Exactly that is what I did in steps.

Here are some basic transformations of the complex plane.

- $\displaystyle f(z) = kz$, where $\displaystyle k$ is a positive real number. If $\displaystyle k>1$ then the shape gets magnified by a factor of $\displaystyle k$ if $\displaystyle k<1$ then the shape gets minimized by a factor of $\displaystyle 1/k$.
- $\displaystyle f(z) = z + \alpha$ where $\displaystyle \alpha$ is a complex number. This is a translation in the direction of $\displaystyle \alpha$.
- $\displaystyle f(z) = e^{i\theta} z$. Where $\displaystyle \theta$ is a real number. This is a counter-clockwise rotation by $\displaystyle \theta$.

When I was studing conformal mappings I came up with a very simple function which will make all problems involving lines and circles really easy. It is $\displaystyle f(z) = (1-z)/(1+z)$.

This function has some really nice properties. Which you might want to try to prove.

- $\displaystyle f(z)$ will map $\displaystyle |z|=1$ onto $\displaystyle \Re (z) = 0$.
- $\displaystyle f(z)$ will map $\displaystyle \Re(z) = 0$ onto $\displaystyle |z|=1$.
- $\displaystyle f(z) = f^{-1}(z)$

Actually we should be a little careful. $\displaystyle f(z)$ is not defined at $\displaystyle z=-1$, at that point on the unit circle. So it is more formal to say $\displaystyle \{ |z| = 1\} - \{ -1\}$ gets mapped to $\displaystyle \Re (z) = 0$. And conversely, $\displaystyle f(z)$ does not really map $\displaystyle \Re (z) = 0$ to $\displaystyle \{ |z| = 1\}$ but rather to $\displaystyle \{ |z| = 1\} - \{ -1\}$. However, this is a minor problem which we can fixed like we did now.