MHB Is This the Correct Laurent Series Expansion for $\frac{1}{z^3-z^4}$?

aruwin
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Hello.
Can you check if my answer is correct please?

For the region ${\{z\inℂ\big|0<|z|<1\}}$, expand $\frac{1}{z^3-z^4}$ that has a center z=0 into Laurent series.

My solution:
$$\frac{1}{z^3(1-z)}=\frac{1}{z^3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}$$
 
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aruwin said:
Hello.
Can you check if my answer is correct please?

For the region ${\{z\inℂ\big|0<|z|<1\}}$, expand $\frac{1}{z^3-z^4}$ that has a center z=0 into Laurent series.

My solution:
$$\frac{1}{z^3(1-z)}=\frac{1}{z^3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}$$

Yes, all right!...

Kind regards

$\chi$ $\sigma$
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
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