# Integrating form over a path on projective algebraic curve

Let $X$ be an algebraic projective curve in $\mathbb{C}P^2$ given by $$X = \left\{ w \in \mathbb{C}P^2 \mid w_0^2 = w_1 w_2 \right\}.$$ I have a differential form on $X$ defined by $$\Omega_{\xi} = \frac{w_1}{w_0^2} \frac{w_2 dw_0}{w_2 - \xi w_0}, \; \xi \neq 0$$ It is holomorphic on $X$ without points, where denominator vanishes. If $w_0 = 0$ then from equation $w_0^2 = w_1 w_2$ we obtain two such points $M = [0:0:1]$ and $N =[0:1:0]$. If $w_2 = \xi w_0$ we have $w_0^2 = w_1 \xi w_0$ and hence $w_0 = \xi w_1$. That gives third point $K_{\xi} = [\xi : 1: \xi^2]$. By theorem of total sum of residues we must have $$\mathop{\mathrm{Res}} (\Omega_{\xi},M) + \mathop{\mathrm{Res}}(\Omega_{\xi},N) + \mathop{\mathrm{Res}}(\Omega_{\xi},K_{\xi}) = 0$$ To compute first residue we can integrate $\Omega_{\xi}$ over a little curve around $M$. We can take as local coordinates $w_0 = \lambda$, $w_1 = \lambda^2$, $w_2 = 1$. Then $dw_0 = d\lambda$ and $$2 \pi i\mathop{\mathrm{Res}} (\Omega_{\xi},M) = \lim\limits_{\varepsilon \to +0}\int\limits_{|\lambda|=\varepsilon} \frac{\lambda^2}{\lambda^2}\frac{ d\lambda}{1 - \xi \lambda} = 0.$$ Near $N$ we can take $w_0 = \lambda$, $w_1 = 1$, $w_2 = \lambda^2$. Then $$2 \pi i\mathop{\mathrm{Res}} (\Omega_{\xi},N) = \lim\limits_{\varepsilon \to +0}\int\limits_{|\lambda|=\varepsilon} \frac{1}{\lambda^2}\frac{ \lambda^2 d\lambda}{\lambda^2 - \xi \lambda} = \lim\limits_{\varepsilon \to +0} \int\limits_{|\lambda| = \varepsilon} \frac{1}{\lambda} \frac{d\lambda}{\lambda - \xi} = -\frac{1}{\xi}.$$ Near $K_{\xi}$ we can take $w_0 = 1$ and hence $dw_0 = 0$ and $\Omega_{\xi} = 0$. But this is a contradiction since the total sum of residues equal zero. Where is a problem? I think I use local coordinates in incorrect manner.

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