Q.25. \text{If } a > 0 \text{ and } z = \frac{(1+i)^2}{a+i}, (i = \sqrt{-1}) has magnitude \frac{2}{\sqrt{5}} then \bar{z} is equal to:
A. -\frac{2}{5} + \frac{4}{5}i
B. \frac{2}{5} - \frac{4}{5}i
C. -\frac{2}{5} - \frac{4}{5}i
D. \frac{2}{5} + \frac{4}{5}i
Answer:- B. \frac{2}{5} - \frac{4}{5}i
Explanation :-
z = \frac{(1+i)^{2}}{a+i} = \frac{2i}{a+i} = \frac{2i(a-i)}{(a+i)(a-i)} = \frac{2+2ai}{a^2+1} \quad....\text{(i)} |z| = \frac{2}{\sqrt{5}} \Rightarrow \frac{4}{(a^2+1)^2} + \frac{4a^2}{(a^2+1)^2} = \frac{4}{5} \Rightarrow 20 + 20a^2 = 4(a^4 + 2a^2 + 1) \Rightarrow 4a^4 - 12a^2 - 16 = 0 \Rightarrow a^4 - 3a^2 - 4 = 0 \Rightarrow (a^2 - 4)(a^2 + 1) = 0 \Rightarrow a^2 = 4 \text{ and } a^2 = -1 \Rightarrow a = 2 \quad\ldots[\because a > 0] \therefore \text{(i)} \Rightarrow z = \frac{2}{5} + \frac{4}{5}i \therefore \bar{z} = \frac{2}{5} - \frac{4}{5}i