Question

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

• A. $\frac{1}{2}$
• B. 1
• C. 2
• D. none of these

Answer 1

Answer: (B)

1

Let $\alpha =x+iy\beta =p+iq$

$\left|\alpha \right|=\sqrt{x^2+y^2}=1$

$\left|\beta \right|=\sqrt{p^2+q^2}=1$

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\left|\frac{p+iq-(x+iy)}{1-(x-iy)(p+iq)}\right|$

$\Rightarrow \frac{\left|\right(p-x)+i(q-y\left)\right|}{|1-(px+qy+i(xq-py)\left)\right|}$

$\Rightarrow \frac{\sqrt{(p-x{)}^2+(q-y{)}^2}}{\sqrt{(px+qy-1{)}^2+(xq-py{)}^2}}$

$\Rightarrow \frac{\sqrt{p^2+q^2+x^2+y^2-2px-2qy}}{\sqrt{p^2{x}^2+q^2{y}^2+1-2px-2qy+2pqxy+x^2{q}^2+p^2{y}^2-2pqxy}}$

$\Rightarrow \sqrt{\frac{1+1-2px-2qy}{x^2(p^2+q^2)+y^2(p^2+q^2)+1-2px-2qy}}$

$=\sqrt{\frac{2[1-px-qy]}{x^2+y^2+1-2px-2qy}}$

$\Rightarrow \sqrt{\frac{2(1-px-qy)}{(2-2px-2qy)}}=\sqrt{\frac{2(1-px-qy)}{2(1-px-qy)}}$

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

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