Convert from rectangular to trigonometric form. (Choose an argument $\theta$ such that $0 \leq \theta

Question

Convert from rectangular to trigonometric form. (Choose an argument $\theta$ such that $0 \leq \theta < 2 \pi$ .) $- \frac { 1 } { 2 } + \frac { 1 } { 2 } \sqrt { 3 } i$
A) $\frac { 1 } { 2 } \left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$
B) $\frac { 1 } { 2 } \left( \sin \frac { 2 \pi } { 3 } + i \cos \frac { 2 \pi } { 3 } \right)$
C) $\left( \cos \frac { 2 \pi } { 3 } - i \sin \frac { 2 \pi } { 3 } \right)$
D) $\left( \sin \frac { 2 \pi } { 3 } - i \cos \frac { 2 \pi } { 3 } \right)$
E) $\left( \cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } \right)$

Quizplus · Accepted Answer

Let $z=-\frac{1}{2}+\frac{1}{2}\sqrt{3}i=r\cos{	heta}+r\sin{	heta}i$. 
We can determine $r$ by finding the magnitude of $z$: 
$|z|=\sqrt{\left(-\frac{1}{2}ight)^2+\left(\frac{1}{2}\sqrt{3}ight)^2} = 1$. 
Since $z$ lies in the second quadrant, we choose $	heta=\frac{2\pi}{3}$. 
Thus, we have $-\frac{1}{2}+\frac{1}{2}\sqrt{3}i = \cos{\frac{2\pi}{3}}+\sin{\frac{2\pi}{3}}i=\boxed{	extbf{(E) }\left(\cos \frac { 2 \pi } { 3 } + i \sin \frac { 2 \pi } { 3 } ight)}.$

Convert from Rectangular to Trigonometric Form θ\thetaθ Such That 0≤θ<2π0 \leq \theta < 2 \pi0≤θ<2π

Convert from Rectangular to Trigonometric Form $\theta$ Such That $0 \leq \theta < 2 \pi$