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| Both sides previous revision Previous revision | Last revision Both sides next revision | ||
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natural_logarithm_of_complex_numbers [2016/05/05 17:06] nikolaj |
natural_logarithm_of_complex_numbers [2016/05/05 17:13] nikolaj |
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| == Series == | == Series == | ||
| + | At least around $z=0$ (I think for $|z|<1$) | ||
| + | |||
| $\ln{\left(\frac{1}{1-z}\right)} = \sum_{n=1}^\infty \frac{z^n}{n}$ | $\ln{\left(\frac{1}{1-z}\right)} = \sum_{n=1}^\infty \frac{z^n}{n}$ | ||
