Question

The modulus-amplitude form of $\sqrt{3}+i$, where $i=\sqrt{-1}$ is

• A. $2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$
• B. $2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$
• C. $4(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$
• D. $4(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$

Answer 1

The correct option is A $2(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3})$

Modulus amplitude of form

$(\sqrt{3}+i)$

$z=r(\cos \theta +i\sin \theta )$

$r\cos \theta =\sqrt{3},r\sin \theta =1$

$\to r^2({\cos }^2\theta +{\sin }^2\theta )=4\Rightarrow r^2=4r=2$

$2\cos \theta =\sqrt{3}2\sin \theta =1$

$\to \cos \theta =\frac{\sqrt{3}}{2}\sin \theta =\frac{1}{2}$

$\theta =\frac{\pi }{6}\theta =\frac{\pi }{6}$

$\therefore z=2(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6})$.