Real part of a complex number 
Compute $\operatorname{Re}\left(\frac{1}{z+1}\right)$ when when $|z| = 1$.

The only way I could think to go about this is to simply go by definitions. If $z\in \Bbb C$, then $z\bar z$ = $|z|^2$. Now $$z=\frac{|z|^2}{\bar z}$$and$$z=\frac{1^2}{\bar z}.$$ So $z$ must be the inverse of the conjugate of $z$ which can be written as $$z=\frac{z}{\bar z z}$$ I don't know how to proceed from here. Is this even a good approach?

Let $z = a+bi$
$$\begin{align}Re\left(\frac{1}{z+1}\right) &= Re\left(\frac{1}{a+1+bi}\right) = Re\left(\frac{1}{a+1+bi}\cdot\frac{a+1-bi}{a+1-bi}\right) = \\ &= Re\left(\frac{a+1-bi}{(a+1)^2-(bi)^2}\right) = Re\left(\frac{a+1-bi}{(a+1)^2+b^2}\right) = \\ &= Re\left(\frac{a+1}{(a+1)^2+b^2} - \frac{b}{(a+1)^2+b^2}i\right) = \frac{a+1}{(a+1)^2+b^2} = \\ &= \frac{Re(z)+1}{(Re(z)+1)^2+Im(z)^2}.\end{align}$$
