Find the value of complex expression $\left(\frac{\sqrt{3}+i}{2}\right)^{69}$

Question

Find the value of

$$\left(\dfrac{\sqrt{3}+i}{2}\right)^{69}.\DeclareMathOperator{\cis}{cis}$$

I tried to solve this complex expression by converting it into polar form. I expressed it in polar form $r\cis(t)$ from rectangular form $x+iy$ where $\cis(t) = \cos(t) + i\sin(t)$. But I am unable to solve further due to the exponent of 69!

Answer 1

$$\left(\dfrac{\sqrt{3}+i}{2}\right)^{69}=\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)^{69}=(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6})^{69}=\cos\dfrac{69\pi}{6}+i\sin\dfrac{69\pi}{6}=-i$$ by De Moivre's formula.

Answer 2

$$\frac{\sqrt3+i}{2}=e^{i\frac{\pi}{6}}$$

$$\big(\frac{\sqrt3+i}{2}\big)^{69}=e^{i\frac{\pi}{6}\cdot69}=e^{i(11\pi+\frac{\pi}{2})}=e^{i(12\pi-\frac{\pi}{2})}=e^{-i\frac{\pi}{2}}=-i$$

I used the fact that $e^{i(2n\pi)}=1$ where $n\in Z$

Answer 3

Converting to polar coordinates is probably the best way to go.

$$r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}=1$$

$$\theta = \arctan(1/\sqrt{3})=\frac{\pi}{6}$$

Then $$(re^{\theta i })^{69}=e^{\frac{69\pi}{6}i}=e^{10\pi i+\frac{3\pi}{2}i}=e^{\frac{3\pi}{2}i}=-i$$