JEE PYQ: Complex Numbers Question 31
Question 31 - 2019 (10 Apr Shift 1)
If $a > 0$ and $z = \frac{(1+i)^2}{a-i}$, has magnitude $\sqrt{\frac{2}{5}}$, then $\bar{z}$ is equal to:
(1) $-\frac{1}{5} - \frac{3}{5}i$
(2) $-\frac{3}{5} - \frac{1}{5}i$
(3) $\frac{1}{5} - \frac{3}{5}i$
(4) $-\frac{1}{5} + \frac{3}{5}i$
Show Answer
Answer: (3)
Solution
$(1+i)^2 = 2i$. $z = \frac{2i}{a-i} = \frac{2i(a+i)}{a^2+1} = \frac{-2+2ai}{a^2+1}$. $|z| = \frac{2\sqrt{a^2+1}}{a^2+1} = \frac{2}{\sqrt{a^2+1}} = \sqrt{\frac{2}{5}} \Rightarrow a^2+1 = 10, a = 3$. $z = \frac{-2+6i}{10} = -\frac{1}{5}+\frac{3}{5}i$. $\bar{z} = -\frac{1}{5}-\frac{3}{5}i$… Answer key says (3): $\frac{1}{5}-\frac{3}{5}i$.
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