Question

If $(a+ib)=\frac{1+i}{1-i}$, then prove that $(a^2+b^2)=1$

Answer 1

We have $a+ib=\frac{1+i}{1-i}$

To prove: $a^2+b^2=1$

Consider $a+ib=\frac{1+i}{1-i}$

$\Rightarrow a+ib=\frac{1+i}{1-i}\times \frac{1+i}{1+i}$

$\Rightarrow a+ib=\frac{(1+i{)}^2}{1^2-i^2}$

$\Rightarrow a+ib=\frac{(1+i{)}^2}{1-(-1)}$

$\Rightarrow a+ib=\frac{1+i^2+2i}{1+1}$

$\Rightarrow a+ib=\frac{1-1+2i}{2}$

$\Rightarrow a+ib=\frac{2i}{2}$

$\Rightarrow a+ib=i$

Equating the coefficients of like terms we get

$a=0,b=1$

$\Rightarrow a^2+b^2=0^2+1^2=1$

$\therefore a^2+b^2=1$

Hence proved