If $n$ is a positive integer and $z \in \mathbb{C}$ with $|z|=1$, prove that:
$$n^2-(\Re(z^2+z^4+...+z^{2n}))^2\geq (\Im(1+z^2+...+z^{2n}))^2$$
Because $1$ is real, $\Im(1)=0$, so I can ignore it and try to prove:
$$n^2-(\Re(z^2+z^4+...+z^{2n}))^2\geq (\Im(z^2+z^4+...+z^{2n}))^2$$
I know that in general $|z|^2=(\Re(z))^2+(\Im(z))^2$, so the inequality is:
$$|z^2+z^4+...+z^{2n}|^2\leq n^2$$
and this I can prove with triangle inequality
$$|z^2+z^4+...+z^{2n}| \le |z^2|+|z^4|+...+|z^{2n}|=n$$
so this should be done. Is this reasoning correct? Are there other ways to prove this inequality?
I think your proof is ok. We can also prove it using the polar form of $z = \cos \theta+i\sin \theta$ and using $z^n= \cos n\theta +i \sin n\theta$, the question can be written as:
$$n^2 - \left(\sum_{k=1}^n \cos 2k\theta\right)^2\geq \left(\sum_{k=1}^n \sin 2k\theta\right)^2$$
or
$$\left(n-\sum_{k=1}^n \cos 2k\theta\right)\left(n+\sum_{k=1}^n \cos 2k\theta\right)\geq \left(\sum_{k=1}^n 2\sin k\theta\cos k\theta\right)^2$$
or
$$\left[n-\sum_{k=1}^n\left(1-2\sin^2 k\theta \right)\right]\left[n+\sum_{k=1}^n \left(2\cos^2k\theta-1\right)\right]\geq 4\left(\sum_{k=1}^n \sin k\theta\cos k\theta\right)^2$$
or
$$\left(\sum_{k=1}^n\sin^2 k\theta\right)\left(\sum_{k=1}^n \cos^2k\theta \right)\geq \left(\sum_{k=1}^n \sin k\theta\cos k\theta\right)^2$$
which follows directly from Cauchy-Schwarz.