Locus in complex z-plane given eqn

Question

I have the question:

$ \text{Find the locus in the complex }z\text{-plane that satisfies the equation: } z-c=\rho\dfrac{1+it}{1-it}, \text{where }c\text{ is complex, }\rho\text{ is real, and }t\text{ is a real parameter that varies in the range }-\infty<t<\infty. $ [1]

But I am unsure how to proceed - I haven't seen such a question with as many variables before; I don't really understand what should be the format of a solution. The only similar questions I can find I understand, but I can't translate that to what I should be doing here.

I realise that $c$ is essentially a transformation, so I assume that could be ignored to begin with, similarly I assume it could be assumed momentarily that $\rho=1$. But what more of an answer am I trying to find? The question already gives a condition for $z$, and wants a solution for $z$ too? Or have I totally misunderstood?

1 - Mathematical Methods for Physics and Engineering - Ex3.4a

Answer 1

Let $\alpha=1+it$. Then

$$z-c=\rho\frac{\alpha}{\bar{\alpha}}.$$

It follows that

$$|z-c|^2=(z-c)(\bar{z}-\bar{c})=\rho\frac{\alpha}{\bar{\alpha}}\rho\frac{\bar{\alpha}}{\alpha}=\rho^2,$$

as $|\alpha|^2\neq 0$.

The desired locus is then

$$\mathcal L=\{z\in\mathbb C ~|~~ |z-c|=|\rho|\}. $$

Answer 2

Hint: The Möbius transformation $\dfrac{1+it}{1-it}$ maps the real line onto the set $\{z \in \mathbb{C} \colon\ |z|=1,\ z \ne -1\}$.