Singularities

Poles of Order r > 2

Function 1/(1–z)r

The function \dfrac{1}{(1-z)^r} \; (z\in\mathbb{C}) has one pole of order r at z=1 .
Its finite value at its pole z=1 is:

\dfrac{1}{(1-1)^r} = (-1)^r \, G_r

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