# Given that $|z| = 1$ ( $z \neq1$) prove show that $\Re\left[\frac{1}{1-z}\right] = \frac{1}{2}$ [closed]

For a complex number $$z$$, whose modulus $$|z|=1$$ (but it's not entirely real $$z\neq1$$), how do I show that

$$\Re\left[\frac{1}{1-z}\right] = \frac{1}{2}$$

I have tried applying the formula $$|z|^2 = z\bar{z}$$ given that i have seen a lot of similar problems solved this way and the fact that $$\Re[x] = \frac{x+\bar{x}}{2}$$ for $$x \in \mathbf{C}$$.

I can't seem to rearrange it properly however. Thanks in advance!

Your idea is fine, indeed using that $$\Re(w)=\frac12 (w+\bar w)$$, we obtain
$$\Re\left[\frac{1}{1-z}\right] =\frac12\left(\frac1{1-z}+\frac1{1-\bar z}\right)=\\=\frac12\left(\frac{(1-\bar z)+(1-z)}{(1-z)(1-\bar z)}\right)=\frac12\left(\frac{2-z-\bar z}{2-z-\bar z}\right)=\frac 12$$
As an alternative, by $$z=e^{i\theta}$$ and Euler's identity
$$\frac{1}{1-z}=\frac{1}{1-e^{i\theta}}\frac{1-e^{-i\theta}}{1-e^{-i\theta}}=\frac{1-e^{-i\theta}}{2-e^{i\theta}-e^{-i\theta}}=\frac{1-\cos \theta+i\sin \theta}{2(1-\cos \theta)}=\frac12+i\frac{\sin \theta}{2(1-\cos \theta)}$$