## Power series expansion

1. The problem statement, all variables and given/known data
Find a power series expansion for log(1-z) about z = 0. Find the residue at 0 of 1/-log(1-z) by manipulation of series, residue theorem and L'Hopitals rule.

2. Relevant equations

3. The attempt at a solution
Is this power series the same as the case for real numbers.
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 I have the power series expansion about z = 0 for log(1-z). -z - z2/2 - z3/3 - ... But how do I find the residues with the methods mentioned
 When I manipulate do I use the power series for log(1-z)

## Power series expansion

My series for -1/log(1-z) is:
1/z - 1/2 - z/12 - z2/24 - ...
So my residue is a-1 = -1/2.
Is that right?
 How do I do it by the residue theorem and L'Hopitals rule.
 Blog Entries: 2 Make me think of the power series of log(1+z) Recall Sir, $$log(1+z) = \sum_{j=1}^\infty} \frac{(-1)^{j+1}}{j}z^{j} = z - \frac{z^2}{2} + \frac{z^3}{3}-\cdots$$ so by very very simply replacing z with -z you get $$-z - \frac{(-z)^2}{2} + \frac{(-z)^3}{3}-\cdots$$ So the power series expansion of log(1-z) Is $$P_{n} = -\sum_{j=1}^{n} \frac{z^n}{n}$$

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