Power of a complex number, find $(\sqrt{3} + i)^{99}$

Question

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 $99$?

Answer 1

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 $2\pi$.

Answer 2

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.

Answer 3

Hint: $\frac{99\pi}6=8\cdot2\pi+\frac\pi2$

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

Answer 4

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