Let the similarity transformation $f(z)=az+b$ have $z_0$ as a fixed point. Show that $f$ can be rewritten in the form $f(z)=z_0+a(z-z_0)$. Deduce from this representation that the geometric effect of $f$ on the complex plane amounts to a rotation about the point $z_0$ in which every ray issuing from $z_0$ is mapped to itself and all distances get multiplied by the factor $\vert a \vert$. .
This is the question as it is written in the book. So since $z_0$ is a fixed point then $f(z_0)=az_0+b=z_0$ and hence $b=z_0-az_0$ and therefore for any $z$ we have $f(z)=az+b=az+z_0-az_0=z_0+a(z-z_0)$ as required. My problem is with the next part of the question which is "Deduce from this representation that the geometric effect of $f$ on the complex plane amounts to a rotation about the point $z_0$ in which every ray issuing from $z_0$ is mapped to itself and all distances get multiplied by the factor $\vert a \vert$. " So I would appreaciate any help with this
Translate your complex plane, so the origin of the new one is exactly $z_0$. Then in the "new" complex plane the function becomes $f(w) = aw$, which exactly means that every point on the complex plane is sent to a point in the same direction and the distance to the "new" origin is multiplied by a factor of $|a|$
Indeed if you write $w = re^{i\theta}$ we have that $f(w) = (ar)e^{i\theta}$, which coresponds to the function keeps the angle relative to $z_0$ same and only changes the distance, by a factor of $|a|$.
Moreover you can skip the explicit translation and write $z= z_0 + re^{i\theta}$ and get the same result, where $r = |z-z_0|$ and the angle $\theta$ is with respect to the fixed point $z_0$.