Is $f: \mathbb{C} \to \mathbb{C}$ with $f(z) = z^2$ surjective? So I know that, the map is surjective if
$$\forall b \in C, \exists a \in C \text{ such that } f(a) = b.$$
The problem I'm encountering is that normally I would try to find the $a$ by doing this
$$b = a^2 \text{ so } a = \sqrt{b}.$$
Then I would say:
$$f(x) = f(\sqrt{b}) = (\sqrt{b})^2 = b.$$
So $\forall b. \exists a \text{ such that } f(a) = b$.
But I was taught that I couldn't take the square root of an imaginary number. So I don't what to do...?
*PS: English is not the language I'm taught in, so I may have used the wrong words like 'map' or I may have used the terms the wrong way. In advance my apologies if I couldn't make myself clear. 
 A: Using $\sqrt{b}$ requires knowing it exists to begin with, which is essentially what you want to prove. (Actually a square root function on the complex number can only defined to a certain extent, but it's not relevant.)
If you already know De Moivre's formulas and the polar representation of complex numbers, it's easy. However it can also be done with the $x+iy$ representation. Given $a+ib\in \mathbb{C}$ (with $a$ and $b$ real), we want to find $x$ and $y$ real such that $(x+iy)^2=a+ib$.
Expanding the square and equating the real and imaginary parts, we get
$$
\begin{cases}
x^2-y^2=a \\[6px]
2xy=b
\end{cases}
$$
The case $b=0$ is the easiest: we must have $x=0$ or $y=0$. Looking at the first equation we get $x=\pm\sqrt{a}$ and $y=0$ if $a\ge0$; $x=0$ and $y=\pm\sqrt{-a}$ if $a<0$.
Note that $\sqrt{t}$ is well defined when $t$ is a nonnegative real number.
Suppose now $b\ne0$. Then we have $y=b/(2x)$ and the first equation gets transformed into
$$
4x^4-4ax^2-b^2=0
$$
The polynomial $4t^2-4at-b^2$ has a positive root, so we obtain
$$
x^2=\frac{\sqrt{a^2+b^2}+a}{2}
$$
from which we get two values for $x$ and the corresponding two values for $y$.
A: The map is surjective anyway as any $z$ can be written as $r(\cos \phi+i\sin\phi)$ with $r\ge 0$ and that is the square of (among others) $\sqrt r(\cos\frac\phi 2+i\sin\frac \phi2)$.
A: Let $\alpha \in \mathbb{C}$. Then $\alpha$ is in the image of $f$ if the polynomial $x^2 - \alpha$ has a solution. Well, the field $\mathbb{C}$ is algebraically closed, so this is the case.
