Question
If $$z = 1 + i\,\tan \,\alpha \left( { - \pi < \alpha < - \frac{\pi }{2}} \right),$$ then polar form of the complex number $$z$$ is :
A.
$$\frac{1}{{\cos \alpha }}\left( {\cos \alpha + i\sin \alpha } \right)$$
B.
$$\frac{1}{{ - \cos \,\alpha }}\left[ {\cos \left( {\pi + \alpha } \right) + i\,\sin \left( {\pi + \alpha } \right)} \right]$$
C.
$$\frac{1}{{ \cos \,\alpha }}\left[ {\cos \left( {2\pi + \alpha } \right) + i\,\sin \left( {2\pi + \alpha } \right)} \right]$$
D.
None of these
Answer :
$$\frac{1}{{ - \cos \,\alpha }}\left[ {\cos \left( {\pi + \alpha } \right) + i\,\sin \left( {\pi + \alpha } \right)} \right]$$
Solution :
$$\eqalign{
& z = 1 + i\,\tan \,\alpha = r\left( {\cos \theta + i\,\sin \theta } \right) \cr
& \Rightarrow \,r\,\cos \,\theta = 1,r\,\sin \,\theta = \tan \,\alpha \cr
& \Rightarrow \,{r^2} = {\sec ^2}\alpha \cr
& \Rightarrow \,r = \left| {\sec \,\alpha } \right| = \frac{1}{{\left| {\cos \,\alpha } \right|}} \cr
& {\text{Since}}, - \pi < \alpha < - \frac{\pi }{2} \cr
& \Rightarrow \,\cos \,\alpha \, < 0 \cr
& \Rightarrow \,\left| {\cos \,\alpha } \right| = - \cos \,\alpha \cr
& \therefore \,r = \frac{1}{{ - \cos \,\alpha }}. \cr
& {\text{Further,}}\,\,{\text{we}}\,{\text{get,}} \cr
& \cos \,\theta = - \cos \,\alpha = \cos \left( {\pi + \alpha } \right) \cr
& {\text{Now}},\, - \pi < \alpha < - \frac{\pi }{2} \cr
& \Rightarrow \,\pi - \pi < \pi + \alpha < \pi - \frac{\pi }{2} \cr
& \Rightarrow \,0 < \pi + \alpha < \frac{\pi }{2}\,\,\,\,\,\left[ {{\text{Converted to principal value}}} \right] \cr
& \therefore \,\cos \,\theta = \cos \left( {\pi + \alpha } \right) \cr
& \Rightarrow \,\theta = \pi + \alpha \cr
& {\text{Hence}},\,\,z = \frac{1}{{ - \cos \,\alpha }}\left[ {\cos \left( {\pi + \alpha } \right) + i\,\sin \left( {\pi + \alpha } \right)} \right] \cr} $$