Mathematics · 2019
JEE Main · 10 April 2019, Shift 1 · Q62
If a >0 and z=((1+i)^2)/(a -i), has magnitude √(2/5), then z̄ is equal to:
If $\displaystyle \mathrm{a}>0$ and $\displaystyle z=\frac{(1+i)^2}{\mathrm{a}-i}$, has magnitude
$\displaystyle \sqrt{\frac{2}{5}}$, then $\displaystyle \bar{z}$ is equal to :
Official answer
(1)
$\displaystyle -\frac{1}{5}-\frac{3}{5} i$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.