Given the mapping $w=\frac{z+1}{z-i}$ evaluate the images on the $w$ plane of:
$$a)\hspace{1cm} x^2 + (y - 1)^2 = 1$$ $$b)\hspace{1cm}y - x = 0$$
I'm completely lost here. Hit a mental roadblock trying to solve for $z$. Kind of learning this on my own, so any pointers on bibliography is appreciated.
(a) Writing $z=a+ib$, equation $x^2+(y-1)^2=1$ is equivalent to $|z-i|=1$, i.e. (a) describes a circle with center $i$ and radius 1. Thus, each such $z$ can be written as $z=i+e^{it}$ for $t\in[0,2\pi)$. Then \begin{align*} \frac{z+1}{z-i}=\frac{i+e^{it}+1}{e^{it}}=ie^{-it}+1+e^{-it}=e^{-it}(i+1)+1. \end{align*} Since $i+1=\sqrt{2}e^{i\pi/4}$ we obtain \begin{align*} \frac{z+1}{z-i}=\sqrt{2}e^{i(\pi/4-t)}+1. \end{align*} And this is a circle with center 1 and radius $\sqrt{2}$.
(b) Writing $z=x+iy$, equation $y-x=0$ is equivalent to $z=x+ix$, which is a straight line through $0$ and $1+i$. Thus, \begin{align*} \left|\frac{z+1}{z-i}-(1+i)\right|^2=\left|\frac{z+1-(1+i)(z-i)}{z-i}\right|^2=\frac{|z+1-z+i-iz-1|^2}{|z-i|^2}=\frac{|1-z|^2}{|z-i|^2}=1, \end{align*} which shows that we again have a circle with center $1+i$ and radius 1.