Question

If$\alpha$ and $\beta$ are different complex numbers with
$\left|\beta \right|=1then\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|$ is equal to

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

Answer: (C)

1

${\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|}^2=\left(\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right)\cdot \left(\frac{\overline{\beta -\alpha }}{1-\overline{\alpha }\beta }\right)$

$=\left(\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right)\left(\frac{\overline{\beta }-\overline{\alpha }}{1-\alpha \overline{\beta }}\right)$

$=\frac{\beta \overline{\beta }-\beta \overline{\alpha }-\alpha \overline{\beta }+\alpha \overline{\alpha }}{1-\alpha \overline{\beta }-\overline{\alpha }\beta +\alpha \overline{\alpha }\beta \overline{\beta }}$

$=\frac{{\left|\beta \right|}^2-\alpha \overline{\beta }-\overline{\alpha }\beta +{\left|\alpha \right|}^2}{1-\alpha \overline{\beta }-\overline{\alpha }\beta +{\left|\alpha \right|}^2{\left|\beta \right|}^2}$

$=\frac{1-\alpha \overline{\beta }-\overline{\alpha }\beta +{\left|\alpha \right|}^2}{1-\alpha \overline{\beta }-\overline{\alpha }\beta +{\left|\alpha \right|}^2}$

${\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|}^2=1$

$Hence,optionCiscorrectanswer.$