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Find the value of $$\frac{(1-\sqrt{3}i)^{53}}{2^{53}}$$

The answer I get to is $-1 + 0i$, which is incorrect.

I start with putting $1-\sqrt{3}i$ into polar form.

I know that given a complex number in form $x + iy$ we have the following: $$ x=r\cos\theta\\ y=r\sin\theta $$ And I know that $r=\left|x+iy\right|$. $$ r=\sqrt{1^2+(-\sqrt3)^2}=\sqrt{4}=2 $$ This all yields, \begin{align*} 1&=2\cos\theta\\ \frac{1}{2}&=\cos\theta\\ \theta&=\frac{\pi}{3}=\frac{5\pi}{3}\quad\text{(for }\cos\theta\text{)}\\ \\\ -\sqrt3&=2\sin\theta\\ \frac{-\sqrt3}{2}&=\sin\theta\\ \theta&=\frac{-\pi}{3}=\frac{5\pi}{3}\quad\text{(for }\sin\theta\text{)} \end{align*} Which gives me a polar form as follows. $$ 1-\sqrt{3}i=2\left(\cos{\frac{5\pi}{3}}+\sin{\frac{5\pi}{3}}\right) $$

De Moivre's Formula tells me that $$ (r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta) $$

So taking the original fraction given, \begin{align*} &\quad\frac{(1-\sqrt{3}i)^{53}}{2^{53}}\\ &= \frac{2^{53}\left(\cos\frac{53\times5\pi}{3}+i\sin\frac{53\times5\pi}{3}\right)}{2^{53}}\\ &= \cos(\frac{159\pi}{3}) + i\sin(\frac{159\pi}{3})\\ &= \cos(\pi) + i\sin(\pi)\quad\text{Since these repeat every }2\pi\\ &= -1 + 0i \end{align*}

I’ve clearly made a faulty assumption somewhere here, but I am not sure where.

Edit: Yup, simple arithmetic error described in comments and accepted answer. Final answer is $\frac{1}{2}+i\frac{\sqrt3}{2}$.

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    $\begingroup$ 53*5=265, isn't it? $\endgroup$ – N74 Mar 9 '17 at 21:29
  • $\begingroup$ Well, $53\times5=265$. $\endgroup$ – Fimpellizieri Mar 9 '17 at 21:29
  • $\begingroup$ Ah, yup, that was the whole issue. $\endgroup$ – Emily Horsman Mar 9 '17 at 21:35
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We have $$53\cdot\frac{5\pi}3=\frac{265\pi}{3}=44\cdot2\pi+\frac\pi3$$

so $\cos\left(53\cdot\frac{5\pi}3\right)=\cos\left(\frac\pi3\right)$ and $\sin\left(53\cdot\frac{5\pi}3\right)=\sin\left(\frac\pi3\right)$.

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Without too much effort or indeed any trig, all you need to really do is notice that

$$\begin{align}\left(\frac12-\frac{\sqrt3}2i\right)^{53}&=\left(\frac12-\frac{\sqrt3}2i\right)^{51}\left(\frac12-\frac{\sqrt3}2i\right)^2\\&=\left(\left(\frac12-\frac{\sqrt3}2i\right)^3\right)^{17}\left(-\frac12-\frac{\sqrt3}2i\right)\\&=\left(-1\right)^{17}\left(-\frac12-\frac{\sqrt3}2i\right)\\&=\frac12+\frac{\sqrt3}2i\end{align}$$

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  • $\begingroup$ @JohnWaylandBales Thanks man $\endgroup$ – Simply Beautiful Art Mar 9 '17 at 22:12

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