Simplifying $\left(-1+\sqrt{-3}\right)^4+\left(-1-\sqrt{-3}\right)^4$

Question

$$\left(-1+\sqrt{-3}\right)^4+\left(-1-\sqrt{-3}\right)^4 = ?$$

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!

Answer 1

$((-1+i\sqrt{3})^2)^2=(1-2i\sqrt{3}-3)^2=4+8i\sqrt{3}-12=-8+8i\sqrt{3}$

$((-1-i\sqrt{3})^2)^2=(1+2i\sqrt{3}-3)^2=4-8i\sqrt{3}-12=-8-8i\sqrt{3} $

Thus,
$(-1+i\sqrt{3})^4+(-1-i\sqrt{3})^4 = -16$

Answer 2

$(-1+i\sqrt{3})^4=(2e^\frac{i2\pi}{3})^4=16e^\frac{i8\pi}{3}=-8+8i\sqrt{3}$

$\bar z^4=-8-8i\sqrt{3}$

The sum is $-16$.

Alternatively, note that $z^3=8$. The sum is $8z+8\bar z$.

Answer 3

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. $$ z^n+\bar{z}^n=z^n+\overline{z^n}=2\,\Re(z^n) $$

Answer 4

Here's some basic complex algebra

$$\begin{align} z=x+iy&=r(\cos \theta + i \sin \theta) \\ &=r \cdot e^{i \theta} \\ &\implies \\ z^n=(x+iy)^n &=r^n(e^{i \theta})^n \\ &=r^n \cdot e^{i n \theta} \\ &=r^n(\cos n \theta + i \sin n \theta) \\ &\implies \\ \overline{z}^n=(x-iy)^{n} &=r^{n}(\cos \theta - i \sin \theta)^{n} \\ &=r^{n}(\cos n \theta - i \sin n \theta) \\ &\implies \\ z^n+\overline {z}^{n}&=2 r^{n} \cos n \theta \\ &(n=4, r=\sqrt{x^2+y^2}=2, \theta=\arctan\frac{y}{x}=\arctan \frac{\sqrt 3}{1}=\arctan \sqrt 3) \implies \\ &=2 \cdot 2^4 \cdot \cos {(4 \arctan \sqrt3)} \\ &=-16 \end{align}$$