Question

Let $z=1-\sin \alpha +i\cos \alpha$ where $a\in (0,\frac{\pi }{2})$, then the modulus and the prinicipal value of the argument of $z$ are respectively:

• A. $\sqrt{2(1-\sin \alpha )},(\frac{\pi }{4}+\frac{\alpha }{2})$
• B. $\sqrt{2(1-\sin \alpha )},(\frac{\pi }{4}-\frac{\alpha }{2})$
• C. $\sqrt{2(1+\sin \alpha )},(\frac{\pi }{4}+\frac{\alpha }{2})$
• D. $\sqrt{2(1+\sin \alpha )},(\frac{\pi }{4}-\frac{\alpha }{2})$

Answer 1

The correct option is A $\sqrt{2(1-\sin \alpha )},(\frac{\pi }{4}+\frac{\alpha }{2})$

We have $z=(1-\sin \alpha )+i\cos \alpha$

$\Rightarrow |z|=\sqrt{(1-\sin \alpha {)}^2+{\cos }^2\alpha }$

$\Rightarrow |z|=\sqrt{1+{\sin }^2\alpha -2\sin \alpha +{\cos }^2\alpha }$

$\Rightarrow |z|=\sqrt{2-2\sin \alpha }$

$\Rightarrow |z|=\sqrt{2(1-\sin \alpha )}$

$\Rightarrow arg\left(z\right)={\tan }^{-1}\left(\frac{\cos \alpha }{1-\sin \alpha }\right)$

$\Rightarrow \cos \alpha =\frac{1-{\tan }^2\frac{\alpha }{2}}{1+ta{n}^2\frac{\alpha }{2}}$

$\Rightarrow 1-\sin \alpha =1-\frac{2\tan \frac{\alpha }{2}}{1+ta{n}^2\frac{\alpha }{2}}$

$\Rightarrow 1-\sin \alpha =\frac{(1-\tan \frac{\alpha }{2}{)}^2}{1+ta{n}^2\frac{\alpha }{2}}$

Therefore, $\Rightarrow \frac{\cos \alpha }{1-\sin \alpha }=\frac{1-{\tan }^2\frac{\alpha }{2}}{(1-\tan \frac{\alpha }{2}{)}^2}=\frac{1+\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}}$

$\Rightarrow arg\left(z\right)={\tan }^{-1}\left(\frac{1+\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}}\right)$

$\Rightarrow arg\left(z\right)={\tan }^{-1}\left(\tan \right(\frac{\pi }{4}+\frac{\alpha }{2}\left)\right)$

$\Rightarrow arg\left(z\right)=\frac{\pi }{4}+\frac{\alpha }{2}$