Raising a number in Rectangular Form What is the value of $(-2 + 3i\sqrt3)^6$?
Answer is $4096$
Convert $(-2 + 3i\sqrt3)^6$ to Polar Form.
$${ (\sqrt{31} \angle 111.05)^6 }$$
I use something called De Moivre's Theorem
$${z^n = r^n( \cos(n\theta) + i\sin(n\theta) ) }$$
$${z^n = (\sqrt{31})^6( \cos(6\cdot 111.05) + i\sin(6\cdot 111.05) ) }$$
Even if I did continue this I know I wouldn't get a whole number sum thing...
$${z^n = 29791( 0.5920 - 8.0593i ) }$$
$${z^n = 17636 + 240009i }$$
What am I doing wrong?
 A: Either the answer you were given was incorrect, or you have copied the problem incorrectly. This is clear if you simply square the original in rectangular form:
$$(-2+3i\sqrt3)^2=-23-12i\sqrt3$$
And follow that with cubing the number:
$$(-23-12i\sqrt3)^3=-\left(12167+19044i\sqrt3-29808-5184i\sqrt3\right)=17641-13860i\sqrt3$$
Notice that the imaginary part doesn't cancel. Further notice that the real part is nowhere near $4096$.
If you wanted a number that WOULD end up at $4096$, you would want numbers of the form: $$4e^{i\frac{k\pi}{3}}$$ for $k\in\{0,1,2,3,4,5\}$
Also, the only thing you are doing "wrong" in your work is rounding too early.
A: First of all we know the following things:
$$a+bi=\left|a+bi\right|e^{\arg\left(a+bi\right)i}=\left|a+bi\right|\left(\cos\left(\arg\left(a+bi\right)\right)+\sin\left(\arg\left(a+bi\right)\right)i\right)$$

$$\left(-2+3i\sqrt{3}\right)^6=$$
$$\left(\left|-2+3i\sqrt{3}\right|e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{\Re\left(-2+3i\sqrt{3}\right)^2+\Im\left(-2+3i\sqrt{3}\right)^2}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{(-2)^2+(3\sqrt{3})^2}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{4+27}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{31}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{31}e^{\left(\pi-\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}\right)^6=$$
$$\sqrt{31}^6e^{6\left(\pi-\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}=$$
$$29791e^{\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}$$
So in the three forms we get:
$$29791e^{\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}=$$
$$29791\left(\cos\left(\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)\right)+\sin\left(\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)\right)i\right)=$$
$$29791\left(\frac{17641}{29791}-\frac{13860\sqrt{3}}{29791}i\right)=17641-13860\sqrt{3}i$$
