Question

If $\omega$ is a complex cube root of unity, evaluate $\left(\frac{\sqrt{3}-i}{-\sqrt{2}+\sqrt{2}i}\right)^{20}$.}

• A. $\omega$
• B. $\omega^2$
• C. $-\omega$
• D. $\omega-\omega^2$

Hint:

Convert complex numbers into polar form for powers.

Answer 1

The Correct Option is B

Step 1: Simplify numerator: \[ \sqrt{3}-i = 2\left(\cos\frac{\pi}{6}-i\sin\frac{\pi}{6}\right)=2e^{-i\pi/6} \]
Step 2: Simplify denominator: \[ -\sqrt{2}+\sqrt{2}i=\sqrt{2}(-1+i)=2e^{i3\pi/4} \]
Step 3: Ratio: \[ \frac{2e^{-i\pi/6}}{2e^{i3\pi/4}}=e^{-i(\pi/6+3\pi/4)}=e^{-i(11\pi/12)} \]
Step 4: Raise to power 20: \[ e^{-i(220\pi/12)}=e^{-i(55\pi/3)} \]
Step 5: Reduce angle modulo $2\pi$: \[ 55\pi/3 = 18\pi + \pi/3 \Rightarrow e^{-i\pi/3} \]
Step 6: \[ e^{-i\pi/3}=\omega^2 \]