Question

\(If a + ib =\frac{(x+1)^2}{2x^2+1},\,prove\, that\, a^2 + b^2 =\frac{(x^2+1)^2}{(2x^2+1)}\)

Answer 1

\(If a + ib =\frac{(x+1)^2}{2x^2+1}\)

\(=\frac{x^2+i^2+2xi}{2x^2+1}\)

\(=\frac{x^2-1+i2x}{2x^2+1}\)

\(=\frac{x^2-1}{2x^2+1}+1(\frac{2x}{2x^2+1})\)

on comparing real and imaginary parts, we obtain

\(a=\frac{x^2-1}{2x^2+1}\,and\,\,b=\frac{2x}{2x^2+1}\)

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

\(=\frac{x^4+1-2x^2+4x^2}{(2x+1)^2}\)

\(\frac{x^2+1+2x^2}{(2x^2+1)^2}\)

\(=\frac{(x^2+1)^2}{(2x^2+1)^2}\)

\(=a^2+b^2=\frac{(x^2+1)^2}{(2x+1)^2}\)

Hence, proved.