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I'm trying to follow Ahlfors's proof that any analytic function defined in an annulus $R_1 < |z-a| < R_2$ will have a Laurent representation. To do this, he defines two functions:

$$f_1(z) = \frac{1}{2\pi i} \int_{|\zeta-a|=r} \frac{f(\zeta) d\zeta}{\zeta-z} \text{ for $|z-a| < r < R_2$ } $$

$$f_2(z) = - \frac{1}{2\pi i} \int_{|\zeta - a|=r} \frac{f(\zeta)d\zeta}{\zeta-z} \text{ for $R_1 < r <|z-a|$}$$

and he says that it follows by Cauchy's integral theorem that $f(z) = f_1(z) + f_2(z)$. I was wondering if someone could explain to me why this is true.

Also, should I assume that $f_1$ is defined to be $0$ for $|z-a| \geq r$ and $f_2=0$ for $|z-a| \leq r$?

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up vote 3 down vote accepted

You can consider the union $C$ of the two circles (where the outer one is oriented one way and the interior the opposite way) as a cycle (i.e. formal sum of paths with boundary zero). Cauchy's theorem says that if the winding number is zero outside the domain in question (that is, outside the annulus), then the integral over $C$ of $\frac{f(\zeta)}{\zeta - z}$ is $2\pi f(z) n_C(z)$. The way the two circles are oriented is going to imply that $C $ has zero winding numbers about every point outside the annulus, so we get $$f(z) = \frac{1}{2\pi i} \int_C \frac{f(\zeta)}{\zeta - z} d \zeta,$$ which is what you want.

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I think I understand what you're saying. If we use this argument, does that mean that when we consider $C$, the circle in the integral of $f_1$ can actually be contained in the circle in the integral of $f_2$? – user1736 Apr 1 '11 at 0:52
@user1736: Dear user1736, Yes. In fact, I think you may have stated this incorrectly in the question: the paths of integration of $f_1, f_2$ should be different. – Akhil Mathew Apr 1 '11 at 3:06
Sorry, but can you explain why they need to be (I'm assuming that by different you mean that the radii of the two circles need to be different)? It seems like all we require is to have radii satisfy the inequality. Then, the fact that they are oriented differently will make it so that the winding number of the union is zero outside of the annulus. – user1736 Apr 1 '11 at 4:30
@user1736: If they are not different, $f_1$ and $f_2$ will be the same! (The expression $f_1 + f_2 \equiv f$ is only true inside the circles around which one integrates to get $f_1, f_2$, because only in that region do we have the expression $n_C(z) = 1$.) – Akhil Mathew Apr 1 '11 at 4:42
Oh, right. Thanks for the explanation! – user1736 Apr 1 '11 at 16:56

This is confusing because there are two different $r$'s with the same name. Let me restate it this way. Take $R_1 < r_2 < r_1 < R_2$, and for $r_2 < |z - a| < r_1$ define

$$ f_1(z) = \frac{1}{2 \pi i} \int_{|\zeta - a|=r_1} \frac{f(\zeta) \, d\zeta}{\zeta - z}$$

$$ f_2(z) = - \frac{1}{2 \pi i} \int_{|\zeta - a| = r_2} \frac{f(\zeta) \, d\zeta}{\zeta - z}$$

To see that $f(z) = f_1(z) + f_2(z)$, draw two radial line segment joining the two circles (not going through $z$), say from $p_1$ to $p_2$ and $q_1$ to $q_2$ where $|p_k-a| = |q_k - a| = r_k$, and consider the following two contours: $\Gamma_1$ from $q_1$ to $p_1$ counterclockwise on the outer circle, then to $p_2$, clockwise on the inner circle to $q_2$, then to $q_1$, and $\Gamma_2$ from $q_1$ to $q_2$, then clockwise on the inner circle to $p_2$, then to $p_1$, then counterclockwise on the outer circle to $q_1$. In this picture $z$ is inside $\Gamma_1$ but not $\Gamma_2$.

enter image description here

By Cauchy's formula, $\int_{\Gamma_1} \frac{f(\zeta)\, d\zeta}{\zeta - z} = f(z)$ while by Cauchy's theorem, $\int_{\Gamma_2} \frac{f(\zeta)\, d\zeta}{\zeta - z} = 0$. Now note that the sum of these is $f_1(z) + f_2(z)$.

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Wow, thanks for the great picture and spelling everything out so nicely! According to Akhil's argument though, it seems like we could also have $r_1=r_2$, which is a case in which the creation of the two contours that you described wouldn't be possible... – user1736 Apr 1 '11 at 0:55
You really do need $r_2 < |z - a| < r_1$ to define $f(z)$ by those contour integrals. However, at the next step after you get the Laurent series coefficients $a_j = \frac{1}{2 \pi i} \oint_{|\zeta - a| = r_1} \frac{f(\zeta)\, d\zeta}{(\zeta-a)^{j+1}}$ for $j \ge 0$ and $a_j = \frac{1}{2 \pi i} \oint_{|\zeta - a| = r_2} \frac{f(\zeta)\, d\zeta}{(\zeta-a)^{j+1}}$ for $j < 0$, you notice that these don't depend on $r_1$ and $r_2$ as long as they are between $R_1$ and $R_2$. – Robert Israel Apr 1 '11 at 4:58
Ah, that's kind of weird to me, but I see what you're saying. Thanks for your help! – user1736 Apr 1 '11 at 16:56

Here's how I think about Laurent series, which might be helpful.

