What is the value of $1 -omega^h + omega^{2h} -...+(-1)^{n-1} omega^{(n-1)h}$ when $omega$ is a root of unity?

I’m reading Ahlfors’ complex analysis book. One of the problems in the book says as follows

What is the value of $1 -omega^h + omega^{2h} -…+(-1)^{n-1} omega^{(n-1)h}$?

where $h$ is some integer and $ omega = cosleft(frac{2pi}{n}right) + i sin left(frac{2 pi}{n}right)$, for some fixed $n in mathbb{N}$, is one of the $n$-th roots of unity.

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The first thing I noticed is that I could write the series in terms of $-omega^h$ as
$$
1 +left(-omega^hright) + left(-omega^hright)^2 +…+left(-omega^hright)^{n-1}
$$
Inspired by this, I separated the problem into 2 cases

1. If $h$ is an integer of the form $ h = frac{n(2k+1)}{2}$ for some $k in mathbb{Z}$, then I get the following
   $$
   -omega^h = -cosleft(frac{2pi}{n}hright) – i sin left(frac{2 pi}{n}hright)= -cosleft(pi + 2pi kright) – i sin left(pi + 2pi kright) = 1
   $$
   which means the sum evaluates to $sum_{j=0}^{n-1} 1 = n$.
2. If $-omega^h neq 1$, then using the fact that the sum in question is a sum of the first $n$ terms in a geometric series, I can write $$
   1 -omega^h + omega^{2h} -…+(-1)^n omega^{(n-1)h} = frac{1 – left(-omega^hright)^n}{1 – left(-omega^hright)} = frac{1 – (-1)^nomega^{nh}}{1 +omega^h}
   $$
   and since $h$ is an integer, I see that
   $$
   omega^{nh} = cosleft(frac{2pi}{n}nhright) + i sin left(frac{2 pi}{n}nhright) = cosleft(2pi hright) + i sin left(2pi hright) = 1
   $$
   which means the sum simplifies to $frac{1 – (-1)^n}{1 +omega^h}$. From here I see that if $n$ is odd the sum will become $0$ because of the numerator, but for the case of $n$ being an even number, I don’t see a way to simplify $frac{2}{1 +omega^h}$ more than it already is.

Is my solution correct? And if so, is this as simplified as I can write the solution, or can it still be simplified further? Thank you very much!