Let S be the set of all (α, β), π Then ∑ _ (α, β) ∈ S (i Z _ αβ + 1...

Given $\pi < \alpha, \beta < 2\pi$

$\dfrac{1 - isin\alpha}{1 + i\left(2sin\alpha\right)}$ is purely imaginary

$⇒ \dfrac{\left(1 - isin\alpha\right)\left(1 - i\left(2sin\alpha\right)\right)}{1 + 4\sin^{2}\alpha}$ is purely imaginary

$⇒ \dfrac{1 - 2\sin^{2}\alpha}{1 + 4\sin^{2}\alpha} = 0$

$⇒ \sin^{2}\alpha = \dfrac{1}{2}$

$⇒ \alpha = \left\{\dfrac{5\pi}{4}, \dfrac{7\pi}{4}\right\}$

Also $\dfrac{1 + icos\beta}{1 + i\left(- 2cos\beta\right)}$ is purely real

$⇒ \dfrac{\left(1 + icos\beta\right)\left(1 + 2icos\beta\right)}{1 + 4\cos^{2}\beta}$ is purely real

$⇒ 3cos\beta = 0$

$⇒ \beta = \dfrac{3\pi}{2}$

Now ${Z}_{\alpha\beta} = \sin 2\alpha + icos2\beta$

$⇒ {Z}_{\alpha\beta} = sin\dfrac{5\pi}{2} + icos3\pi = 1 - i$

or ${Z}_{\alpha\beta} = sin\dfrac{7\pi}{2} + icos3\pi = - 1 - i$

$\mathop{∑}_{\left(\alpha, \beta\right) \in S}\left(i{Z}_{\alpha\beta} + \dfrac{1}{i\overset{¯}{Z}\alpha\beta}\right)$= \left[i\left(1 - i\right) + \dfrac{1}{i\left(1 + i\right)}\right] + \left[i\left(- 1 - i\right) + \dfrac{1}{i\left(- 1 + i\right)}\right]$</p><p>$= \left[i + 1 - \dfrac{i\left(1 - i\right)}{2}\right] + \left[- i + 1 + \dfrac{i\left(1 + i\right)}{2}\right]$</p><p>$= 2 - 1 = 1