The curve $C_{r}$ coincides with the closure of the image of the map $\varphi\colon\mathbb{T}\to\mathbb{P}^{r}$ given by $\varphi(t)=(1:t:t^{2}:\dots:t^{r})$. With the notation in Corollary 8.12, this map correspond to $m_{i}=i$ and $p_{i}=1$, for $i=1,\dots,r$. Then $q_{v}=\sum_{j=0}^{r}z^{j}$ for all $v\in\mathfrak{M}_{\mathbb{Q}}$. Consider the primitive $(r+1)$-th root of unity $\omega=\operatorname{e}^{\frac{2\pi i}{r+1}}$. The polynomial $q_{v}$ is separable and its set of roots is $\{\omega^{l}\}_{l=1,\dots,r}$. Since $|\omega^{l}|_{v}=1$ for all $v$, Corollary 8.12 implies that

(8.15)

$$\operatorname{h}_{{\overline{L}}}(Y)=\frac{r}{2}+\frac{1}{2}\sum_{l<j}\frac{\omega^{l}+\omega^{j}}{\omega^{l}-\omega^{j}}(\log(-\omega^{l})-\log(-\omega^{j}))\\ =\frac{r}{2}+\frac{1}{2}\sum_{l\neq j}\frac{\omega^{l}+\omega^{j}}{\omega^{l}-\omega^{j}}\log(-\omega^{l}).$$

We have that

$$\sum_{j=1}^{r}\frac{\omega^{j}+1}{\omega^{j}-1}=\sum_{j=1}^{r}\frac{\omega^{j}}{\omega^{j}-1}+\sum_{j=1}^{r}\frac{1}{\omega^{j}-1}=\sum_{j=1}^{r}\frac{1}{1-\omega^{-j}}+\sum_{j=1}^{r}\frac{1}{\omega^{j}-1}=0.$$

This implies that, for $l=1,\dots,r$,

$$\sum_{1\leq j\leq r,j\neq l}\frac{\omega^{l}+\omega^{j}}{\omega^{l}-\omega^{j}}=-\frac{\omega^{l}+1}{\omega^{l}-1}=i\cot\Big(\frac{\pi l}{r+1}\Big)$$

Hence,

$$\frac{1}{2}\sum_{l\neq j}\frac{\omega^{l}+\omega^{j}}{\omega^{l}-\omega^{j}}\log(-\omega^{l})=-\frac{i}{2}\sum_{l=1}^{r}\cot\Big(\frac{\pi l}{r+1}\Big)\log(-\omega^{l})\\ =\pi\sum_{l=1}^{\lfloor r/2\rfloor}\cot\Big(\frac{\pi l}{r+1}\Big)\Big(1-\frac{2l}{r+1}\Big),$$

since $\cot(\frac{\pi(r+1-l)}{r+1})\log(-\omega^{r+1-l})=\cot(\frac{\pi l}{r+1})\log(-\omega^{l})$ for $l=1,\dots,\lfloor r/2\rfloor$ and $\log(-\omega^{\frac{r+1}{2}})=0$ whenever $r$ is odd. The statement follows from this calculations together with (8.15). ∎