Question

If $z$ is a complex number and $\alpha =\frac{z+i}{iz+2}$, where $i=\sqrt{-1}$. Suppose $\alpha$ is a real number then

• A. $|z-\frac{i}{2}|=\frac{3}{2}$
• B. $|z-\frac{1}{2}|=\frac{3}{2}$
• C. $|z-i|=\frac{3}{2}$
• D. $|z-\frac{1}{4}|=\frac{9}{4}$

Answer 1

The correct option is A $|z-\frac{i}{2}|=\frac{3}{2}$

Let $z=x+iy$
$\alpha =\frac{z+i}{iz+2}=\frac{x+i(y+1)}{(2-y)+ix}\Rightarrow \alpha =\frac{x(2-y)-i{x}^2+i(y+1)(2-y)+x(y+1)}{(2-y{)}^2+x^2}$
It is given that $\alpha$ is a real number then the imaginary part will be $=0$
$x^2-(y+1)(2-y)=0\Rightarrow x^2+y^2-y-2=0\Rightarrow x^2+(y-\frac{1}{2}{)}^2=\frac{9}{4}$

$\therefore |z-\frac{i}{2}|=\frac{3}{2}$