Question

Let $\alpha ,\beta ,\gamma$ and $a,b,c$ are the complex numbers such that $\frac{\alpha }{a}+\frac{\beta }{b}+\frac{\gamma }{c}=1+i$ and $\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=0.$ If $\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}+\frac{{\gamma }^2}{c^2}=p+iq$, where $p,q\in R$, then the value of $p+q$ is

Answer 1

Given : $\frac{\alpha }{a}+\frac{\beta }{b}+\frac{\gamma }{c}=1+i$ and $\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=0$
Now,
$\frac{\alpha }{a}+\frac{\beta }{b}+\frac{\gamma }{c}=1+i$
Squaring on both sides, we get
${(\frac{\alpha }{a}+\frac{\beta }{b}+\frac{\gamma }{c})}^2=(1+i{)}^2\Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}+\frac{{\gamma }^2}{c^2}+2\frac{\alpha }{a}\cdot \frac{\beta }{b}+2\frac{\beta }{b}\cdot \frac{\gamma }{c}+2\frac{\gamma }{c}\cdot \frac{\alpha }{a}=2i\Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}+\frac{{\gamma }^2}{c^2}+2\frac{\alpha \beta \gamma }{abc}(\frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma })=2i\Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}+\frac{{\gamma }^2}{c^2}=2i=0+2i\Rightarrow p+qi=0+2i\Rightarrow p=0,q=2$
So,
$p+q=0+2=2$