Question

A complex number is defined as $z=x+iy$. If $Im\left(\frac{iz-a}{z-i}\right)=|\frac{iz-a}{z-i}|$, where $a\in R^{+}$ then

• A. $z$ is a purely real number
• B. $z$ is a purely complex number
• C. No such complex number is possible
• D. $z$ can be any complex number

Answer 1

The correct option is B $z$ is a purely complex number

Let $\beta$ be a complex number denoted by,
$\beta =\frac{iz-a}{z-i}\Rightarrow \beta =\frac{i(x+iy)-a}{x+iy-i}\Rightarrow \beta =\frac{-(y+a)+ix}{x+i(y-1)}$
$\beta =\frac{-(y+a)+ix}{x+i(y-1)}\times \frac{x-i(y-1)}{x-i(y-1)}$
$\Rightarrow \beta =\frac{-x(1+a)+i\left[\right(y-1\left)\right(y+a)+x^2]}{x^2+(y-1{)}^2}$

If $Im\left(\beta \right)=|\beta |$ then,
$Re\left(\beta \right)=0$
$\Rightarrow Re\left(\beta \right)=-\frac{x(1+a)}{x^2+(y-1{)}^2}=0\Rightarrow x(1+a)=0$
$a\ne -1,\because a\in R^{+}$
$\therefore x=0$

Hence, the complex number is $z=iy$, which is purely complex.