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?
 A: All you have stated is that
$$n+(n-1)i^2=n+(n-1)(-1)=1$$
$$n+(n+1)i^2=n+(n+1)(-1)=-1$$
has infinitely many solutions (which is trivially true). But if we were to solve the equation this way, we would need to substitute $x=a+bi$ in our original equation getting
$$(a+bi)^4=a^4-6a^2b^2+b^4+(4a^3b-4ab^3)i=1$$
which (comparing real and imaginary parts) is equivalent to the simultaneous equations
$$a^4-6a^2b^2+b^4=1$$
$$4a^3b-4ab^3=0$$
You should find this only has the solutions $(a,b)\in\{(1,0),(-1,0),(0,1),(0,-1)\}$. Hence there are only $4$ solutions for $x$ namely
$$x=\pm1,\pm i$$
A: Actually, in $\mathbb C$ we have: $$x^4=1\Leftrightarrow x\in\sqrt[4]1,$$
where $$\sqrt[4]1=\{1,-1,i,-i\}$$
A: 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: 
    $$z_k=\sqrt[n]{||w||}\cdot\left( \cos\left(\frac{\arg(w)+2k\pi}{n}\right) + i\cdot\sin\left(\frac{\arg(w)+2k\pi}{n}\right) \right),~k=0,1,\cdots,n-1$$


A: The exponential form of a complex number is $x = r\cdot e^{~i~\varphi} $. The $r$ is the length of the segment line connecting the origin of the complex plain and the point representing the complex number $x$ in the complex plane, so $r=\sqrt{\mathcal{Re}^2(x)+\mathcal{Im}^2(x)}$, where $\mathcal{Im}(x)$ is the imaginary part (the number beside the $i$) of the complex number $x$ and where $\mathcal{Re}(x)$ is the real part of the complex number $x$. The $\varphi$ is defined as $\tan(\varphi)=\mathcal{Im}(x)/\mathcal{Re}(x)$. This $x = r\cdot e^{~i~\varphi} $ is a nicer way of describing a complex number than $x = a+i~b = \mathcal{Re}(x)+i~\mathcal{Im}(x)$, because with the euler formula $e^{~i~\varphi} = \cos(\varphi) + i~\sin(\varphi)$ we can think of the complex numbers as "rotating" vectors in the complex plane. we can also think of $1$ as  a vector in the complex plane, because $1 = e^{~i~0} = \cos(0) + i~\sin(0)  $ (and because $\cos(0)=1$ and $\sin(0)=0$). 
With that in mind, the equation $x^4 = 1$ gets a new meaning. It means that we are looking for vectors in the complex plane, that can be rotated to the vector $e^{~i~0} = 1$. That would mean that the length $r$ of the vectors we are looking for are all $r=1$, because we are only rotating (and not scaling for example). This means we are trying to solve $e^{~i~0}=e^{~4~i~\varphi_x}$ instead of $1=x^4$ or we are trying to solve $$ \cos(0) + i~\sin(0)= \cos(4~\varphi_x) + i~\sin(4~\varphi_x)$$ 
This equation is fullfilled for $4~\varphi_x = 2~\pi~ k$ for $k\in\mathbb{Z}$ or for $\varphi_x = k~(\pi/2)$, because of the definitions of the $\sin$- and $\cos$- functions. So the solutions for $x^4 = 1$ are numbers of the form $$x = e^{i~k~(\pi/2)} = \cos(k~(\pi/2))+i\sin(k~(\pi/2)), ~~~ k = 0,1,2,3$$
The first solution for the equation $x^4=1$ is for $k=0$ the following:
$$x = e^{i~0} = \cos(0)+i\sin(0)= (1)+i~(0)= +1$$
The second solution for the equation $x^4=1$ is for $k=1$ the following:
$$x = e^{i~(\pi/2)} = \cos(\pi/2)+i\sin(\pi/2) = (0)+i~(1) = +i$$
The third solution for the equation $x^4=1$ is for $k=2$ the following:
$$x = e^{i~\pi} = \cos(\pi)+i\sin(\pi) = (-1)+i~(0) = -1$$
The fourth solution for the equation $x^4=1$ is for $k=3$ the following:
$$x = e^{i~(3\pi/2)} = \cos(3\pi/2)+i\sin(3\pi/2) = (0)+i~(-1)=-i$$
After this derivation we could generalize the result, because what we did just means that the solutions for $y^n = 1$ where $n \in\mathbb{N}$ and $n > 1$ and $y \in \mathbb{C}$ and $k \in\mathbb{Z}$ and $k \in [0,n-1]$ are
$$y = e^{i~(2~k~\pi)/n} = \cos((2~k~\pi)/n)+i\sin((2~k~\pi)/n)$$
Look at this visualisation from the Principal square root of a complex number and at the roots of unity (de Moivre numbers)
