Mathematics - Complex Number Question with Solution | TestHub

Given $\pi <\alpha ,\beta <2\pi$

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

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

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

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

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

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

$\Rightarrow \frac{\left(1+i\cos \beta \right)\left(1+2i\cos \beta \right)}{1+4{\cos }^2\beta }$ is purely real

$\Rightarrow 3\cos \beta =0$

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

Now $Z_{\alpha \beta }=\sin 2\alpha +i\cos 2\beta$

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

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

$\sum _{\left(\alpha ,\beta \right)\in S}\left(i{Z}_{\alpha \beta }+\frac{1}{i{\overline{Z}}_{\alpha \beta }}\right)$$=\left[i\left(1-i\right)+\frac{1}{i\left(1+i\right)}\right]+\left[i\left(-1-i\right)+\frac{1}{i\left(-1+i\right)}\right]$

$=\left[i+1-\frac{i\left(1-i\right)}{2}\right]+\left[-i+1+\frac{i\left(1+i\right)}{2}\right]$

$=2-1=1$