Complex number inequality $\left|\sum_{k=1}^n\Im(z^k)\right| < \sqrt{\frac{1}{2}\Re\left(n-\sum_{k=1}^n\frac{z^{2k}}{|z|^{2k}}\right)}$

Question

If $z\in \mathbb{C}^*$ with $|z|< \frac{\sqrt{2}}{2}$ and $n$ is a positive integer, prove that:

$$\left|\sum_{k=1}^n\Im(z^k)\right| < \sqrt{\frac{1}{2}\Re\left(n-\sum_{k=1}^n\frac{z^{2k}}{|z|^{2k}}\right)}$$

I coould not prove this inequality. It is from an old preparation test at my school (2001). I tried with polar form $z=|z|(\cos a+i\sin a)$ and I arrive at:

$$\left(\sum_{k=1}^n|z|^k\sin(ka)\right)^2 < \frac{1}{2}\left(n-\sum_{k=1}^n\cos(2ka)\right)$$

but no further success.

Answer 1

Proof: We have \begin{align} \Big(\sum_{k=1}^n |z|^k \sin k a\Big)^2 &\le \Big(\sum_{k=1}^n |z|^k |\sin k a|\Big)^2\\ &\le \sum_{k=1}^n |z|^{2k} \cdot \sum_{k=1}^n |\sin ka|^2 \tag{1}\\ &\le \sum_{k=1}^n \frac{1}{2^k} \cdot \sum_{k=1}^n \sin^2 ka\tag{2}\\ &< \sum_{k=1}^n \sin^2 ka \tag{3}. \end{align} Explanation: in (1), we have used Cauchy-Bunyakovsky-Schwarz inequality; in (2), we have used $|z| < \frac{1}{\sqrt{2}}$; in (3), we have used $\sum_{k=1}^n \frac{1}{2^k} = 1 - \frac{1}{2^n} < 1$.

On the other hand, we have \begin{align} \frac{1}{2}\Big(n - \sum_{k=1}^n \cos 2k a\Big) &= \frac{1}{2}\Big(n - \sum_{k=1}^n (1 - 2\sin^2 ka)\Big)\\ &= \sum_{k=1}^n \sin^2 ka. \end{align}

We are done.