Let S be the set of all $(\mathrm{\alpha }, \mathrm{\beta }), \mathrm{\pi }<\mathrm{\alpha }, \mathrm{\beta }<2\mathrm{\pi }$, for which the complex number $\frac{1-\mathrm{isin}\mathrm{\alpha }}{1+2\mathrm{isin}\mathrm{\alpha }}$ is purely imaginary and $\frac{1+\mathrm{icos}\mathrm{\beta }}{1-2\mathrm{icos}\mathrm{\beta }}$ is purely real. Let ${\mathrm{Z}}_{\mathrm{\alpha \beta }}=\sin 2\mathrm{\alpha }+\mathrm{icos}2\mathrm{\beta }, (\mathrm{\alpha }, \mathrm{\beta })\in \mathrm{S}.$

Then $\sum _{(\mathrm{\alpha }, \mathrm{\beta })\in \mathrm{S}} \left({\mathrm{iZ}}_{\mathrm{\alpha \beta }}+\frac{1}{\mathrm{i}{\overline{\mathrm{Z}}}_{\mathrm{\alpha \beta }}}\right)$ is equal to :

$\begin{array}{l}\mathrm{\pi }<\mathrm{\alpha }, \mathrm{\beta }<2\mathrm{\pi }\\ \frac{1-\mathrm{isin}\mathrm{\alpha }}{1+\mathrm{i}(2\sin \mathrm{\alpha })}=\text{ Purely imaginary }\\ \Rightarrow \frac{(1-\mathrm{isin}\mathrm{\alpha })(1-\mathrm{i}(2\sin \mathrm{\alpha }\left)\right)}{1+4{\sin }^2\mathrm{\alpha }}=\text{ Purely imaginary }\\ \Rightarrow \frac{1-2{\sin }^2\mathrm{\alpha }}{1+4{\sin }^2\mathrm{\alpha }}=0\\ \Rightarrow {\sin }^2\mathrm{\alpha }=\frac{1}{2}\\ \Rightarrow \mathrm{\alpha }=\left\{\frac{5\mathrm{\pi }}{4}, \frac{7\mathrm{\pi }}{4}\right\}\\ \& \frac{1+\mathrm{icos}\mathrm{\beta }}{1+\mathrm{i}(-2\cos \mathrm{\beta })}=\text{ Purely real }\\ \Rightarrow \frac{(1+\mathrm{icos}\mathrm{\beta })(1+2\mathrm{icos}\mathrm{\beta })}{1+4{\cos }^2\mathrm{\beta }}=\text{ Purely real }\\ \Rightarrow 3\cos \mathrm{\beta }=0\\ \Rightarrow \mathrm{\beta }=\frac{3\mathrm{\pi }}{2}\\ \Rightarrow {\mathrm{Z}}_{\mathrm{\alpha \beta }}=\sin \frac{5\mathrm{\pi }}{2}+\mathrm{icos}3\mathrm{\pi }=1-\mathrm{i}\end{array}$

or

$\begin{array}{l}{\mathrm{Z}}_{\mathrm{\alpha \beta }}=\sin \frac{7\mathrm{\pi }}{2}+\mathrm{icos}3\mathrm{\pi }=-1-\mathrm{i}\\ \text{ Required value }=\left[\mathrm{i}(1-\mathrm{i})+\frac{1}{\mathrm{i}(1+\mathrm{i})}\right]+\left[\mathrm{i}(-1-\mathrm{i})+\frac{1}{\mathrm{i}(-1+\mathrm{i})}\right]\\ =\mathrm{i}(-2\mathrm{i})+\frac{1}{\mathrm{i}}\frac{2\mathrm{i}}{(-2)}\Rightarrow 2-1=1\end{array}$