The question:

Show that $\displaystyle Re(\frac{1 - z}{1 + z}) = 0$ for any complex number z with |z| = 1.

My attempt:

I tried finding the the result of the division, hoping to get a value of 0 as the real part, to confirm the problem. This is what I got:

Let z = a+ib

$\displaystyle \frac{1 - (a + ib)}{1 + (a + ib)} = \frac{(1 - a) - ib)}{(a + 1) + ib}$

$\displaystyle \frac{(1 - a) - ib)}{(a + 1) + ib} . \frac{(a + 1) - ib)}{(a + 1) - ib} = \frac{[(1 - a) - ib)][(a + 1) - ib]}{(a + 1)^2 + b^2}$

$\displaystyle \frac{(1 - a^2) - 2ib -b^2)}{(a + 1)^2 + b^2} = \frac{(1 - a) - 2ib -b^2}{(a + 1) + b^2}$

$\displaystyle \frac{(1 - a) - b^2 - 2ib}{(a + 1) + b^2}$

Therefore the real part is:

$\displaystyle \frac{(1 - a) - b^2}{(a+1) + b^2}$

Now I'm not sure where to go from here. I think what I've done so far is correct. Any help would be great, thanks!