Question

If α and β are different complex numbers with |β|=1, then find \(| \frac{β-α}{1-\bar{α}β} |.\)

Answer 1

Let α=a+ib and β²=x+iy

It is given that, |β|=1

\(∴\sqrt{x^2+y^2=1}\)

\(⇒x^2+y^2=1....(i)\)

\(=|\frac{β-α}{1-\bar{α}β}|=|\frac{(x_iy)-(a+ib)}{1-(a-ib)(x+iy)}|\)

\(=|\frac{(x-a)+i(y-b)}{1-(ax+aiy-ibx+by)}|\)

\(=|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}|\)

\(=|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}|\)  \([|\frac{z_1}{z_2}|=|\frac{z_1}{z_2}|]\)

\(=\sqrt\frac{(x-a)^2+i(y-b)^2}{(1-ax-by)^2+i(bx-ay)^2}\)

\(=\frac{\sqrt x^2+a^2-2ax+y^2+b^2-2by}{\sqrt 1+a^2x^2+b^2y^2-ax+2abxy-2by+b^2x^2+a^2y^2-2abxy}\)

\(\frac{\sqrt (x^2+y^2)+a^2+b^2-2ax-2by}{\sqrt 1+a^2(x^2+y^2)+b^2(y^2+x^2)-2ax-2by}\)

\(=\frac{\sqrt 1+a^2+b^2-2ax-2by}{\sqrt1+a^2+b^2-2ax-2by}\)      \([Using\,(1)]\)

\(∴|\frac{β-α}{1-\bar{α}β}|=1\)