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

[aruwin]

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

 

[chisigma]

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

$\chi$ $\sigma$