Question

If $z=x+iy, \, \forall x,y\in R, \, i^{2}=-1, \, xy\neq 0$ and $\left|z\right|=2,$ then the imaginary part of $\frac{z + 2}{z - 2}$ cannot be

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

Given, $\left|z\right|=2\Rightarrow x^{2}+y^{2}=4\&xy\neq 0$
Let, $z=x+iy=2\left(cos \theta + i sin ⁡ \theta \right)$
So, $\frac{z + 2}{z - 2}=\frac{\left(x + 2\right) + i y}{\left(x - 2\right) + i y}$
$=\frac{\left(x^{2}+y^{2}-4\right)+i(y(x-2)-y(x+2))}{(x-2)^{2}+y^{2}}=\frac{-4 y}{8-4 x} i$
So, the imaginary part $=-\frac{4 y}{8 - 4 x}=\frac{y}{x - 2}$
$=\frac{- 2 sin \theta }{2 - 2 cos ⁡ \theta }=-cot⁡\frac{\theta }{2},$ $\theta \in \left(0 , 2 \pi \right)$
Now, $\theta \neq \frac{\pi }{2},\pi ,\frac{3 \pi }{2}\Rightarrow cot \frac{\theta }{2}\neq -1,0,1$