- #1
aruwin
- 208
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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:
$$\frac{1}{z^3(1-z)}=\frac{1}{z^3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}$$
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:
$$\frac{1}{z^3(1-z)}=\frac{1}{z^3}\sum_{n=0}^{\infty}z^n=\sum_{n=0}^{\infty}z^{n-3}$$