Question

Let $a=Im\left(\frac{1+z^{2}}{2iz}\right),$ where $z$ is any non-zero complex number.
The set $A=\left\{a:\left|z\right|=1\,and\,z\ne\pm1\right\}$ 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$
$\Rightarrow x^{2}+y^{2}=1$
$\therefore a=\frac{x\left(1+1\right)}{2\times1}=x$
Also $z\ne1$
$ \Rightarrow x+iy\ne1$
$\therefore A=\left(-1,1\right)$