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