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?
 A: 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 $z=-1.$
A: Assume $z=e^{i\theta}$ therefore
$$\frac{1+z}{1+\bar z}=\frac{1+e^{i\theta}}{1+e^{-i\theta}}=\frac{e^{i\theta}(1+e^{i\theta})}{e^{i\theta}(1+e^{-i\theta})}=\frac{e^{i\theta}(1+e^{i\theta})}{e^{i\theta}+1}=e^{i\theta}$$
which is valid for $z\neq -1$.
A: Note that for any
$\chi \in \Bbb C \tag 1$
we have
$\vert \chi \vert = \vert \bar \chi \vert; \tag 2$
this is easily seen by writing
$\chi = a + bi; \tag 3$
then
$\bar \chi = a - bi; \tag 4$
thus,
$\vert \chi \vert = \sqrt{a^2 + b^2} = \vert \bar \chi \vert; \tag 5$
or, if one prefers polars,
$\chi = re^{i\theta}, \; \bar \chi = re^{-i\theta}, \tag 6$
and so
$\vert \chi \vert = \vert re^{i\theta} \vert = r \vert e^{i\theta} \vert = r = r \vert e^{-i\theta} \vert = \vert re^{-i\theta} \vert = \vert \bar \chi \vert; \tag 7$ 
so if 
$\chi \ne 0, \tag 8$
$\left \vert \dfrac{\chi}{\bar \chi} \right \vert = \dfrac{\vert \chi \vert}{\vert \bar \chi \vert} = 1; \tag 8$
we finish off by setting
$\chi = 1 + z \tag 9$
(assuming of course $z \ne -1$), whence
$\bar \chi = \overline{1 + z} = 1 + \bar z, \tag{10}$
whence
$\left \vert \dfrac{1 + z}{{1 + \bar z}} \right \vert = \left \vert \dfrac{1 + z}{\overline{1 + z}} \right \vert = 1.  \tag{11}$
Finally, if we write
$1 + z = re^{i\theta}, \tag{12}$
we find
$1 + \bar z = \overline{1 + z} = re^{-i\theta}, \tag{13}$
and so
$\left \vert \dfrac{1 + z}{\overline{1 + z}} \right \vert = \left \vert \dfrac{re^{i\theta}}{re^{-i\theta}} \right \vert = \vert e^{2i\theta} \vert = 1. \tag{14}$
With
$\vert z \vert = 1, \tag{15}$
$z = e^{i\theta}, \; \bar z = e^{-i\theta}, \tag{16}$
so
$\dfrac{1 + z}{1 + \bar z} = \dfrac{1 + e^{i\theta}}{1 + e^{-i\theta}} = \dfrac{e^{i\theta}}{e^{i\theta}}  \dfrac{1 + e^{i\theta}}{1 + e^{-i\theta}}$
$= e^{i\theta} \dfrac{1 + e^{i\theta}}{e^{i\theta} + 1} = e^{i\theta} = z, \tag{17}$
stipulating of course that $\theta$ is not an odd multiple of $\pi$.
