If $\alpha $ and $\beta $ are different complex number with $\left| \alpha \right| = 1$, then what is $\left| {\dfrac{{\alpha - \beta }}{{1 - \alpha \overline \beta }}} \right|$ equal to?
A. $\left| \beta \right|$
B. 2
C. 1
D. 0

Hint: Here we will use the conjugate of the given complex numbers to solve.

Complete step-by-step answer:
Multiplying by $\overline \alpha $ on numerator and denominator, we get
$\left| {\dfrac{{(\alpha - \beta )\overline \alpha }}{{(1 - \alpha \overline \beta )\overline \alpha }}} \right| = \left| {\dfrac{{(\alpha - \beta )\overline \alpha }}{{\overline \alpha - \alpha \overline \alpha \overline \beta }}} \right|$
We know that
$
  z.\overline z = {\left| z \right|^2} \\
  \overline \alpha \alpha = {\left| \alpha \right|^2} = 1 \\
  \left| {\dfrac{{(\alpha - \beta )\overline \alpha }}{{(\overline \alpha - \overline \beta )}}} \right| = \left| {\dfrac{{(\alpha - \beta )}}{{(\overline \alpha - \overline \beta )}}} \right|\left| {\overline \alpha } \right| \\
$
As we know $\left| z \right| = \left| {\overline z } \right|$
Therefore $\left| {\alpha - \beta } \right| = \left| {\overline {\alpha - \beta } } \right|$
So it gets cancel out,
$\left| {\overline \alpha } \right| = \left| \alpha \right| = 1$

Note: For modulus type questions in complex numbers, we have to simplify using conjugate and using property of modulus.