Consider the equation ##\mathcal{F}(\lambda)=0\ \ \ \forall\ \lambda = \lambda_{n},\ n \in \mathbb{N}##.(adsbygoogle = window.adsbygoogle || []).push({});

I understand that the expression ##\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}## has poles of order 1 exactly at ##\lambda_{n}## because ##\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'(\lambda)}}{(\lambda- \lambda_{1})...(\lambda-\lambda_{n})}##.

I wonder how I might expand the expression ##\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}## about ##\lambda_{n}## to find out that the residue of the poles is 1.

Any ideas?

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# Finding the residues of poles

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