Why is the domain D being simply connected important for defining the antiderivative $F(z)$ via a contour integral?

Conditions

• The domain D is simply connected.
• The function f is holomorphic on D.
• The antiderivative $F(z)$ is defined as a contour integral from a fixed base point $z_0$ to z.

Reasoning, step by step

1. Identify the definition of the antiderivative $F(z)$ as the contour integral of $f(w)$ along a path $\gamma_z$ from $z_0$ to $z$.
2. Recognize that in the complex plane, there are generally multiple paths between two points.
3. Apply the property of a simply connected domain: any two paths from $z_0$ to $z$ can be continuously deformed into each other without leaving the domain.
4. Invoke Cauchy's Theorem, which states that the integral of a holomorphic function over any closed contour in a simply connected domain is zero.
5. Conclude that the integral along any path from $z_0$ to $z$ is the same, ensuring $F(z)$ is a single-valued, well-defined function.

Example

The lecturer explains that because the domain is simply connected, 'it doesn't matter which way we choose... the integral along every way from z naught to z is the same as the integral along every other way'.

Common misconceptions

• Assuming that contour integrals are always path-independent regardless of the domain's topology.
• Believing that a holomorphic function on a domain with holes (not simply connected) will always have a well-defined single-valued antiderivative defined by path integration.

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Explore next

• What is the explicit formula for the antiderivative of a holomorphic function used in this proof?
• How does the openness of a domain help in calculating derivatives?