Exercise: Find the Taylor series of the function $\log(z)$ on the disk $|z-1|<1$ of his principal branch. Then, continue analytically the function along the curve $\gamma:z(t)=e^{it},\, 0\leq t \leq 2\pi$.

My attempt: I've already find the Taylor series of $\log(z)$, which is

$$ \log(z) = \sum_{n=1}^{\infty} \frac{f^{(n)}(1)}{n!}(z-1)^n = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(z-1)^n}{n}, \, |z-1|<1. $$

Now, i've been looking all the theory about analytic continuation along curves but still don't understand, i've never seen any example at all. I know that if

$$D_1=\{z\in \mathbb{C}:|z-1|=1\}, \quad D_0=\{z\in \mathbb{C}:|z|=1\} $$

the curve $\gamma$ has the intersection point $a=(1/2,\sqrt{3}/2)$ of $D_0,D_1$. Now, i can try to find an analytic function $f_1$ on the disk $D_{a}$ centered at $a$ of radius $1/2$ such that $f_1$ is analytic and $f_1 \equiv log(z)$ on $D_0 \cap D_a$. Should i do this with every circle centered on a point of $\gamma$? I still can't understand the process of this type of exercises (then, i've to do the same with the function $\sqrt{z}$).

Thank you for your time

  • 1
    $\begingroup$ In general you have to cover your curve by a sequence of open balls such that two successive balls have a non empty intersection and the function coincides on this intersection.. But here it is easier, since where it is analytic, the derivative of $\log(z)$ is $1/z$ which is meromorphic (having only one possible analytic continuation). So all you have to do is looking at $\int_a^z \frac{1}{s}ds$. And as you probably know $\int_a^b \frac{1}{z}dz = \ln |b/a| + i \text{ arg}(b-a)+2i k \pi$ for some $k\in \mathbb{Z}$ $\endgroup$ – reuns Nov 26 '16 at 0:57
  • $\begingroup$ @user1952009 i don't understand, if $\log(z)$ is analytic on $\mathcal{R}$ (his Riemann surface) why i can't use $f(z)= \log(z)$ as the analytic continuation for all the balls centered at a point of $\gamma$ of radius 1? $\endgroup$ – Jeybe Nov 29 '16 at 14:51
  • 1
    $\begingroup$ Nobody said you can't. But at first you have to define $\log(z)$, this is what I did : as the local primitive of $1/z$ with specified value $\ln|a|+i \text{ arg}(a)+2i k \pi$ at $z=a$ $\endgroup$ – reuns Nov 29 '16 at 15:24

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