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Question: let $$z = \sqrt{3} + i. $$ Find $$z^{99}.$$ Put your answer in $$a + bi$$ form.

My work:
$$r = 2$$ $$z = r(\cos \theta + i \sin \theta) = (2(\cos(\pi/6) + \sin(\pi/6))^{99} = 2^{99}(\cos(99\pi/6) + i \sin(99\pi/6))$$

Now to get it back to the $$a + bi$$ form, what do I do with the $99$?

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  • $\begingroup$ An essential duplicate of this $\endgroup$ – Jyrki Lahtonen Apr 8 '18 at 18:07
  • $\begingroup$ And also of this. $\endgroup$ – Jyrki Lahtonen Apr 8 '18 at 18:09
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Note that $\displaystyle \frac{99\pi}{6} = \frac{33\pi}{2} = \left( 16+\frac12 \right) \pi$, so $$z^{99} = 2^{99}(\cos(99\pi/6) + i\sin(99\pi/6)) = 2^{99} (\cos(\pi/2) + i\sin(\pi/2)) = 2^{99}i,$$ which is pure imaginary.

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  • $\begingroup$ thank you! Is this universal for all questions like this? $\endgroup$ – user433562 Apr 7 '18 at 23:09
  • $\begingroup$ @P.Layton Yes, in a finding powers for complex numbers, you first calculate $|z|$ and $\theta = \arg z$, then apply De Moivre's Formula $${\displaystyle {\big (}\cos(x)+i\sin(x){\big )}^{n}=\cos(nx)+i\sin(nx).}$$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 23:12
  • $\begingroup$ I meant, for the part where you bring in the $$n$$ value. I tend to get stuck on that part. $\endgroup$ – user433562 Apr 7 '18 at 23:13
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    $\begingroup$ @P.Layton As the other two answers suggest, take "mod $2\pi$". i.e. substract it by integral multiples of $2\pi$ to that the angle is between $0$ and $2 \pi$, then calculate the trigo function $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 7 '18 at 23:17
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Hint: $\frac{99\pi}6=8\cdot2\pi+\frac\pi2$

And $\sin$ and $\cos$ are periodic of period $2\pi$...

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Alt. hint:  for a shortcut in this particular case, note that:

$$\require{cancel} \left(\sqrt{3}+i\right)^3= \bcancel{\left(\sqrt{3}\right)^3}+3\cdot\left(\sqrt{3}\right)^2i + \bcancel{3 \cdot \sqrt{3}i^2}+i^3=9 \cdot i-i = 8i $$

Then of course $\;\left(\sqrt{3}+i\right)^{99}=\left(\left(\sqrt{3}+i\right)^3\right)^{33}=\ldots\,$

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Leave the $99$ in the exponent of $2$, and calculate which of the twelve angles ($0,\frac\pi6,\dots,\frac{11\pi}6$) the $\frac{99\pi}6$ falls modulo $2\pi$.

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