• # question_answer If $(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i$, then A) $p=x,q=y$ B) $p={{x}^{2}},\,\,q={{y}^{2}}$ C) $x=q,y=p$ D) None of these

$(x+iy)(p+iq)=({{x}^{2}}+{{y}^{2}})i$ Þ   $(xp-yq)+i(xq+yp)=({{x}^{2}}+{{y}^{2}})i$ Þ   $xp-yq=0,xq+yp={{x}^{2}}+{{y}^{2}}$ Þ  $\frac{x}{q}=\frac{y}{p}$and$xq+yp={{x}^{2}}+{{y}^{2}}$ Let $\frac{x}{q}=\frac{y}{p}=\lambda$. then $x=\lambda q,y=\lambda p$ \$xq+yp={{x}^{2}}+{{y}^{2}}$Þ $\lambda ={{\lambda }^{2}}$Þ $\lambda =1$ \ $x=q,y=p$.