Naively, we "know" there are two primitive 6-th roots of unity, and they are
$ \frac{1}{2} \pm i \frac{\sqrt{3}}{2} $
and so
$ 1 - \left(\frac{1}{2} \pm i \frac{\sqrt{3}}{2} \right) = \frac{1}{2} \mp i \frac{\sqrt{3}}{2}$
Done! But does this make sense in the ring $\mathbb{Z} / p \mathbb{Z}$?
Well first, observe that, since $p \neq 2,3$ it makes sense in the algebraic closure of the field $\mathbb{Z} / p \mathbb{Z}$, since such a thing does have a square root of $-1$ as well as a square root of $3$, and so it's true there. And if the primitive sixth root of unity happens to be a member of $\mathbb{Z} / p \mathbb{Z}$, then this relationship holds there too.
But maybe we don't want to go that far. Can we stick within $\mathbb{Z} / p \mathbb{Z}$? Well, it turns out we don't need square roots of $3$ and of $-1$: we just need a square root of $-3$, since the primitive 6th roots of unity are
$ \frac{1}{2} \pm \frac{\sqrt{-3}}{2} $
Does this make sense in $\mathbb{Z} / p \mathbb{Z}$ if it has a primitive 6-th root of unity $x$? Well, we just need to check that $-3$ has a square root (and that $2 \neq 0$). If it does, then the sixth roots of unity have the same form as they do in the complexes, and the original argument applies.
There are a number of ways to do this. We know that $p \equiv 1 \pmod 6$ (because $\varphi(p)$ is divisible by 6). By quadratic reciprocity,
$ \left( \frac{-3}{p} \right) = \left( \frac{p}{-3} \right) = \left( \frac{6k+1}{-3} \right) = \left( \frac{1}{-3} \right) = 1 $
Or, we could compute the square root of -3: if the equation
$ x = \frac{1}{2} \pm \frac{\sqrt{-3}}{2} $
really does make sense, then
$ 2\left(x - \frac{1}{2} \right) = \pm \sqrt{-3} $
If we check
$ (2x - 1)^2 + 3 = 0 $
then we know the previous formula computes a square root of 3. We can check this with polynomial arithmetic. Because we know $x^6 = 1$, we know that $x$ is a root of
$ t^6 - 1 = (t-1) (t+1) (t^2 + t + 1) (t^2 - t + 1) $
The factors are polynomials whose roots are 1, square roots of unity, cube roots of unity, and sixth roots of unity, respectively. $x$ must be a root of the last factor then, so we know
$ x^2 - x + 1 = 0$
The equation we need to check, then, is
$(2x-1)^2 + 3 = 4x^2 - 4x + 4 = 4(x^2 - x + 1) = 0$
The four factors of $t^6 - 1$ listed above are cyclotomic polynomials. If you were very familiar with such things, then the very first idea that popped into your head might have been the fact $x$ has to be a root of $t^2 - t + 1$. The problem is actually fairly simple starting from this fact, since the two roots of this polynomial must add to 1.