Complex Numbers And Quadratic Equations question 702
Question: If $ z=1+i\tan \alpha ( -\pi <\alpha <-\frac{\pi }{2} ), $ then polar form of the complex number z is:
Options:
A) $ \frac{1}{\cos \alpha }(\cos \alpha +i\sin \alpha ) $
B) $ \frac{1}{-\cos \alpha }[\cos (\pi +\alpha )+i\sin (\pi +\alpha ) $
C) $ \frac{1}{\cos \alpha }[\cos (2\pi +\alpha )+i\sin (2\pi +\alpha )] $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ z = 1 + i tan \alpha = r ( cos\theta +i sin\theta ) $
$ \Rightarrow r cos \theta = 1, r sin \theta = tan a \Rightarrow r^{2} = sec^{2}\alpha $
$ \Rightarrow r=| \sec \alpha |=\frac{1}{| \cos \alpha |} $ Since, $ - \pi < \alpha <-\frac{\pi }{2} $
$ \Rightarrow \cos \alpha < 0 \Rightarrow | cos \alpha | =-cos\alpha $
$ \therefore r=\frac{1}{-\cos \alpha }. $ Further, we get $ \cos \theta = - cos \alpha = cos ( \pi + \alpha ) $ Now, $ -\pi <\alpha <-\frac{\pi }{2}\Rightarrow \pi -\pi <\pi +\alpha <\pi -\frac{\pi }{2} $
$ \Rightarrow 0<\pi +\alpha <\frac{\pi }{2} $ [Converted to principal value]
$ \therefore cos \theta = cos ( \pi + a ) \Rightarrow \theta =\pi +\alpha $ Hence $ z = \frac{1}{-\cos \alpha }[ cos ( \pi + \alpha ) + i sin( \pi + \alpha ) ] $
BETA
