Why does $\int_{C_2-C_1} {f(w)-f(z)\over w-z}dw =0$ where $f$ is analytic in an annulus?

Question

In my book the writer wants to prove: If $f$ is analytic in the annulus $A:R_1<|z|<R_2$, then $f$ has a Laurent expansion $f(z)=\sum_{k=-\infty}^\infty a_k z^k$.

Let $C_1, C_2$ represent circles centered at $0$ of radii $r_1,r_2$ respectively, with $R_1<r_1<r_2<R_2$. Fix $z$ with $R_1<|z|<R_2$. Then $$ g(w)={f(w)-f(z)\over w-z } $$ is analytic in $A$, and by Cauchy's theorem $$ \int_{C_1-C_2}g(w)dw=0 .$$

I'm trying to figure out why does $\int_{C_1-C_2}g(w)dw=0$.

ATTAMPT: I know that the following is true (according to Cauchy's theorem) where $f$ is analytic in $D$ and $C_1,C_2\subset D$. But why is it true on our case:

$$ \int_{C_2-C_1}g(w)dw= \\ \int_{C_2}{f(w)\over w-z}dw-2\pi i f(z) -\int_{c_1} {f(z)\over w-z} dw +2\pi i f(z)= \\ = \int_{C_2}{f(w)\over w-z}dw-\int_{c_1} {f(z)\over w-z} dw $$ Here I can use $\int_{c_j} {f(z)\over w-z} dw =2\pi i f(z)$ for $j=1,2$ and that's finish the proof.

Answer 1

The paths $C_1$ and $C_2$ are homotopic in the domain of $\frac{f(z)}{z-w}$. Therefore, by the homotopic version of Cauchy's integral theorem,$$\int_{C_1-C_2}\frac{f(z)}{w-z}\,\mathrm dw=\int_{C_1}\frac{f(z)}{w-z}\,\mathrm dw-\int_{C_2}\frac{f(z)}{w-z}\,\mathrm dw=0.$$

Answer 2

Hint: Show $\int_{C_1-C_2}g(w)\,dw = 0$ for any $g$ holomorphic in the annulus. Then note the function $w\to \dfrac{f(w)-f(z)}{w-z}$ is holomorphic in the annulus.