Proof that analytic functions defined in an annulus have a Laurent representation 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 $|z-a| \leq r$?
 A: 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.
A: 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 $\Gamma_2$.

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 $f_1(z) + f_2(z)$.  
A: 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.
