How do I find the image of the domain under the map $f(z)=z^2$? I am trying to understand what the image of the regions a)$|z|\leq 1$ and b) $Re(z)\ge 1$ under the map $f(z)=z^2$ look like.
My progress till now(for $|z|\leq 1$: 
Let $z=a+ib$. Then $f(z)=z^2$ and so $f(z)= (a^2-b^2)+2iab$. Now, $|f(z)|=a^2+b^2=|z|^2$ and so $0\leq |f(z)| \leq 1$. That just means $0\leq a,b \leq 1$ but I can't see anything beyond this. 
So, what does $f$ map the region $|z|\leq 1$ to?
Those are two separate regions; sorry for any confusion.
 A: The region $|z|\le 1$ is a circle of radius $1$ centered in the origin, and the complex numbers in this circle are $ z=\rho e^{i \theta}$ with $\rho\le 1$, so $z^2= \rho^2 e^{i2\theta}$ and, since $\rho \le 1 \Rightarrow \rho^2\le 1$, this numbers are in the same circle of radius $1$.
A: For understanding the behavior of the squaring function (or any power function, for that matter), it’s extremely useful to have in mind the polar representation of a number, $z=r(\cos\theta+i\sin\theta)$. Here $r$ is the distance from the origin, equal to the absolute value of $z$, and $\theta$ is the argument, an angle measured ccw from the positive real axis to the line joining the origin to $z$. So, $-2+2i$ has $r=2\sqrt2$ and $\theta=135^\circ$, or $3\pi/4$ if you want to be mathematical. 
What the squaring function does is square the value of $r$ and double the angle $\theta$. In the easy case of the interior of the unit circle, you see immediately what’s happening.
In the case of the half-plane $\Re(z)\ge1$, the polar representation is less useful, and it’s helpful to think of the boundary, $\Re(z)=1$, the line $x=1$. Consider a point on this line, $1+bi$, and square it. You get a nice relation between the new $x$-coordinate and the new $y$-coordinate.
If all fails, take explicit points (like $(1,3)\leftrightarrow1+3i$) and square them.
