Let a= Im ( 1+z^2 2 i z ), where z is any non-zero complex number. The set A=a:|z|=1 and z ≠ ± 1 is equal to:

Let $z=x+i y \Rightarrow z^2=x^2-y^2+2 i x y$z=x+iy⇒z2=x2−y2+2ixy Now, $$\begin{aligned} & \frac{1+z^2}{2 i z}=\frac{1+x^2-y^2+2 i x y}{2 i(x+i y)}=\frac{\left(x^2-y^2+1\right)+2 i x y}{2 i x-2 y} \\ & =\frac{\left(x^2-y^2+1\right)+2 i x y}{-2 y+2 i x} \times \frac{-2 y-2 i x}{-2 y-2 i x} \\ & =\frac{y\left(x^2+y^2-1\right)+x\left(x^2+y^2+1\right) i}{2\left(x^2+y^2\right)} \\ & \end{aligned}$$​2iz1+z2​=2i(x+iy)1+x2−y2+2ixy​=2ix−2y(x2−y2+1)+2ixy​=−2y+2ix(x2−y2+1)+2ixy​×−2y−2ix−2y−2ix​=2(x2+y2)y(x2+y2−1)+x(x2+y2+1)i​​ $$a=\frac{x\left(x^2+y^2+1\right)}{2\left(x^2+y^2\right)}$$a=2(x2+y2)x(x2+y2+1)​ Since, $|z|=1 \Rightarrow \sqrt{x^2+y^2}=1$∣z∣=1⇒x2+y2​=1 $$\begin{aligned} & \Rightarrow x^2+y^2=1 \\ & \therefore a=\frac{x(1+1)}{2 \times 1}=x \\ & \text { Also } z \neq 1 \Rightarrow x+i y \neq 1 \\ & \therefore \mathrm{A}=(-1,1) \end{aligned}$$​⇒x2+y2=1∴a=2×1x(1+1)​=x Also z=1⇒x+iy=1∴A=(−1,1)​