If $z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$
show that $z^{3n}+z^{3m+1}+z^{3k+2}=0$ where $k,m,n$ are natural numbers.
I'm not allowed to use Euler's formula. I reached this $$\text{cis}(-\pi n)+\text{cis}(-\pi m - \frac{\pi}{3})+\text{cis}(-\pi k - \frac{2\pi}{3})$$ and there I got stuck.
If you are not allowed to use Euler's formula or the polar representation of a complex number, then simply perform the computation on the Cartesian form:
$$z = -\frac{1}{2} - \frac{\sqrt{3}}{2} i = -\frac{1}{2}(1 + \sqrt{3} i)$$ implies $$z^2 = \left(-\frac{1}{2}\right)^2 (1 + \sqrt{3}i)^2 = \frac{1}{4} (1 + 2\sqrt{3} i + 3i^2) = \frac{1}{4}(-2 + 2 \sqrt{3}i) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = -\frac{1}{2}(1-\sqrt{3}i),$$ and $$z^3 = z\cdot z^2 = \left(-\frac{1}{2}\right)^2(1 + \sqrt{3}i)(1 - \sqrt{3}i) = \frac{1}{4}(1 - 3i^2) = 1.$$
Therefore, for any integer $n$, $$z^{3n} = (z^3)^n = 1^n = 1,$$ and it follows that $$z^{3n} + z^{3m+1} + z^{3k+2} = 1 + z + z^2 = 1 - \frac{1}{2} - \frac{\sqrt{3}}{2} i - \frac{1}{2} + \frac{\sqrt{3}}{2} i = 0.$$
Is it tedious? Yes. Is it inefficient? Yes. Is it an answer that meets the problem-solving criteria? Yes.
$$z=-\frac{1}{2}-\frac{\sqrt 3}{2}i\Rightarrow (2z+1)^2+3=0\tag{1}$$
The last equation simplifies to $$z^2+z+1=0\tag{2}$$
Multiplying $(2)$ with $z-1$ yields $$z^3=1\tag{3}$$ Now your result follows since $$\begin{align}0&=z^2+z+1\\&=z^2\cdot 1+z\cdot 1+1\cdot 1\\&=z^2\cdot (z^3)^n+z\cdot (z^3)^m+1\cdot (z^3)^k\\&=z^{3n+2}+z^{3m+1}+z^{3k}\end{align}$$
First, show that $z^3 = 1$. If you can't use Euler's or DeMoivre's formula, just expand it the hard way using $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Then,
$$z^{3n}+z^{3m+1}+z^{3k+2}$$ $$=(z^3)^n+ (z^3)^mz + (z^3)^kz^2$$ $$= 1^n + 1^mz + 1^kz^2$$ $$= 1 + z + z^2$$
You'll also need to show that $z^2 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i $. (Or this can be done as an intermediate step towards calculating $z^3$.) Then,
$$ 1 + z + z^2$$ $$= 1 -\frac{1}{2} - \frac{\sqrt{3}}{2}i -\frac{1}{2} + \frac{\sqrt{3}}{2}i$$ $$= 0$$
Use the fact that $~(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.$
So,
$$z^3 = \left[ ~\left( ~\frac{-1}{2} ~\right) + \left( ~\frac{-i\sqrt{3}}{2} ~\right) ~\right]^3$$
$$= ~\left( ~\frac{-1}{2} ~\right)^3 + 3\left( ~\frac{-1}{2} ~\right)^2\left( ~\frac{-i\sqrt{3}}{2} ~\right) + 3\left( ~\frac{-1}{2} ~\right)\left( ~\frac{-i\sqrt{3}}{2} ~\right)^2 + \left( ~\frac{-i\sqrt{3}}{2} ~\right)^3 $$
$$= ~\left( ~\frac{-1}{8} ~\right) + 3\left( ~\frac{1}{4} ~\right)\left( ~\frac{-i\sqrt{3}}{2} ~\right) + 3\left( ~\frac{-1}{2} ~\right)\left( ~\frac{-3}{4} ~\right) + \left( ~\frac{i3\sqrt{3}}{8} ~\right) $$
$$= \left( ~\frac{-1}{8} - \frac{-9}{8} ~\right) + i\left( ~\frac{-3\sqrt{3}}{8} + \frac{3\sqrt{3}}{8} ~\right) = (1) + i(0) = 1.$$
Also, use the fact that $~(a+b)^2 = a^2 + 2ab + b^2.$
So,
$$z^2 = \left[ ~\left( ~\frac{-1}{2} ~\right) + \left( ~\frac{-i\sqrt{3}}{2} ~\right) ~\right]^2$$
$$= ~\left( ~\frac{-1}{2} ~\right)^2 + 2\left( ~\frac{-1}{2} ~\right)\left( ~\frac{-i\sqrt{3}}{2} ~\right) + \left( ~\frac{-i\sqrt{3}}{2} ~\right)^2 $$
$$= ~\left( ~\frac{1}{4} ~\right) + \left( ~\frac{i\sqrt{3}}{2} ~\right) + \left( ~\frac{-3}{4} ~\right) $$
$$= \left( ~\frac{-1}{2} ~\right) + \left( ~\frac{i\sqrt{3}}{2} ~\right).$$
So, since $~z^3 = 1,~$ you have that for all $~n,m,k \in \Bbb{Z^+},~$ that :
$\displaystyle z^{3n} = 1^{n} = 1.$
$\displaystyle z^{3m+1} = 1^{m} \times z = z = \left( ~\frac{-1}{2} ~\right) + \left( ~\frac{-i\sqrt{3}}{2} ~\right).$
$\displaystyle z^{3k+2} = 1^{k} \times z^2 = z^2 = \left( ~\frac{-1}{2} ~\right) + \left( ~\frac{i\sqrt{3}}{2} ~\right).$
Therefore,
$$z^{3n} + z^{3m+1} + z^{3k+2}$$
$$= 1 + z + z^2 $$
$$= 1 + \left[ ~\left( ~\frac{-1}{2} ~\right) + \left( ~\frac{-i\sqrt{3}}{2} ~\right) ~\right] + \left[ ~\left( ~\frac{-1}{2} ~\right) + \left( ~\frac{i\sqrt{3}}{2} ~\right) ~\right] = 0.$$
$\underline{\text{Addendum}}$
I wasn't sure whether you would be comfortable with the following shortcut.
Since
$z^3 = 1,~$
and $~(1 + z + z^2) \times (1 - z) = 1 - z^3,~$
you can conclude, with some of the already shown math omitted, that
$$1 + z + z^2 = \frac{1 - z^3}{1 - z} = 0.$$
