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Question

If f(z) is analytic in a simply connected domain D, then for every closed path C in D:

The correct answer is \(\mathop \smallint \limits_c^{} f\left( z \right)dz = 0\)

Cauchy's Integral Theorem and Analytic Functions

The question revolves around a fundamental concept in complex analysis known as Cauchy's Integral Theorem. To understand this theorem, let's first clarify a few key terms mentioned in the question: an analytic function and a simply connected domain.

Analytic Functions Explained

An analytic function (also known as a holomorphic function) is a complex-valued function of a complex variable that is differentiable at every point within its domain. This means that at each point \(z_0\) in the domain, the derivative \(f'(z_0)\) exists. Functions that are analytic are "smooth" in the complex plane, similar to how differentiable functions are "smooth" in real calculus. Their derivatives exist and are continuous.

Simply Connected Domain

A simply connected domain is a region in the complex plane that has no "holes" and consists of a single piece. More formally, a domain \(D\) is simply connected if every simple closed curve (a loop) within \(D\) can be continuously shrunk to a single point within \(D\) without ever leaving the domain. This property is crucial for many theorems in complex analysis, including Cauchy's Integral Theorem.

Cauchy's Integral Theorem Application

Cauchy's Integral Theorem states the following: If a function \(f(z)\) is analytic at all points inside and on a simple closed contour \(C\) in a simply connected domain \(D\), then the contour integral of \(f(z)\) along \(C\) is zero.

In mathematical notation, this is expressed as:

\[ \mathop \smallint \limits_C^{} f\left( z \right)dz = 0 \]

The question explicitly states that \(f(z)\) is analytic in a simply connected domain \(D\), and it asks for the value of the integral of \(f(z)\) for every closed path \(C\) in \(D\). Given the conditions perfectly match those required by Cauchy's Integral Theorem, the result must be zero.

Therefore, for any closed path \(C\) within the simply connected domain \(D\), where \(f(z)\) is analytic, the integral \(\mathop \smallint \limits_C^{} f\left( z \right)dz\) will always be equal to 0.

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