Show that there is a strict inequality $|\int_{|z|=R} \frac{z^n}{z^m-1}| < \frac{2\pi R^{n+1}}{R^m-1}$

Question

I need to prove this strict inequality holds.

$\Big \lvert \int_{|z|=R} \frac{z^n}{z^m-1} \Big \rvert < \frac{2\pi R^{n+1}}{R^m-1}$ where $R>1, m \geq 1, n \geq 0$

Letting $f(z)= \frac{z^n}{z^m-1}$, using the ML estimate, I know that the inequality '$\leq$' holds, but how can I prove that the RHS is strictly larger than LHS? Assuming the equality holds, I got $\cos{m\theta} = 1$ but that does not imply anything much.

Any help would be grateful.

Answer 1

Note that \begin{align} \Bigg \lvert \int_{|z|=R}\frac{z^n}{z^m - 1}dz ~\Bigg\rvert &= \Bigg \lvert \int_{0}^{2\pi}\frac{R^ne^{in\theta}}{R^m e^{im\theta} - 1}iRe^{i\theta}d\theta ~\Bigg\rvert \\ &\leq \int_{0}^{2\pi} \Bigg \lvert \frac{R^ne^{in\theta}}{R^m e^{im\theta} - 1}iRe^{i\theta} \Bigg\rvert d\theta \\ & \leq \int_{0}^{2\pi} \frac{R^{n+1}}{R^m -1 } d\theta \\&=\frac{2\pi R^{n+1}}{R^m - 1} \end{align}

In particular, $$ \int_{0}^{2\pi} \Bigg \lvert \frac{R^ne^{in\theta}}{R^m e^{im\theta} - 1}iRe^{i\theta} \Bigg\rvert d\theta \leq \int_{0}^{2\pi} \frac{R^{n+1}}{R^m -1 } d\theta $$

Define $g: [0, 2\pi] \to \mathbb R$ by $$g(\theta)= \frac{R^{n+1}}{R^m -1 } - \Bigg \lvert \frac{R^ne^{in\theta}}{R^m e^{im\theta} - 1}iRe^{i\theta} \Bigg\rvert $$ This is a continuous function of $\theta$ with $g(\theta) \geq 0$ for all $\theta \in [0, 2\pi]$.

If $$ \Bigg \lvert \int_{|z|=R}\frac{z^n}{z^m - 1}dz ~\Bigg\rvert = \frac{2\pi R^{n+1}}{R^m - 1}$$ holds, then $\int_{0}^{2\pi} g(\theta) d\theta = 0$, whence $g(\theta) = 0$ for all $\theta$; this is not the case since $$g(\alpha) = \frac{R^{n+1}}{R^m -1 } - \frac{R^{n+1}}{R^m + 1 }\neq 0$$ where $m\alpha = \pi$.