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Mathematics Question on Complex Numbers and Quadratic Equations

Ifa+ib=(x+1)22x2+1, prove that a2+b2=(x2+1)2(2x2+1)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

Ifa+ib=(x+1)22x2+1If a + ib =\frac{(x+1)^2}{2x^2+1}

=x2+i2+2xi2x2+1=\frac{x^2+i^2+2xi}{2x^2+1}

=x2−1+i2x2x2+1=\frac{x^2-1+i2x}{2x^2+1}

=x2−12x2+1+1(2x2x2+1)=\frac{x^2-1}{2x^2+1}+1(\frac{2x}{2x^2+1})

on comparing real and imaginary parts, we obtain

a=x2−12x2+1 and  b=2x2x2+1a=\frac{x^2-1}{2x^2+1}\,and\,\,b=\frac{2x}{2x^2+1}

a2+b2=(x2−12x2+1)+(2x2x2+1)2a^2+b^2=(\frac{x^2-1}{2x^2+1})+(\frac{2x}{2x^2+1})^2

=x4+1−2x2+4x2(2x+1)2=\frac{x^4+1-2x^2+4x^2}{(2x+1)^2}

x2+1+2x2(2x2+1)2\frac{x^2+1+2x^2}{(2x^2+1)^2}

=(x2+1)2(2x2+1)2=\frac{(x^2+1)^2}{(2x^2+1)^2}

=a2+b2=(x2+1)2(2x+1)2=a^2+b^2=\frac{(x^2+1)^2}{(2x+1)^2}

Hence, proved.