# Showing that choice of point on a curve is irrelevant to finding $f(z_0)$ in the curve

Before I start I apologize for the horrible title but I have no idea how to title this.

So the problem is as follows:

Let $f(z)$ be analytic in and on a simple closed curve $\Gamma$, and let $f(z)$ have no zeros in or on $\Gamma$. Now let $z_0$ be a point in $\Gamma$, and $z_1$, $z_2$ be two points on $\Gamma$. Let $\gamma_1$ & $\gamma_2$ be the two curves connecting $z_1$ & $z_2$ to $z_0$ respectively. Where the integraion is towards $z_0$ both times show that

$$f(z_1)e^{\int_{\gamma_1}\frac{f'(x)}{f(x)}dx}=f(z_2)e^{\int_{\gamma_2}\frac{f'(x)}{f(x)}dx}$$

And then show that both sides are equal to $f(z_0)$.

I'm given a hint to represent $f(z_2)/f(z_1)$ as an integral along the part of $\Gamma$ connecting these two points.

This is where the problem begins and thats where I would start, but I'm struggling with that.

One idea I had is that since $f(z_0)$ doesn't equal $0$, I could use that I could make a disk large enough centered at $z_0$ such that $e^{w(z)}=f(z)$, where $w(z)$ is essentially the two integral above, just flip the sign.

Another idea I think is better. If I solve the hint, then I could make a new curve $\gamma_3$ using the arc connecting $z_1$ and $z_2$ and $\gamma_1$ and $\gamma_2$, and then use Cauchy's theorem, so that $\int_{\gamma_3}f(z)=0$, and I would split this integral up into the three parts forming this curve if I could figure it out.

Any help is appreciated, thanks in advance.

• Are you familiar with holomorphic logarithm branches? – Blake Oct 17 '15 at 1:38
• @Blake I am not familiar with the term holomorphic but I do know about the logarithm branches but I am not an expert. Any argument using this though I'm sure I could follow – Jeff Oct 17 '15 at 2:53
• Holomorphic functions are what you probably know as analytic functions. "Analytic" and "holomorphic" are interchangeable in complex analysis. It's kind of an interesting history. It was quickly realized that every complex-differentiable function is analytic in the sense that it can be locally represented as a power series, so the term analytic came to mean complex differentiable. Then I suppose holomorphic came about to distinguish the term from real-analytic functions, which are different. – Blake Oct 17 '15 at 2:55
• They are what I know as analytic because that's the definition I would use for analytic, the more you know, thanks. – Jeff Oct 17 '15 at 3:03

This may be naive of me, but $f'/f$ has an anti derivative (some branch of the logarithm, call it $\log f$) defined everywhere in and on the curve, so we could just evaluate each expression at the endpoints:

\begin{align*} f(z_1)e^{\int_{\gamma_1}\frac{f'(z)}{f(z)}dz}&=f(z_1)e^{\log f(z_0)-\log f(z_1)}=f(z_1)\big(f(z_0)/f(z_1)\big)=f(z_0),\\ f(z_2)e^{\int_{\gamma_2}\frac{f'(z)}{f(z)}dz}&=f(z_2)e^{\log f(z_0)-\log f(z_2)}=f(z_2)\big(f(z_0)/f(z_2)\big)=f(z_0). \end{align*} Is there some technical point I'm missing here?

Edit: The previous procedure can be used without having to assume a branch of the logarithm exists on the entire curve--just assume it exists in simply connected neighborhoods of $\gamma_1,\gamma_2$. Also, Here's a method that uses the hint:

Let $\gamma_3$ is a simple curve beginning at $z_1$ and travelling to $z_2$. $f'/f$ is analytic ($f\neq 0$) in neighborhood (which we can assume--through shrinking if necessary--is simply-connected) of the image of $\gamma_3$, so there exists a holomorphic branch of the logarithm, $\log f$ in just this neighborhood. Thus

$$\frac{f(z_2)}{f(z_1)}=e^{\int_{\gamma_3}\frac{f'(z)}{f(z)}dz}.$$

Then

$$f(z_1)e^{\int_{\gamma_{1}}\frac{f'(z)}{f(z)}\,dz}=f(z_2)e^{\int_{\gamma_{1}}\frac{f'(z)}{f(z)}\,dz-\int_{\gamma{3}}\frac{f'(z)}{f(z)}\,dz}=f(z_2)e^{\int_{\gamma_2}\frac{f'(z)}{f(z)}\,dz}$$ This is true because integrating over "$\gamma_1-\gamma_2-\gamma_3$" gives zero by by Cauchy's theorem since $f(z)\neq 0$ everywhere (draw a picture) so it follows that integrating over "$\gamma_1-\gamma_3$" is the same as integrating over $\gamma_2$. Hopefully this (not really that different) method will lend some insight.

• I don't think you missed anything here although it's possible I just missed a technical point as well but it looks good to me. It's essentially saying that the function only depends on the point chosen inside the curve, which is the goal. Thank you very much for that part of the question it helps a lot, I'll have to remember this technique – Jeff Oct 17 '15 at 3:00
• @Jeff No problem! Note the fact that a branch of the logarithm exists here might require some explanation, since you need more than the fact that $f\neq 0$ everywhere. (For example the image of $f$ could be some punctured neighborhood or annulus about the origin; certainly it's nonvanishing in the region but no branch of the logarithm can be defined.) – Blake Oct 17 '15 at 3:08
• That makes sense, I'm sure I could show that a branch of a logarithm exists here with some work, and the example helps it shows me why I have to show it's defined. Thanks once again – Jeff Oct 17 '15 at 3:13
• Let $A$ be the bounded open region whose boundary is $\Gamma$. For each $z\in \Gamma$ there is an open disc $D(z)$ centered at $z$ with $D(z)\subset dom (f)$ and such that $f\ne 0$ on $D(z)$. Now $E=A\cup \Gamma \cup (\cup \{D(z) : z\in \Gamma\})$ is a connected simply-connected open set and $f\ne 0$ on $E$ so there is a branch of $\log$ such that $\log f$ is analytic on $E$. – DanielWainfleet Oct 19 '15 at 2:58
• My previous comment needs an amendment Or a retraction..Simply-connectedness may not hold even if we have only finitely many $D(z)$ covering $\Gamma$. – DanielWainfleet Oct 19 '15 at 3:20

Notice that:

$\displaystyle\frac{f(z_{2})}{f(z_{1})}=\exp({\int_{C}\frac{f'(z)}{f(z)}d\mathrm{z}})$

Where $C$ is the arc going from $z_{1}$ to $z_{2}$ in the curve $\Gamma$. You can now expand the terms and use the theorem of Cauchy to get the result.