why $x^{4} = 1 : x \in \mathbb{C}$ doesn't have infinite solutions? I know that an equation has a number of roots equal to the degree of the equation, but when I was solving this equation:
$$x^{4} = 1$$
I did that:
$$x^{4} = 1 \iff x = \pm \sqrt[4]{1} = \pm 1 \implies x = 1 \text{ or } x = -1$$
then knowing that this equation must have at most 4 distinct roots I replaced $x$ with $a + bi$:
$$(a + bi) = \pm 1$$
And here I realised that $(n + (n - 1)i^{2}) = 1$ and $(n + (n + 1)i^{2}) = -1$.
This means that there are infinite solutions, so where did I go wrong?

This is a good approach:
Every non zero complex number can be determined by an angle and a length as follows:
$$z=r(\cos\theta+i\sin\theta)~~~~~~~~~~~(\ast)$$
So, you need to find a number $z=x+iy$ (or a set of numbers) such that
$$z^4=1=1+0\cdot i=\cos(2\pi)+i\cdot\sin(2\pi)$$
But $$cos(2\pi)=\cos(4\pi)=\cos(6\pi)=\cdots \mbox{ and } \sin(2\pi)=\sin(4\pi)=\sin(6\pi)=\dots$$
For this reason, the angle for our complex number will be only from $[0;2\pi\rangle$. 
Then, as we said we need a lenght and a angle, in complex plane those properties are called module and argument:
If we need a number $z$ such that: 
$$z^4=1$$
it must satisfies $$\arg(z^4)=\arg(1)\mbox{ and } ||z^4||=||1||$$

Finding the argument:
$\arg(z^4)=4\cdot\arg(z)$ but multiplying a number of the range $[0;2\pi\rangle$ four times may lead that it escapes from the interval $[0;2\pi\rangle$, so we need to add a correction factor of $2\pi$ several times ($k$ times) because $\cos,\sin$ has a period $2\pi$:$~~\arg(z^4)=4\cdot\arg(z)-2k\pi,~k\in\mathbb{Z}$
So we have: $$4\times\arg(z)-2k\pi = \arg(1)=0\Rightarrow\arg(z)=\frac{2k\pi}{4}$$

Finding the module:
This is easiest:
$$||z^4||=||1||=1\Rightarrow||z||^4=1\Rightarrow ||z||=1 \mbox{ (normal real numbers root)}$$

therefore, the complex $z$ number we are looking for, has a module $1$ ands its argument has to be of the form $\dfrac{2k\pi}{4}$. Since the argument must to be in the range $[0,2\pi\rangle$ we see that if $k>4\Rightarrow\dfrac{2k\pi}{4}>2\pi$ and since $\frac{2k\pi}{4}$ is increasing function respect $k$, the only values for $k$ are $0,1,2\mbox{ and }3$.
our numbers $z$ using $(\ast)$ are:


*

*$k=0: z_1=\cos(0)+i\sin(0)=1$

*$k=1: z_2=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})=0+1\cdot i=i$

*$k=2: \cos(\pi)+i\sin(\pi)=-1+0i=-1$

*$k=3: \cos(\frac{3\pi}{2})+i\sin(\frac{3\pi}{2})=0-1\cdot i=-i $


We conclude: If we are looking for all solutions for the equation $$z^n=w$$
    then, those solutions are: 
    $$z_k=\sqrt[n]{||w||}\cdot\left( \cos\left(\frac{\arg(w)+2k\pi}{n}\right) + i\cdot\sin\left(\frac{\arg(w)+2k\pi}{n}\right) \right),~k=0,1,\cdots,n-1$$


