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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!

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

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  • $\begingroup$ Thanks Nosrati, I didn't know about De Moivrer's formula. $\endgroup$ – Shubh Khandelwal Aug 17 '18 at 2:45
  • $\begingroup$ You are welcome dear! $\endgroup$ – Nosrati Aug 17 '18 at 2:46
  • $\begingroup$ @Shubh Khandelwal Basically complex multiplication has the rule that the norm of the product is the product of norms, and the argument of product is the sum of arguments. $\endgroup$ – MonkeyKing Aug 17 '18 at 3:38
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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 $$(re^{\theta i })^{69}=e^{\frac{69\pi}{6}i}=e^{10\pi i+\frac{3\pi}{2}i}=e^{\frac{3\pi}{2}i}=-i$$

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  • $\begingroup$ this is same as my answer which I published minutes ago $\endgroup$ – Deepesh Meena Aug 17 '18 at 2:52
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    $\begingroup$ @James Yes, and? Great minds think alike, no? That being said, this answer is not the same answer that you published. It is more detailed in terms of explaining the conversion from rectangular to exponential form. This is neither good nor bad, just different. $\endgroup$ – Xander Henderson Aug 17 '18 at 3:25
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$$\frac{\sqrt3+i}{2}=e^{i\frac{\pi}{6}}$$

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

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

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Bruteforcing: $$\begin{align}\left(\dfrac{\sqrt{3}+i}{2}\right)^{69}&=\left[\left(\dfrac{\sqrt{3}+i}{2}\right)^{2}\right]^{34}\cdot \frac{\sqrt{3}+i}{2}=\\ &=\left[\left(\frac{1+\sqrt{3}i}{2}\right)^2\right]^{17}\cdot \frac{\sqrt{3}+i}{2}=\\ &=\left[\left(\frac{-1+\sqrt{3}i}{2}\right)^2\right]^{8}\cdot \frac{-1+\sqrt{3}i}{2}\cdot \frac{\sqrt{3}+i}{2}=\\ &=\left[\left(\frac{-1-\sqrt{3}i}{2}\right)^2\right]^{4}\cdot \frac{-\sqrt{3}+i}{2}=\\ &=\left[\left(\frac{-1+\sqrt{3}i}{2}\right)^2\right]^{2}\cdot \frac{-\sqrt{3}+i}{2}=\\ &=\left[\frac{-1-\sqrt{3}i}{2}\right]^2\cdot \frac{-\sqrt{3}+i}{2}=\\ &=\frac{-1+\sqrt{3}i}{2}\cdot \frac{-\sqrt{3}+i}{2}=\\ &=-i.\end{align}$$ Alternatively: $$E=\left(\frac{\sqrt{3}+i}{2}\right)^{3\cdot 23}=\left(\frac{3\sqrt{3}+9i-3\sqrt{3}-i}{8}\right)^{23}=i^{23}=(i^2)^{11}\cdot i=-i.$$

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  • $\begingroup$ Bruteforcing is not the best way,but thank you. $\endgroup$ – Shubh Khandelwal Aug 24 '18 at 4:11

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