Analytic continuation of holomorphic function along clockwise/counterclockwise path

Question

"Write down (say, as a power series) a holomorphic function $f(z)$ on $D(1, 1)$ which satisfies $f(z)^5 = z$ and $f(1) = 1$. What is the result of analytically continuing $f$ along a path which travels once counterclockwise around the origin, returning to the point $1$? What about if you go $N$ times counterclockwise around the origin, where $N$ is an integer?"

For the analytic continuation, I understand how to do it in the way where I explicitly write down square root functions in successive disks around the origin in terms of polar coordinates. Is there a more general/"abstract" way to do it though?

Thanks in advance!

Answer 1

Is this the more $``$abstract$"$ way you are looking for...?

Let $f$ be any holomorphic function defined in a neighborhood of $1$ which satisfies $f(z)^5 = z$ in that neighborhood. Consider the map $h(z) = z^5$ from $\mathbb{C}$ to $\mathbb{C}$, as well as the path $\gamma: [0,1] \to \mathbb{C}$ given by$$\gamma(t) = \exp(2\pi i \cdot t/5) \cdot f(1).$$$($So, $\gamma$ travels counterclockwise at constant speed around the circle of radius one, starting at $f(1)$ and ending once it has swept out an angle of $2\pi/5$.$)$

Since the values of $\gamma$ all lie in the region $\mathbb{C}\, \backslash\, \{0\}$ where $h$ has nonzero derivative, it follows from the Inverse Function Theorem that for every $t \in [0, 1]$ there is a holomorphic function $f_t$ defined in an open disk around $\gamma(t)^5$ such that $$f_t(z)^5 = z, \text{ }f_t(\gamma(t)^5) = \gamma(t).$$ Now, we claim that we can even guarantee something stronger, namely $f_t(\gamma(s)^5) = \gamma(s)$ for any $s$ such that $\gamma(s)^5$ is in the disk where $f_t$ is defined. Indeed, the quotient $f_t(\gamma(s)^5)/\gamma(s)$ gives a continuous function from the $s$-interval to the finite set of all fifth roots of unity in $\mathbb{C}$; thus this function must be constant. Since it is equal to $1$ when $s=t$, it is therefore identically $1$.

Because of this stronger claim, we deduce that any two of these functions $f_z$ must agree wherever they are both defined. Indeed, we need only apply the uniqueness claim of the Inverse Function Theorem to any point in the common domain of definition to see that these functions agree in a neighborhood of some point; then we use the identity principle to see that they agree wherever both are defined.

By the same reasoning we see that $f_0$ and our original function $f$ agree in an open disk around $\gamma(0)$. It follows that the collection of functions $\{f_t\}$ provides analytic continuation of $f$ along the path $h \circ \gamma$, which goes once counterclockwise around the origin. $($We could pass to a finite subcollection using the compactness of the unit interval.$)$ However, by definition $$f_1(1) = e^{2\pi i/5} \cdot f(1);$$ therefore, by the same kind of uniqueness argument given above, we in fact have$$f_1 = e^{2\pi i/5} \cdot f.$$Thus the result of analytic continuation of $f$ along a circle going counterclockwise around the origin is that $f$ gets multiplied by $e^{2\pi i/5}$. Iterating this, we see that the result of going around $N$ times is to multiply by $$e^{2\pi iN/5}.$$

Answer 2

Let $F_0(z) = z^{1/5} = e^{\log(z)/5}$ (using the principal branch, with branch cut on the negative real axis), and $F_j(z) = e^{2 j \pi i/5} F_0(z)$ for integers $j$. The analytic continuation of $F_j$ across the branch cut (counterclockwise) is $F_{j+1}$.