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

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SUMMARY

The correct Laurent series expansion for the function $\frac{1}{z^3 - z^4}$ in the region ${\{z \in \mathbb{C} \mid 0 < |z| < 1\}}$ is derived as follows: $\frac{1}{z^3(1-z)}$ can be expressed using the geometric series expansion, resulting in $\sum_{n=0}^{\infty} z^{n-3}$. This confirms that the series is valid for the specified annular region with a center at z=0.

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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

$\chi$ $\sigma$
 
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