Adding a complex and its conjugate, both raised to a degree of 4? $(-1+\sqrt{-3})^4+(-1-\sqrt{-3})^4$ = ?
I deduced the complex form:
$z=(-1+i\sqrt{3})$ 
I can see that the question is basically $z^4+\overline z^4$
Now, if we add a complex number and its conjugate the imaginary part gets eliminated and we are left with adding the real part.
I also tried solving this through binomial theorum but I am not able to get the right answer.
I would appreciate any help and hints in solving this, thank you!
 A: $(-1+i\sqrt{3})^4=(2e^\frac{i2\pi}{3})^4=16e^\frac{i8\pi}{3}=-8+8i\sqrt{3}$
$\bar z^4=-8-8i\sqrt{3}$
The sum is $-16$.
Alternatively, note that $z^3=8$. The sum is $8z+8\bar z$.
A: Since $$z^4=-8+i\,8\sqrt3$$ then
$$
z^4+\bar{z}^4=z^4+\overline{z^4}=2\,\Re(z^4)=-16
$$
This generalizes, i.e.
$$
z^n+\bar{z}^n=z^n+\overline{z^n}=2\,\Re(z^n)
$$
A: Here's some basic complex algebra 
$$\begin{align}
z=x+iy&=r(\cos \theta + i \sin \theta) \\
&=r \cdot e^{i \theta} \\
&\implies \\
z^n=(x+iy)^n &=r^n(e^{i \theta})^n \\
&=r^n \cdot e^{i n \theta} \\
&=r^n(\cos n \theta + i \sin n \theta) \\
&\implies \\
\overline{z}^n=(x-iy)^{n} &=r^{n}(\cos  \theta - i \sin  \theta)^{n} \\
&=r^{n}(\cos n \theta - i \sin n \theta) \\ 
&\implies \\
z^n+\overline {z}^{n}&=2 r^{n} \cos n \theta \\
&(n=4, r=\sqrt{x^2+y^2}=2, \theta=\arctan\frac{y}{x}=\arctan \frac{\sqrt 3}{1}=\arctan \sqrt 3) \implies \\
&=2 \cdot 2^4 \cdot \cos {(4 \arctan \sqrt3)} \\
&=-16
\end{align}$$
A: $((-1+i\sqrt{3})^2)^2=(1-2i\sqrt{3}-3)^2=4+8i\sqrt{3}-12=-8+8i\sqrt{3}$
$((-1-i\sqrt{3})^2)^2=(1+2i\sqrt{3}-3)^2=4-8i\sqrt{3}-12=-8-8i\sqrt{3}  $
Thus,
$(-1+i\sqrt{3})^4+(-1-i\sqrt{3})^4 = -16$
