Cauchy's Integral Formula Introduction

This problem requires us to evaluate a complex contour integral using Cauchy's Integral Formula. Cauchy's Integral Formula is a powerful tool in complex analysis for evaluating integrals of analytic functions over closed contours.

The general form of Cauchy's Integral Formula states that if \(f(z)\) is analytic inside and on a simple closed contour \(C\), oriented counterclockwise, and \(a\) is any point inside \(C\), then:

\[ \oint_C \frac{f(z)}{z-a} dz = 2\pi i f(a) \] Therefore, if we need to find the value of \(\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-a} dz\), it simply equals \(f(a)\).

Function \(f(z)\) Analysis

We are given the function \(f(z)=\frac{1}{z+1}-\frac{2}{z+3}\). To evaluate the integral \(\frac{1}{2\pi i}\int_c f(z)dz\), we can split the integral into two parts due to the linearity of integration:

\[ \frac{1}{2\pi i}\int_c f(z)dz = \frac{1}{2\pi i}\left(\int_c \frac{1}{z+1}dz - \int_c \frac{2}{z+3}dz\right) \]

We need to identify the singularities (poles) of each term in the function:

• For the term \(\frac{1}{z+1}\), the singularity is at \(z = -1\).
• For the term \(\frac{2}{z+3}\), the singularity is at \(z = -3\).

Contour \(C\) Definition

The contour \(C\) is described as a counterclockwise path in the z-plane such that \(|z + 1| = 1\).

• This equation represents a circle in the complex plane.
• The center of the circle is \(z = -1\).
• The radius of the circle is \(R = 1\).

Integral Evaluation Step-by-Step

Now, let's evaluate each integral separately based on whether its singularity lies inside or outside the contour \(C\).

Evaluating the First Integral: \(\int_c \frac{1}{z+1}dz\)

• The singularity is at \(z = -1\).
• The contour \(C\) is a circle centered at \(z = -1\) with radius 1.
• Since the singularity \(z = -1\) is the center of the circle and therefore lies *inside* the contour \(C\), we can apply Cauchy's Integral Formula.
• In this case, we can consider \(g(z) = 1\) (which is analytic everywhere) and \(a = -1\).
• Applying the formula: \[ \int_c \frac{1}{z - (-1)} dz = 2\pi i \cdot g(-1) = 2\pi i \cdot 1 = 2\pi i \]

Evaluating the Second Integral: \(\int_c \frac{2}{z+3}dz\)

• The singularity is at \(z = -3\).
• Let's check if this singularity lies inside the contour \(C\). The contour is \(|z + 1| = 1\). The distance from the center of the contour (\(-1\)) to the singularity (\(-3\)) is \(|-3 - (-1)| = |-3 + 1| = |-2| = 2\).
• Since the distance (2) is greater than the radius (1), the singularity \(z = -3\) lies *outside* the contour \(C\).
• According to Cauchy's Integral Theorem, if a function is analytic inside and on a simple closed contour, its integral over that contour is zero.
• The function \(\frac{2}{z+3}\) is analytic everywhere inside and on \(C\) because its only singularity at \(z=-3\) is outside \(C\).
• Therefore: \[ \int_c \frac{2}{z+3}dz = 0 \]

Final Calculation

Now, we combine the results of the two integrals:

\[ \frac{1}{2\pi i}\int_c f(z)dz = \frac{1}{2\pi i}\left(\int_c \frac{1}{z+1}dz - \int_c \frac{2}{z+3}dz\right) \]

Substitute the values we found:

\[ \frac{1}{2\pi i}\int_c f(z)dz = \frac{1}{2\pi i}(2\pi i - 0) \]

\[ \frac{1}{2\pi i}\int_c f(z)dz = \frac{1}{2\pi i}(2\pi i) \]

\[ \frac{1}{2\pi i}\int_c f(z)dz = 1 \]

Thus, the value of the given integral is 1.