Question

If $\alpha$ and $\beta$ are two different complex numbers such that $|\alpha |=1,|\beta |-1$, then the expression $\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|$ equals :

• A. $1/2$
• B. $1$
• C. $2$
• D. $Noneofthese$

Answer 1

The correct option is B $1$

$\alpha =ai+b$ $|\alpha |=1$

$\beta =ci+d$ $|\beta |=1$

$\left|\frac{\beta -\alpha }{1-\alpha \beta }\right|=?$

$\beta =\sqrt{c^2+d^2}=1\Rightarrow c^2+d^2=1c=0,d=1$ or

$c=1,d=0$

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

$\beta -\alpha =(a-c)+(b-d)i\Rightarrow |(\beta -\alpha )|=\sqrt{(a-c{)}^2+(b-d{)}^2}$

$|1-\overrightarrow{\alpha }.\beta |=\sqrt{1-(\alpha +bd)+i(ad-bc)}$

$=\sqrt{1-(ac+bd{)}^2+(ad+bc{)}^2}$

$\frac{|(\beta -\alpha )|}{|(1-\overrightarrow{\alpha }\beta )|}=\frac{\sqrt{(a-c{)}^2+(b-d{)}^2}}{\sqrt{(1-(ac+bd){)}^2+(ad-bc{)}^2}}$

consider $c=0d=1$

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