Question

If $(x+iy)(p+iq)=(x^2+y^2)i$, prove that $x=q,y=p$.

Answer 1

$(x+iy)(p+iq)=(x^2+y^2)i$

$xp+xqi+ypi+yq{i}^2=(x^2+y^2)i$

$(xp-yq)+(xq+yp)i=(x^2+y^2)i$

Comparing real parts, we have

$xp-yq=0$

$x=\frac{yq}{p}..........\left(1\right)$

Compaaring imaginary parts, we have

$xq+yp=x^2+y^2..........\left(2\right)$

Substituting the value of x from ${eq}^n\left(1\right)$ in ${eq}^n\left(2\right)$, we have

$\frac{yq}{p}q+yp={\left(\frac{yq}{p}\right)}^2+y^2$

$\Rightarrow \frac{y{q}^2}{p}+yp=\frac{y^2{q}^2}{p^2}+y^2$

$\Rightarrow \frac{y(q^2+p^2)}{p}=\frac{y^2(q^2+p^2)}{p^2}$

$\Rightarrow \frac{y}{p}=1$

$\Rightarrow y=p$

Substituting the value of y in ${eq}^n\left(1\right)$, we have

$x=\frac{pq}{p}$

$\Rightarrow x=q$

Hence, it is proved that $x=q$ and $y=p$.