Question

Let $a=Im\left(\frac{1+z^2}{2iz}\right),$ where $z$ is any non-zero complex number.
The set $A=\left\{a:\left|z\right|=1andz\ne \pm 1\right\}$ is equal to:

• A. $(-1,1)$
• B. $[-1,1]$
• C. $[0,1)$
• D. $(-1,0]$

Answer 1

Answer: (A)

Let $z=x+iy\Rightarrow z^2=x^2-y^2+2ixy$
Now,
$\frac{1+z^2}{2iz}=\frac{1+x^2-y^2+2ixy}{2i\left(x+iy\right)}=\frac{\left(x^2-y^2+1\right)+2ixy}{2ix-2y}$
$=\frac{\left(x^2-y^2+1\right)+2ixy}{-2y+2ix}\times \frac{-2y-2ix}{-2y-2ix}$
$=\frac{y\left(x^2-y^2-1\right)+x\left(x^2+y^2+1\right)i}{2\left(x^2+y^2\right)}$
$a=\frac{x\left(x^2+y^2+1\right)}{2\left(x^2+y^2\right)}$
Since, $\left|z\right|=1$
$\Rightarrow \sqrt{x^2+y^2}=1$
$\Rightarrow x^2+y^2=1$
$\therefore a=\frac{x\left(1+1\right)}{2\times 1}=x$
Also $z\ne 1$
$\Rightarrow x+iy\ne 1$
$\therefore A=\left(-1,1\right)$