Question
What is the argument of $$\left( {1 - \sin \theta } \right) + i\,\cos \theta \, ?$$
A.
$$\frac{\pi }{2} - \frac{\theta }{2}$$
B.
$$\frac{\pi }{2} + \frac{\theta }{2}$$
C.
$$\frac{\pi }{4} - \frac{\theta }{2}$$
D.
$$\frac{\pi }{4} + \frac{\theta }{2}$$
Answer :
$$\frac{\pi }{4} + \frac{\theta }{2}$$
Solution :
Given complex number is
$$\eqalign{
& \left( {1 - \sin \theta } \right) + i\,\cos \theta \equiv a + ib \cr
& {\text{Argument}} \equiv {\text{tan}}\,\theta = \frac{b}{a} \cr
& \Rightarrow \,\tan \theta = \frac{{\cos \theta }}{{1 - \sin \theta }} \cr
& = \,\frac{{{{\cos }^2}\frac{\theta }{2} - {{\sin }^2}\frac{\theta }{2}}}{{{{\sin }^2}\frac{\theta }{2} + {{\cos }^2}\frac{\theta }{2} - 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}} \cr
& = \,\frac{{\left( {\cos \frac{\theta }{2} - \sin \frac{\theta }{2}} \right)\left( {\cos \frac{\theta }{2} + \sin \frac{\theta }{2}} \right)}}{{{{\left( {\sin \frac{\theta }{2} - \cos \frac{\theta }{2}} \right)}^2}}} \cr
& = \,\frac{{\cos \frac{\theta }{2} + \sin \frac{\theta }{2}}}{{\cos \frac{\theta }{2} - \sin \frac{\theta }{2}}} \cr
& = \,\frac{{1 + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\theta }{2}}} = \frac{{\tan \frac{\pi }{4} + \tan \frac{\theta }{2}}}{{1 - \tan \frac{\pi }{4}\tan \frac{\theta }{2}}} \cr
& \tan \,\theta = \tan \left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) \cr
& {\text{Hence, argument}} = \frac{\pi }{4} + \frac{\theta }{2} \cr} $$