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

$\frac{1-i \sin \alpha}{1+i(2 \sin \alpha)}=\text { Purely imaginary }$

$\Rightarrow \frac{(1-i \sin \alpha)(1-i(2 \sin \alpha))}{1+4 \sin ^{2} \alpha}=\text { Purely imaginary }$

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

$\sin ^{2} \alpha=\frac{1}{2}$

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

$\frac{1+i \cos \beta}{1+i(-2 \cos \beta)}=\text { Purely real }$

$\frac{(1+i \cos \beta)(1+2 i \cos \beta)}{1+4 \cos ^{2} \beta}=\text { Purely real }$

$3 \cos \beta=0$

$\beta=\frac{3 \pi}{2}$

$Z_{\alpha \beta}=\sin \frac{5 \pi}{2}+i \cos 3 \pi=1-i$

$Z_{\alpha \beta}=\sin \frac{7 \pi}{2}+i \cos 3 \pi=-1-i$

Required value $=\left[i(1-i)+\frac{1}{i(1+i)}\right]+\left[i(-1-i)+\frac{1}{i(-1+1)}\right]$

$=i(-2 i)+\frac{1}{i} \frac{2 i}{(-2)} \Rightarrow 2-1=1$