- #1

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[tex]z\in\mathbb{Z}[/tex]

[tex]\frac{1}{1-(-z)}=\sum_{n=0}^{\infty}(-z)^n[/tex]

[tex]\frac{1}{(z+2)^2}=\frac{d}{dz} \frac{-1}{1-(1-z)} = \frac{d}{dz} (1 + (1-z) + (1-z)^2+\cdots = 0 -1 -2(1-z)-3(1-z)^2 - \cdots = \sum_{n=0}^{\infty} ???[/tex]

Not to sure about the second one.

[tex]\frac{1}{1-(-z)}=\sum_{n=0}^{\infty}(-z)^n[/tex]

[tex]\frac{1}{(z+2)^2}=\frac{d}{dz} \frac{-1}{1-(1-z)} = \frac{d}{dz} (1 + (1-z) + (1-z)^2+\cdots = 0 -1 -2(1-z)-3(1-z)^2 - \cdots = \sum_{n=0}^{\infty} ???[/tex]

Not to sure about the second one.

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