Question

If $\frac{|(a+ib){|}^2}{(a-ib)}-\frac{|(a-ib){|}^2}{(a+ib)}=x+iy$, then find the value of $x$.

Answer 1

Now,

$\frac{|(a+ib){|}^2}{(a-ib)}-\frac{|(a-ib){|}^2}{(a+ib)}=x+iy$

or, $\frac{(a+ib).\overline{(a+ib)}}{(a-ib)}-\frac{(a-ib).\overline{(a-ib)}}{(a+ib)}=x+iy$ [$z.\overline{z}=|z{|}^2$ for complex number $z$]

or, $\frac{(a+ib).(a-ib)}{(a-ib)}-\frac{(a-ib).(a+ib)}{(a+ib)}=x+iy$

or, $(a+ib)-(a-ib)=x+iy$

or, $2ib=x+iy$

or, $x+iy=0+i\left(2b\right)$

or, $x=0$.