Question

What is ${\left[\frac{\sin \frac{\pi }{6}+i(1-\cos \frac{\pi }{6})}{\sin \frac{\pi }{6}-i(1-\cos \frac{\pi }{6})}\right]}^3$ where $i=\sqrt{-1}$, equal to?

• A. $1$
• B. $-1$
• C. $i$
• D. $-i$

Answer 1

Answer: (D)

$-i$

${\left[\frac{sin\frac{\pi }{6}+i(1-cos(\frac{\pi }{6}\left)\right)}{sin\frac{\pi }{6}-i(1-cos(\frac{\pi }{6}\left)\right)}\right]}^3={\left[\frac{\frac{1}{2}+i(1-\frac{\sqrt{3}}{2})}{\frac{1}{2}-i(1-\frac{\sqrt{3}}{2})}\right]}^3$

$={[\frac{\frac{1}{2}+i(1-\frac{\sqrt{3}}{2})}{\frac{1}{2}-i(1-\frac{\sqrt{3}}{2})}\times \frac{\frac{1}{2}-i(1-\frac{\sqrt{3}}{2})}{\frac{1}{2}-i(1-\frac{\sqrt{3}}{2})}]}^3$

$={\left[\frac{\frac{1}{4}+{(1-\frac{\sqrt{3}}{2})}^2}{\frac{1}{4}-{(1-\frac{\sqrt{3}}{2})}^2-i(1-\frac{\sqrt{3}}{2})}\right]}^3$

$={\left[\frac{2}{\sqrt{3}-i}\right]}^3$

$={\left(\frac{-1}{i}\right)}^3$

$=-i$