Question

If $Z$ is a complex number with $I Z I = 1$ and $Z + \frac{1}{Z} = x + iy$, then $xy =$

• A. $0$
• B. $1$
• C. $2$
• D. $cannot\, be\, found$

Answer 1

Answer: (A)

$0$

Answer 2

We have, $|Z| = 1$ and $Z + \frac{1}{Z} = x + iy$
Let $Z = P + iQ$
$\therefore \:\:\:\: \sqrt{P^2 + Q^2} = 1\:\:\: \Rightarrow \:\: P^2 + Q^2 = 1$
Now $Z + \frac{1}{Z} = x + iy$
$\therefore\:\:\:\: P +iQ + \frac{1}{P +iQ} = x + iy$
$\Rightarrow P +iQ + \frac{P - iQ}{P^{2} +Q^{2}} =x +iy$
$\Rightarrow P+iQ + P - iQ = x + iy \left(\because\:\:\: P^{2} +Q^{2} = 1\right)$
$\Rightarrow 2P = x + iy$
Equating coefficients of real and imaginary parts, we get
$x = 2P$ and $y = 0$
$\therefore \:\: xy = 0$