Complex number reduction problem

Question

Given the complex number $ z = x + yi; x, y \in \!R$. Also, let $ w = \frac{z-1-i}{z+1+i}$. Assuming $\mid w \mid = 1$, we can affirm that

a) $x + y = 1$

b) $x + y = -1$

c) $x - y = 1$

d) $x + y = 0$

e) $xy = 1$

SOLUTION

I tried to solve this problem by replaying $z$ in $w$ and expand from, but it seems a dead-end, I ended up with a very long expression with $x$ and $y$.

Answer 1

$|w|=1$ means $|z-1-i|=|z+1+i|$ so $z$ is equidistant from $1+i$ and $-1-i$ What does that locus look like?

Answer 2

hint

$$|w|=1 \implies$$ $$ |z-1-i|^2=|z+1+i|^2$$

$$\implies $$ $$|(x-1)+i(y-1)|^2=$$ $$|(x+1)+i(y+1)|^2$$