$Q.14)$ If $\alpha,\beta$ are complex numbers where $\alpha\ne \beta$ and $|\alpha| = 1$. Prove that $\displaystyle\left|\frac{\alpha\overline\beta-1}{\alpha-\beta}\right|=1$. Image of question.
For $Q.14$ I tried separating the expression into real and imaginary components and then solving it but it does not lead to anywhere. What approach should I take to solve this question? Thanks.
Best Answer
For question (14), $\frac{|\alpha\beta'-1|}{|\alpha-\beta|}$=|$\alpha $| $\frac{|\beta'-\alpha'|}{|\beta-\alpha|}$(since, |$\alpha $|=1),= $\frac{|\beta-\alpha|'}{|\beta-\alpha|} $ =1.(as ${|\beta-\alpha|'} $ =|$\beta-\alpha$|) (I use prime notation instead of bar notation!)