Verification involving modulus and complex numbers Q: If $\alpha$ and $\beta$ are different complex numbers with $ |\beta|=1$, then find $ \bigl|\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\bigr|$   [correct answer is $1$]

My proof:
Let $ \alpha = a+ \iota b$ and $\ \beta = c + \iota d$
Since $|\beta| = 1 \implies \beta = 1,-1$ [opening modulus]
 Required: $ \cfrac{|\beta - \alpha|}{|1- \bar \alpha \beta|}$ 
 Case 1: When $ \beta =1$ 

$$ \implies \cfrac{|(1) - \alpha|}{|1- \bar \alpha (1)|} $$
$$ \implies \cfrac{|(1) - (a+\iota b)|}{|1- (a-\iota b)|} = \cfrac{|(1) - a -\iota b|}{|1- a+ \iota b|} = \cfrac{|(1 - a) -(\iota b)|}{|(1- a) + (\iota b)|}  $$
Since modulus of a complex number of the form $z = x+ \iota y$ is $|z| = \sqrt{x^2+y^2} $

$$\implies \require{cancel} \frac{\cancel{\sqrt{(1-a)^2+(-b)^2}}}{\cancel{\sqrt{(1-a)^2+(-b)^2}}} = 1$$

 Case 2: When $\beta = -1$ 

$$ \implies \cfrac{|(-1) - \alpha|}{|1- \bar \alpha (-1)|} $$
$$ \implies \cfrac{|(-1) - (a+ \iota b)|}{|1 + (a - \iota b)|} = \cfrac{|(-1 - a) - (\iota b)|}{|(1 + a) - (\iota b)|}  $$
[ the below part is where I think that my solution is wrong ]
$$ \implies \cfrac{\sqrt{(-1)(1+a)^2+(-b)^2}}{\sqrt{(a-1)^2+(b)^2}} $$
$$ \implies \cfrac{ \sqrt{ 1+a^2-2a+b^2 } }{ \sqrt{ 1+a^2-2a+b^2 } } $$ 

If my solution is wrong, how can it be corrected?
 EDIT:
 one error I identified when opening the identities in numerator and denominator: 
 Numerator: $ \bigl((-1)^2 + a^2 -2(-1)(a) \bigr) +b^2 = 1+a^2+2a+b^2$
 Denominator: $ \bigl( (a)^2 + (-1)^2 + 2(-1)(a) \bigr) + b^2 = 1+a^2 -2a +b^2 $
 I don't understand why this way can't work.
 A: In $\mathbb{C}$ if $|\beta|=1$ then $\beta$ can be anywhere on the unit circle centred at $0$. In particular,
$$\vert{\beta}\vert=1 \iff \beta\in \{e^{it}:t\in[0,2\pi)\}$$
A general trick when working with $\vert z \vert$ is to use the fact that $\vert z \vert^2=z \bar{z}$. So:
$$\begin{align}
\left\lvert \cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right\rvert^2 
&= \left(\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right) \overline{\left(\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right)}
\\ &= \left(\cfrac{\beta - \alpha}{1- \bar \alpha \beta}\right) \left(\cfrac{\bar{\beta} - \bar{\alpha}}{1- \alpha \bar{\beta}}\right)
\end{align}$$
By expanding this bracket and cancelling, you should find your solution!
A: So, you have a couple of things wrong here
First, $\beta \in \mathbb{C}$, so we can't simply assume $\beta \in \{1, -1\}$, what about $\beta \in \{ z \in \mathbb{C}: \lvert z\rvert=1\}$ which contains all of the complex numbers of the form $z = 1*[cos\theta+i*sin\theta]$
As far as the proof, it is quite simple.
You should look up "modulus properties" and "complex number properties"
Things I used:
$\lvert z \rvert = \lvert \bar{z} \rvert$
$\lvert z \rvert \lvert w \rvert = \lvert zw \rvert$
$z*\bar{z} = \lvert z \rvert ^2$
Let me get you started.
We know $\lvert \beta \rvert = 1$, so we can also know that $\lvert \bar{\beta} \rvert = 1$ by property 1.
We use this to take:
$ \dfrac{\lvert\beta-\alpha\rvert}{\lvert1-\bar{\alpha}\beta\rvert}*1 = \dfrac{\lvert\beta-\alpha\rvert}{\lvert1-\bar{\alpha}\beta\rvert}*\lvert\bar{\beta}\rvert $
Then we use property 2 to get $\lvert \bar{\beta} \rvert$ into our numerator.
$\dfrac{\lvert\beta\bar{\beta}-\alpha\bar{\beta}\rvert}{\lvert1-\bar{\alpha}\beta\rvert}$
Finally, it is up to you to find why this is $1$
