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

The Correct Option is A

The correct answer is (A) : $( - 1, 1)$
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 \(z\ne1\)
\(\Rightarrow x+iy\ne1\)
\(\therefore A=\left(-1,1\right)\)