How many different complex numbers $z$ satisfying equation $z^4 = 1$ are there?

Question

I'm new to the complex numbers and i can't answer a simple question. Any help and explanation how to solve it?

Answer 1

Fundamental Theorem of Algebra:

An $n^{\text{th}}$ order polynomial has exactly $n$ solutions, taking into consideration the multiplicity of each distinct solution.

$$\begin{align}z^4&=1\\z^4&=e^{2i\pi k}\\z&=e^{\frac12i\pi k}\\&=\cos\left(\dfrac12\pi k\right)+i\sin\left(\dfrac12\pi k\right)\end{align}$$

Putting $k=0,1,2,3...$ $$\begin{align}z&=\cos0+i\sin 0&=1\\z&=\cos\dfrac\pi2+i\sin\dfrac\pi2&=i\\z&=\cos\dfrac{3\pi}2+i\sin\dfrac{3\pi}2&=-i\\z&=\cos\pi+i\sin\pi&=-1\end{align}$$

Another method could be.... $$\begin{align}z^4&=1\\z^4-1&=0\\(z^2+1)(z^2-1)&=0\\z&=\pm i,\pm 1\end{align}$$

Answer 2

hint : $z^4-1=(z^2-1)(z^2+1)=(z-1)(z+1)(z+i)(z-i)$