If the imaginary part of the expression  $\frac{z-1}{e^{i\theta }}+\frac{e^{i\theta }}{z-1}$  be zero, then locus of z is a circle of radius-------

Let u $=\frac{\mathrm{z}-1}{{\mathrm{e}}^{\mathrm{\theta i}}}\Rightarrow \frac{{\mathrm{e}}^{\mathrm{\theta i}}}{\mathrm{z}-1}=\frac{1}{\mathrm{u}}$
Given that imaginary part of u  $+\frac{1}{\mathrm{u}}$ is zero.
$\begin{array}{l}\left(\mathrm{u}+\frac{1}{\mathrm{u}}\right)-\left(\overline{\mathrm{u}}+\frac{1}{\overline{\mathrm{u}}}\right)=0\\ (\mathrm{u}-\overline{\mathrm{u}})+\frac{\overline{\mathrm{u}}-\mathrm{u}}{\mathrm{u}\overline{\mathrm{u}}}=0\end{array}$
$(\mathrm{u}-\overline{\mathrm{u}})\left(1-\frac{1}{\mathrm{u}\overline{\mathrm{u}}}\right)=0\Rightarrow 1-\frac{1}{|\mathrm{u}{|}^2}=0\Rightarrow |\mathrm{u}{|}^2=1$
$\Rightarrow {\left|\frac{\mathrm{z}-1}{{\mathrm{e}}^{\mathrm{i\theta }}}\right|}^2=1\Rightarrow |\mathrm{z}-1|=1$ which is a circle.