This isn't quite homework, but it might as well be. I'm working through Needham's Visual Complex Analysis and I'm finding myself stuck on the following problem:
Explain geometrically why the the locus of z such that $\mathrm{arg}(\frac{z-a}{z-b})=\theta$ is an arc of a certain circle passing through the fixed points a and b.
My first thought was to think about what $\mathrm{arg}(z)=\theta$ means, and I think it should be a line starting at the origin and going in the direction of $\theta$. I also understand that, supposing that we were able to find one z that satisfies this condition, we could find another z by moving the vector in a way that increases the argument of $z-a$ while decreasing the argument if $z-b$.
I'm looking for a hint. Can anyone subtly nudge me in the right direction? I'm having trouble seeing why this has to be the arc of a circle. Also, how can the arc pass through b? Would the fraction $\frac{z-a}{z-b}$ be undefined for z = b? How can it pass through a? Wouldn't the fraction be the zero vector for z = a ?