If $f$ is complex-differentiable on $R_1<|z-a|<R_2$, then any (positively oriented) circle $C_r$ of radius $r$ inside the annulus (i.e. such that $R_1 < r < R_2$) defines a function $f_r(z)=\frac1{2\pi i}\oint_{C_r}\frac{f(\zeta)}{\zeta-z}d\zeta$ analytic everywhere on the disk $|z-a|<R_2$ except the circle $C_r$ itself where the integral is undefined. Because these circles cannot be deformed to a single point it is not necessarily true that $f_r(z)$ equals $f(z)$ for points $z$ inside the circle $C_r$.

Now, if we think about a bigger circle $C_{r'}$ of radius $r'$, i.e. such that $r<r'<R_2$, then $f_{r'}(z)=f_r(z)$ whenever $z$ is inside the smaller circle $C_r$ since we can deform $C_{r'}$ onto $C_r$ without passing through such a $z$, i.e. the two circles are homotopic on the annulus minus $z$, which means that the integrals $\oint_{C_r}\frac{f(\zeta)}{\zeta-z}d\zeta$ and $\oint_{C_{r'}}\frac{f(\zeta)}{\zeta-z}d\zeta$ are equal since $\frac{f(\zeta)}{\zeta-z}$ is complex-differentiable function (of $\zeta$!) everywhere on the annulus except the specific point $z$.

Hence, the power series of $f_r(z)$ around $a$ for any $0<r<R_2$ converges on the whole disk $|z-a|<R_2$ and defines an analytic function $f_1$ on that disk $|z-a|<R_2$. The Taylor Series $f(z)=\sum_{i=0}^\infty a_i(z-a)^i$ in fact gives the the non-negative power part of the Laurent series for $f(z)$ on that annulus, and $f_2$, once we've defined it, will give the negative power part.

So what can we say about $f(z)-f_1(z)$, which is definitely complex-differentiable on that annulus? Well, we can say that $\oint_{C_r}\frac{f(\zeta)-f_1(\zeta)}{\zeta-z}d\zeta=\oint_{C_r}\frac{f(\zeta)}{\zeta-z}-\frac{f_1(\zeta)}{\zeta-z}d\zeta=f_1(z)-f_1(z)=0$ for every $z$ inside $C_r$, which means we cannot use the same trick as before.

But wait! We have forgotten half of the informations given to us by the $f_r(z)$ since $f_1(z)=f_r(z)$ only when $|z-a|<r$ -- what happens when $|z-a|>r$ (recall that $f_r(z)$ is analytic everywhere except the circle $C_r$ given by $|z-a|=r$)? Well, just as before the deformations tell us that $f_r(z)=f_{r'}(z)$ for $|z-a|>r'>r$ so the $f_r(z)$ also define an analytic function $f_2$ on the disk-complement $|z-a|>R_1$ (while $f_1$ was defined on $|z-a|

Note that in particular for $|z-a|>r$ we have $|f_2(z)|<\frac1{2\pi i} \oint_{C_r}\frac{|f(\zeta)|}{|\zeta-z|}=\frac 1{2\pi i}\frac MR$ where $M$ is the largest value $f$ takes on $C_r$ and $R$ is the smallest distance between $z$ and a point on $C_r$. Hence $\lim_{z\to\infty}f_2(z)=0$, so in some sense $f_2$ is in fact analytic at infinity. The sense in which $f_2$ is analytic at infinity is this. The fractional linear transformation $\phi(z)=\frac1{z-a}+a$ is complex-differentiable on the whole complex plane except at $a$, and what it does to a point $a+re^{i\theta}$ is that it sends it to $a+\frac1r e^{i-\theta}$, thus it sends the function $f_2$ analytic outside $C_{R_1}$ to a function $g$ analytic on the inside of $C_{\frac1{R_1}}$, where $g$ is given by $g(z)=f_2(\phi(z))$ for $z\neq a$. Then if the power series expansion of $f_2(\phi(z))=f(\frac1{z-a}+a)$ at $z=a$ is $\sum_{j=1}^\infty b_j(z-a)^j$, we have that in fact $f_2(z)=\sum_{j=1}^\infty b_j(z-a)^{-j}$.

Hence, we can now see that $f_1$ and $f_2$ are simply the unique functions analytic respectively inside the disk $|z-a|<R_2$ and outside (including $\infty$) the disk $|z-a|>R_1$ determined by the values of $f$ on the annulus, with power expansions $f_1(z)=\sum_{i=0}^\infty a_i(z-a)^i$ and $f_2(z)=\sum_{j=1}^\infty b_j(z-a)^{-j}$ respectively the non-negative and negative power part of the Laurent expansion of $f$ in that annulus.

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