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) $\sqrt { 3 }$
B) $1 + \sqrt { 3 } \mathrm { i }$
C) $\sqrt { 3 } i$
D) $\sqrt { 3 } + i$

Question

Quizplus · Accepted Answer

To solve this, we first convert both the numerator and the 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 $\theta = \tan^{-1}(\frac{2}{2\sqrt{3}}) = \frac{\pi}{6}$. Thus, the numerator in trigonometric form is $4(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6})$.Similarly, the denominator $\sqrt{3} - i$ has $r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2$ and $\theta = \tan^{-1}(\frac{-1}{\sqrt{3}}) = -\frac{\pi}{6}$. So, the denominator in trigonometric form is $2(\cos (-\frac{\pi}{6}) + i\sin (-\frac{\pi}{6}))$.Dividing these in trigonometric form, we get $\frac{4}{2}(\cos(\frac{\pi}{6} - (-\frac{\pi}{6})) + i\sin(\frac{\pi}{6} - (-\frac{\pi}{6}))) = 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$.Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, the quotient in rectangular form is $2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 1 + \sqrt{3}i$.

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 $\frac { 2 \sqrt { 3 } + 2 i } { \sqrt { 3 } - i }$