Find the quotient and write in rectangular form. First convert the numerator and denominator to trigonometric form.
-$\frac { 2 \sqrt { 3 } + 2 i } { \sqrt { 3 } - i }$
A) $\sqrt { 3 } \mathrm { i }$
B) $\sqrt { 3 } + i$
C) $1 + \sqrt { 3 } i$
D) $\sqrt { 3 }$

Question

Quizplus · Accepted Answer

To solve this, we first convert both the numerator and denominator into trigonometric form. The numerator $2\sqrt{3} + 2i$ can be represented in polar form as $r(\cos\theta + i\sin\theta)$, where $r = \sqrt{(2\sqrt{3})^2 + (2)^2} = 4$ and $\theta = \tan^{-1}\left(\frac{2}{2\sqrt{3}}\right) = \frac{\pi}{6}$. Thus, the numerator in trigonometric form is $4(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})$.The denominator $\sqrt{3} - i$ can also be converted similarly. Its magnitude is $r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2$, and its angle $\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}$. So, the denominator in trigonometric form is $2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}))$.Dividing these two gives us $\frac{4}{2}(\cos(\frac{\pi}{6} - (-\frac{\pi}{6})) + i\sin(\frac{\pi}{6} - (-\frac{\pi}{6}))) = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, the quotient in rectangular form is $2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 + \sqrt{3}i$.

Find the Quotient and Write in Rectangular Form 23+2i3−i\frac { 2 \sqrt { 3 } + 2 i } { \sqrt { 3 } - i }3​−i23​+2i​

Find the Quotient and Write in Rectangular Form $\frac { 2 \sqrt { 3 } + 2 i } { \sqrt { 3 } - i }$