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$\frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} = z$

${\mathop{\rm Im}\nolimits} (z) = 0$

$\theta = ?$

$z = \frac{{1 - i\sin \theta }}{{1 + i\cos \theta }} \times \frac{{1 - i\cos \theta }}{{1 - i\cos \theta }}$

$ = \frac{{1 - i(\cos \theta  + \sin \theta ) + {i^2}\sin \theta \cos \theta }}{{1 - {i^2}{{\cos }^2}\theta }}$

$ = \frac{{1 - \sin \theta \cos \theta  - i(\cos \theta  + \sin \theta )}}{{1 + {{\cos }^2}\theta }}$

$\because {\mathop{\rm Im}\nolimits} (z) = 0,\cos \theta  + \sin \theta  = 0$

$\Rightarrow \tan \theta = -1 $

$\Rightarrow \theta = n\pi - \frac {\pi}{4}$,$n\in I$

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