Question

If $a,b,c,p,q,r$ are thre non-zero complex numbers such that $\frac{p}{a}+\frac{q}{b}+\frac{r}{c}=1+i$ and $\frac{a}{p}+\frac{b}{q}+\frac{c}{r}=0,$ then value of $\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}$ is

• A. $0$
• B. $-1$
• C. $2i$
• D. $-2i$

Answer 1

The correct option is C $2i$

We have, ${(1+i)}^2={(\frac{p}{a}+\frac{q}{b}+\frac{r}{c})}^2$

$\Rightarrow 1-1+2i=\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}+2(\frac{qr}{bc}+\frac{rp}{ca}+\frac{pq}{ab})$

$\Rightarrow 2i=\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}+\frac{2abc}{pqr}(\frac{a}{p}+\frac{b}{q}+\frac{c}{r})$

$=\frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}+\frac{2abc}{pqr}\left(0\right)$

$\Rightarrow \frac{p^2}{a^2}+\frac{q^2}{b^2}+\frac{r^2}{c^2}=2i$