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Let \(f(z)\) be analytic inside and on a simple closed contour \(C\) except for a finite number of poles inside \(C .\) Denote the zeros by \(z_{1}, \ldots, z_{n}\) (none of which lies on \(C\) ) and the poles by \(w_{1}, \ldots, w_{m} .\) If \(g(z)\) is analytic inside and on \(C\), prove that $$ \frac{1}{2 \pi i} \int_{C} g(z) \frac{f^{\prime}(z)}{f(z)} d z=\sum_{j=1}^{n} g\left(z_{j}\right)-\sum_{j=1}^{m} g\left(w_{j}\right) $$ where each zero and pole occurs as often in the sum as is required by its multiplicity.

Short Answer

Expert verified
The given equation is proven using the residue theorem and the properties of a contour integral in complex analysis. The crucial part in this proof is calculating the residues at the zeros and poles of \( f(z) \).
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Step 1: Calculation of Contour Integral

Let's consider the function \( g(z) \times \frac{f^{\prime}(z)}{f(z)} \). Using the residue theorem we obtain that the contour integral of this function around \( C \) equals \( 2 \pi i \) times the sum of the residues of this function at its poles. These poles are the zeros and poles of \( f(z) \), i.e., the points \( z_1, \ldots, z_n \) and \( w_1, \ldots, w_m \).

Step 2: Evaluating the Residues

The residue of the function \( g(z) \times \frac{f^{\prime}(z)}{f(z)} \) at any pole or zero of \( f \) is the limit of \( (z - z_j) g(z) \) or \( (z - w_j) g(z) \), respectively, as \( z \) approaches \( z_j \) or \( w_j \). Therefore, the residue is \( g(z_j) \) or \( -g(w_j) \) at a zero or pole, respectively.

Step 3: Substituting Residues into Residue Theorem

Substituting the values of residues from Step 2 into the result of Step 1 gives: \( 2 \pi i \times (\sum_{j=1}^{n} g(z_j) - \sum_{j=1}^{m} g(w_j)) \).

Step 4: Simplify the Equation

To make both sides of the given equation to match, we divide both sides by \( 2 \pi i \), which gives: \( \frac{1}{2 \pi i} \int_{C} g(z) \frac{f^{\prime}(z)}{f(z)} dz = \sum_{j=1}^{n} g(z_j) - \sum_{j=1}^{m} g(w_j) \), exactly as given in the exercise.

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