Confusions about the definition of "residue" in Ahlfors' Complex Analysis

Question

On the third edition of Ahlfors' Complex Analysis, page 149 it states: (i) Draw a circle $C_j$ about $a_j$ of radius $<\delta_j$, and let $P_j=\int_{C_j}f(z)dz$ be the corresponding period of $f(z)$. (ii) The particular function $1/(z-a_j)$ has the period $2\pi i$. (iii) Therefore, if we set $R_j=P_j/2\pi i$, the combination $f(z)-\frac{R_j}{z-a_j}$ has a vanishing period. The constant $R_j$ which produces this result is called the residue of $f(z)$ at the point $a_j$.

I don't understand statements of (i), (ii) and (iii), which describe something about "period".

(i) What does it mean $P_j=\int_{C_j}f(z)dz$ is the corresponding period of $f(z)$? Does it mean that $f(z+\int_{C_j}f(z)dz)=f(z)$ for all $z$? Why is $f(z+\int_{C_j}f(z)dz)=f(z)$ true?

(ii) Why does $1/(z-a_j)$ have the period $2\pi i$?

(iii) Why does $f(z)-\frac{R_j}{z-a_j}$ have a vanishing period?

Answer 1

If you look back a couple of pages to p.147 you'll find this:

The numbers $P_i=\int_{\gamma_i}f\,dz$ depend only on the function, and not on $\gamma$. They are called modules of periodicity of the differential $f\,dz$, or, with less accuracy, the periods of the indefinite integral. [...] The vanishing of the periods is a necessary and sufficient condition for the existence of a single-valued indefinite integral.

And this is the definition Ahlfors is using on page 149.

In an act of outright sadism, he has put two entries for "period" in the book's index, neither of which points to page 147 and both of which refer to a quite different meaning of the word.

So, to answer your questions more explicitly:

(i) Here he's just recapitulating the definition of "period" (in the homological sense, not the periodic-function sense). What it "corresponds" to is just the particular little circle he's thinking about integrating around.

(ii) $1/(z-a_j)$ has period $2\pi i$ because if you integrate that around the same circle, that's what you get.

(iii) $f(z)-\frac{R_j}{z-a_j}$ has a vanishing period -- i.e., the period is 0 -- i.e., the integral is 0 -- because $R_j$ was chosen to be the value that makes the integral be zero. (The period-in-this-sense of $f(z)-\frac{R_j}{z-a_j}$ is just the difference between the periods of $f(z)$ and $\frac{R_j}{z-a_j}$: the value of an integral is a linear function of the integrand.)