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natural_logarithm_of_complex_numbers [2015/06/26 22:07] 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 $|z|<1$) | ||
| - | $\ln{\left(1+z\right)}=-\sum_{n=1}^\infty \frac{(-z)^n}{n}$ | + | $\ln{\left(\frac{1}{1-z}\right)} = \sum_{n=1}^\infty \frac{z^n}{n}$ |
| - | and so, from this | + | or |
| - | $\ln{\left(\frac{1}{1-z}\right)}=\sum_{n=1}^\infty \frac{z^n}{n}$ | + | $\ln{\left(1+z\right)} = -\sum_{n=1}^\infty \frac{(-z)^n}{n}$ |
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| + | From this series in $(-z)^n$, if we always constitutive positive and negative germs, we get | ||
| + | |||
| + | $\ln{\left(1+z\right)} = z - \sum_{n=1}^\infty \left(\dfrac{1}{2n}-\dfrac{z}{2n+1}\right) z^{2n}$ | ||
| Note that this implies that $\exp$ maps the series on the right to $\frac{1}{1-z}=1+\sum_{n=1}^\infty z^n$. | Note that this implies that $\exp$ maps the series on the right to $\frac{1}{1-z}=1+\sum_{n=1}^\infty z^n$. | ||
