Let $a=\operatorname{Im}\left(\frac{1+z^2}{2 i z}\right)$, where $z$ is any non-zero complex number. The set…
Let $a=\operatorname{Im}\left(\frac{1+z^2}{2 i z}\right)$, where $z$ is any non-zero complex number. The set $\mathrm{A}=\{a:|z|=1$ and $z \neq \pm 1\}$ is equal to:
$(-1,1)$
$[-1,1]$
$[0,1)$
$(-1,0]$
Solution
Let $z=x+i y \Rightarrow z^2=x^2-y^2+2 i x y$ Now,
$
\begin{aligned}
& \frac{1+z^2}{2 i z}=\frac{1+x^2-y^2+2 i x y}{2 i(x+i y)}=\frac{\left(x^2-y^2+1\right)+2 i x y}{2 i x-2 y} \\
& =\frac{\left(x^2-y^2+1\right)+2 i x y}{-2 y+2 i x} \times \frac{-2 y-2 i x}{-2 y-2 i x} \\
& =\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)} \\
&
\end{aligned}
$
$
a=\frac{x\left(x^2+y^2+1\right)}{2\left(x^2+y^2\right)}
$
Since, $|z|=1 \Rightarrow \sqrt{x^2+y^2}=1$
$
\begin{aligned}
& \Rightarrow x^2+y^2=1 \\
& \therefore a=\frac{x(1+1)}{2 \times 1}=x \\
& \text { Also } z \neq 1 \Rightarrow x+i y \neq 1 \\
& \therefore \mathrm{A}=(-1,1)
\end{aligned}
$