JEE Main 2019 (Online) 10th April Morning ShiftQuestion 1
If $a>0$ and $z=\frac{(1+i)^{2}}{a-i}$ has magnitude $\sqrt{\frac25}$, then $\bar z$ is equal to :
Complex Numbers
Solution
$|z|=\frac{2}{\sqrt{a^2+1}}=\sqrt{\frac25}$ gives $a=3$. $z=\frac{2i}{3-i}=\frac{2i(3+i)}{10}=-\frac15+\frac35i$, so $\bar z=-\frac15-\frac35i$.