Prove that if $z$ is uni modular then $\frac{1+z}{1 + \bar z}$ is equal to $z$. The expression can be written as 
$$\frac{1+z}{\overline {1+z}}$$
Since $z \cdot \overline z=|z|^2$
$$\overline{1+z}= \frac{1}{1+z}$$
As $|z|=1$
So it will become $(1+z)^2$
But the answer is $z$. What am I doing wrong?

You are asserting that if $z\cdot \overline z=1$ then $(1+\overline z)(1+z)=1,$ too. That is not true. A simple case is $z=1.$
Being unimodular means $z\cdot \overline{z}=1.$ Then $\overline z = z^{-1}.$
So $$\frac{1+z}{1+\overline{z}}=\frac{1+z}{1+z^{-1}}=\frac{1+z}{1+z^{-1}}\cdot \frac z z=\frac{z(1+z)}{z+1}=z$$
Note, the left side is not defined in the case $z=-1.$
