If $u+i v=\frac{3 i}{x+i y+2}$, then $y=$
If $u+i v=\frac{3 i}{x+i y+2}$, then $y=$
- $\frac{9 u}{u^2+v^2}$
- $\frac{3 u}{u^2+v^2}$
- $\frac{6 u}{u^2+v^2}$
- $\frac{12 u}{u^2+v^2}$
Solution
$
\begin{aligned}
& \text {We have, } u+i v=\frac{3 i}{x+i y+2} \\
& \Rightarrow x+i y+2=\frac{3 i}{u+i v} \\
& \Rightarrow(x+2)+i y=\frac{3 i}{u+i v}+\frac{u-i v}{u-i v}=\frac{3 u i-3 v\left(i^2\right)}{u^2-(i v)^2} \\
& \Rightarrow(x+2)+i y=\frac{3 u i+3 v}{u^2+v^2} \\
& \Rightarrow(x+2)+i y=\frac{3 v}{u^2+v^2}+\frac{3 u}{u^2+v^2} i
\end{aligned}
$
by comparing Imaginary parts
$
y=\frac{3 u}{u^2+v^2}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 1)
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