How do I calculate the power series of the complex logarithm? How do I calculate the power series of the complex logarithm?
$f(z)=\log(z)$
Next, some derivatives:
$f'(z)=1/z$
$f''(z)=-1/z^2$
$f'''(z)=2/z^3$
$f^{(4)}(z)=-6/z^4$
 A: Around a general point $z=a$, the general formula for Power series is, as @Alex Provost said in a comment, $$f(z) = \sum_n \dfrac{f^{(n)}(a)}{n!}(z-a)^n$$
Try to notice a pattern with your derivatives: $$f'(z)=+\frac{1}{z}$$ $$f''(z)=-\frac{1}{z^2}$$ $$f'''(z)=+\dfrac{2\cdot1}{z^3}$$ $$f^{(4)}(z)=-\frac{3\cdot2\cdot1}{z^4}$$ $$f^{(5)}(z)=+\frac{4\cdot3\cdot2\cdot1}{z^5}$$
Notice how the $n^{\text{th}}$ derivative looks very similar to a factorial of some sort and there are alternating signs. Can you come up for a general formula for $f^{(n)}(z)$ (and hence $f^{(n)}(a)$)? Once you've found that, just put it in the formula and you're practically done (unless you want the radius of convergence or something like that).
A: If $a \neq 0$, then
$$
\log' z = \frac{1}{z} = \frac{1}{a + (z - a)} = \frac{1}{a} \cdot \frac{1}{1 + \frac{z - a}{a}}.
$$
This is a geometric series in $u = -\frac{z - a}{a}$. If $|u| < 1$, i.e., $|z - a| < |a|$, then
$$
\log' z = \sum_{k=0}^{\infty} \frac{(-1)^{k} (z - a)^{k}}{a^{k+1}}.
$$
A series for (a branch of) $\log$ may be found by termwise antidifferentiation:
$$
\log z - \log a
= \int_{a}^{z} \log' t\, dt
= \sum_{k=0}^{\infty} \frac{(-1)^{k} (z - a)^{k+1}}{(k + 1)a^{k+1}},
\qquad |z - a| < |a|.
$$
