EDIT: Its done

Hi there, I have to prove this two sentences . I think I've solved the first, but I'm quiet stuck with the second.

The first says:

1) Demonstrate that the equation of a line or a circumference in the complex plane can be written this way: $\displaystyle \alpha z . \bar{z}+\beta z+\bar{\beta} \bar{z}+\gamma=0$, with $\displaystyle \alpha,\gamma\in{R},\beta\in{C}$

So I called z and beta:

$\displaystyle z=x+iy,\beta=u+iv$

Then developing the products I get:

$\displaystyle \alpha(x^2+y^2)+2(ux-vy)+\gamma=0$

And making alpha equal zero I get the equation for a line, right? (for u and v fixed).

Then completing the square and reordering:

$\displaystyle \left(x+\displaystyle\frac{u}{\alpha}\right )^2+\left(y-\displaystyle\frac{v}{\alpha}\right )^2=\displaystyle\frac{u^2}{\alpha^2}+\frac{v^2}{\ alpha^2}-\frac{\gamma}{\alpha}$

This is the equation for the circle, is this right?

In the other hand I got:

2) Prove that the geometrical place for the points $\displaystyle z\in{C}$ that verifies $\displaystyle \left\|{\displaystyle\frac{z-1}{z+1}}\right\|=k$ is a circumference ($\displaystyle k\in{R},1\neq{k}>0$).

I couldn't make much for this. I called z=x+iy again:

$\displaystyle \left\|{\frac{z-1}{z+1}}\right\|=k\Rightarrow{\left\|{\frac{(x+iy-1)(x-iy+1)}{(x+iy+1)(x-iy+1)}}\right\|=k}\Rightarrow{\left\|{\frac{x^2+y^ 2+2iy-1}{(x+1)^2+y^2}}\right\|=k}$

I don't know what to do from there.

Bye.

(Sorry if this topic doesn't belong here, please move it, I didn't know where to post so I posted here, and sorry too if it bothers you that I made two questions in the same topic, but I thought it was quiet trivial and it didn't deserved two different topics, but I can split it if necessary).