Intermediate Algebra Basic Complex Question

The question implies that the real part of  $\frac{1}{1-z}$  is the same for all nonreal complex values of  z  such that  $|z| = 1$ . So assuming that's true, we can pick any nonreal complex  z  such that  $|z| = 1$  and find the real part of  $\frac{1}{1-z}$ .

Let's pick  z  =  0 + 1i

Then...

${\frac{1}{1-z}}\ =\ \frac{1}{1-(0+1i)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{1-1i}\cdot\frac{1+1i}{1+1i}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-i^2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{1-(-1)}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1+1i}{2}\\~\\ \phantom{\frac{1}{1-z}}\ =\ \frac{1}{2}+\frac{1}{2}i$

The real part is  $\boxed{\frac12}$

We can check a few more cases to see that it is 1/2 for those as well:

z = cis( pi/6 )

z = cis( pi/4 )

It seems like no matter what the angle is, the real part is  1/2