Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | Next revision Both sides next revision | ||
|
natural_logarithm_of_complex_numbers [2015/06/26 22:07] nikolaj |
natural_logarithm_of_complex_numbers [2016/05/05 17:06] nikolaj |
||
|---|---|---|---|
| Line 12: | Line 12: | ||
| == Series == | == Series == | ||
| + | $\ln{\left(\frac{1}{1-z}\right)} = \sum_{n=1}^\infty \frac{z^n}{n}$ | ||
| - | $\ln{\left(1+z\right)}=-\sum_{n=1}^\infty \frac{(-z)^n}{n}$ | + | or |
| - | and so, from this | + | $\ln{\left(1+z\right)} = -\sum_{n=1}^\infty \frac{(-z)^n}{n}$ |
| - | $\ln{\left(\frac{1}{1-z}\right)}=\sum_{n=1}^\infty \frac{z^n}{n}$ | + | 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 $\frac{1}{1-z}=1+\sum_{n=1}^\infty z^n$. | ||
