Help with tough proof involve complex numbers Let $z,w \in \mathbb{C}$
Prove that 

$|1-\bar z|^2 - |z-w|^2 = \left(1-|z|^2\right)\left(1-|w|^2\right)$

Then with help from the above proof, prove the following: 

If $|z|<1$ and $|w|<1$ then $|1-z\bar w|\neq0$ and $\big|\frac{z-w}{1-z\bar w}\big|$ 

 A: 
Prove that 
  $ |1-\bar{z}|^2 -|z-w|^2=(1-|z|^2)(1-|w|^2) $

Take $z=1$ and $w=0$ then
$ |1-\bar{z}|^2 -|z-w|^2\ne(1-|z|^2)(1-|w|^2) $
Also taking $z=a+ ib$ and $w=c+id$
The LHS:
$|1-\bar{z}|^2 -|z-w|^2=(1 - a)^2 + b^2 - (a - c)^2 - (b - d)^2$
and RHS:
$(1-|z|^2)(1-|w|^2)=(1 - a^2 - b^2) (1 - c^2 - d^2)$
You can instantly see that they will not equate since the RHS has the $b^2d^2$ term which you cannot obtain from the LHS. So it would seem that the first equation does not hold. However,
$$|1-z\bar{w}|^2-|z-w|^2 = (b c - a d)^2 + (1 - a c - b d)^2 - (a - c)^2 - (b - d)^2$$
$$ =1 - a^2 - b^2 - c^2 + a^2 c^2 + b^2 c^2 - d^2 + a^2 d^2 + b^2 d^2$$
$$=(1 - a^2 - b^2) (1 - c^2 - d^2)=(1-|z|^2)(1-|w|^2)$$

Then with help from the above proof, prove the following: 
  If $|z|<1$ and $|w|<1$ then $|1-z\bar w|\ne 0$ and $\left|\frac{z-w}{1-z \bar w}\right|$

If $|z|<1$ and $|w|<1$ this implies $(1-|z|^2)(1-|w|^2)> 0$ because $|z|,|w| \geq 0$
So $|1-z\bar{w}|^2-|z-w|^2 > 0$
Now since $|z-w|^2 \geq 0$
$|1-z\bar{w}|^2>0$ strictly i.e. $|1-z\bar{w}| \neq 0$

I'm guessing you need to prove 
  $\left|\frac{z-w}{1-z \bar w}\right|<1$

Since  $|1-z\bar{w}|^2-|z-w|^2 > 0$ we have,
$$\left|\frac{z-w}{1-z\bar{w}}\right|^2\leq \frac{|z-w|^2}{|1-z\bar{w}|^2}<1$$
$$\Rightarrow \left|\frac{z-w}{1-z\bar{w}}\right|<1$$
A: Hint: 
Break up $z$ and $w$ into their complex components respectively, $(a + bi)$ form. And then work from there. You should have four constants $z_a$, $z_b$, $w_a$, $w_b$.
Respond back if you want me to show the work.
