.

Solution 3:

$$z^4=-1=e^{\pi i+2k\pi i}=e^{\pi i(1+2k)}\Longrightarrow z=e^{\frac{\pi i}{4}(1+2k)}\,\,,\,k=0,1,2,3$$