What is the image of the disc $|z|<\frac{1}{2}$ under the complex function $f(z)=\frac{z^2+1}{2z}=\frac{1}{2}\left( z+\frac{1}{z} \right)$?
I guess that a first step is to write: $2f(x+iy) = x+iy+\frac{1}{x^2+y^2}\left( x-iy \right)$ or $2f(re^{i\theta}) =re^{i\theta}+\frac{1}{r}e^{-i\theta} $. Now, whenever $|z|$ is constant (for example $=c$), we can write $f(z)=\frac{c^2 e^{i\theta}+e^{-i\theta}}{c}$, which is something I do not really know how to specify.
Only when $c=1$, I obtain $f(z)=e^{i\theta}+e^{-i\theta}=2\cos (\theta)$ which is just part of the real line.
So to understand what is the image of the disc $|z|<\frac{1}{2}$, I guess I would have to understand what is the image of every circle $|z|=c$ ($0<c<0.5$), or what is the image of every ray $\theta=c, 0<r<0.5$. How should I do it?
You have if $z = ce^{-i\theta}$ and $f(z) = x + iy$ then,
$$x = \frac{1+c^2}{2c}\cos \theta, \quad y = \frac{1-c^2}{2c}\sin \theta$$
this is an ellipse of big axis $a = \frac{1+c^2}{2c}$ and small axis $b = \frac{1-c^2}{2c}$. Both axes are decreasing with the value of $c$ so the image that you are looking for is outside the ellipse:
$$x = \frac{5}4\cos \theta, \quad y = \frac34\sin \theta$$