Understanding the proof of Laurent's theorem

Question

I am reading the proof of Laurent's theorem from the book ''A first course in complex analysis with applications'' by Dennis G. Zill. Line 3 of the proof says the introduction of a crosscut between $C_2$ and $C_1$. I want to know why we have to introduce crosscut to prove this theorem? I am also not able to visualize this fact. I would be really grateful if someone can help me to understand this fact. Also, if it is possible, please help me in visualizing the need of crosscut to prove this theorem.

Answer 1

Recall, Cauchy's integral formula is valid when $C$ is a closed, rectifiable curve.

Since we need a closed curve, we connect $C_1$ and $C_2$ via a curve $C_3$ (we may think of a segment connecting $C_1$ and $C_2$).

A crosscut between $C_1$ and $C_2$ is a curve from a point $A$ of $C_1$ to a point $B$ in $C_2$.

We obtain a closed curve by

• starting at $A$, walking along $C_3$ to $B$,

• following $C_2$ counterclockwise back to $B$,

• walking back $C_3$ to $A$

• and finally going clockwise along $C_1$ till we finally stop in $A$ again.

                                                  

With a crosscut $C_3$ between the concentric circles $C_1$ and $C_2$ we can walk along a closed curve $C$ via \begin{align*} C:=C_3+C_2-C_3-C_1=C_2-C_1 \end{align*} and the function $f(z)$ is analytic in the interior of $C$, so that Cauchy's integral formula is applicable.