Find the value of


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!


3 Answers 3


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

  • $\begingroup$ Thanks Nosrati, I didn't know about De Moivrer's formula. $\endgroup$ Aug 17, 2018 at 2:45
  • $\begingroup$ You are welcome dear! $\endgroup$
    – Nosrati
    Aug 17, 2018 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, 2018 at 3:38

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

  • $\begingroup$ this is same as my answer which I published minutes ago $\endgroup$ Aug 17, 2018 at 2:52
  • 2
    $\begingroup$ @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, 2018 at 3:25



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


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