Find the trigonometric form of the complex number shown below. $- 6 i$
A) $6 ( \cos 0 + i \sin 0 )$
B) $6 ( \cos \pi + i \sin \pi )$
C) $6 \left( \cos \frac { 3 \pi } { 2 } + i \sin \frac { 3 \pi } { 2 } \right)$
D) $6 \left( \cos \frac { \pi } { 4 } + i \sin \frac { \pi } { 4 } \right)$
E) $6 \left( \cos \frac { \pi } { 2 } + i \sin \frac { \pi } { 2 } \right)$

Question

Quizplus · Accepted Answer

To find the trigonometric form of a complex number, we need to convert it to the form $a+bi$ and then use the formulas $r=\sqrt{a^2+b^2}$, $\cos	heta=\frac{a}{r}$, and $\sin	heta=\frac{b}{r}$ to find the modulus $r$ and argument $	heta$. Here, we have $-6i=0-6i$, so $a=0$ and $b=-6$. Thus, $r=\sqrt{0^2+(-6)^2}=6$ and $\cos	heta=\frac{0}{6}=0$ while $\sin	heta=\frac{-6}{6}=-1$. Therefore, $	heta=\frac{3\pi}{2}$ since $\cos	heta=0$ and $\sin	heta=-1$ in the third quadrant. Hence, the trigonometric form of $-6i$ is $$6\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}ight)=\boxed{	extbf{(C)}}.$$

Find the Trigonometric Form of the Complex Number Shown Below −6i- 6 i−6i

Find the Trigonometric Form of the Complex Number Shown Below $- 6 i$