Mathematics - Complex Number Question with Solution | TestHub

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