Problem 10.6_2 Use Cauchy's integral formula to evaluate the following integral. $₁21-1 -dz sino (z) J|2|=1 (Z - π/6)3

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**Problem 10.6_2**

Use Cauchy's integral formula to evaluate the following integral.

\[
\oint_{|z|=1} \frac{\sin^6(z)}{(z - \pi/6)^3} \, dz
\]

**Explanation:**

This problem involves evaluating a contour integral using Cauchy's integral formula. The integral is taken over a path where \(|z| = 1\). The integrand is \(\frac{\sin^6(z)}{(z - \pi/6)^3}\), which has a pole of order 3 at \(z = \pi/6\).

**Cauchy’s Integral Formula**: 
Cauchy’s integral formula allows us to evaluate integrals of functions over closed contours in the complex plane. It states that for a function \(f(z)\) that is analytic inside and on some closed contour \(C\), and for a point \(a\) inside \(C\),

\[
f(a) = \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z - a} \, dz.
\]

The problem here requires the application of a generalized version of Cauchy’s formula for derivatives, due to the presence of a higher-order pole.
Transcribed Image Text:**Problem 10.6_2** Use Cauchy's integral formula to evaluate the following integral. \[ \oint_{|z|=1} \frac{\sin^6(z)}{(z - \pi/6)^3} \, dz \] **Explanation:** This problem involves evaluating a contour integral using Cauchy's integral formula. The integral is taken over a path where \(|z| = 1\). The integrand is \(\frac{\sin^6(z)}{(z - \pi/6)^3}\), which has a pole of order 3 at \(z = \pi/6\). **Cauchy’s Integral Formula**: Cauchy’s integral formula allows us to evaluate integrals of functions over closed contours in the complex plane. It states that for a function \(f(z)\) that is analytic inside and on some closed contour \(C\), and for a point \(a\) inside \(C\), \[ f(a) = \frac{1}{2\pi i} \oint_{C} \frac{f(z)}{z - a} \, dz. \] The problem here requires the application of a generalized version of Cauchy’s formula for derivatives, due to the presence of a higher-order pole.
Expert Solution
Step 1

By using Cauchy's integral formula we have to evaluate the following integral : 

|z|=1sin6(z)(z-π6)3dz

We know the Cauchy integral formula is given by : 

|z|=rf(z)(z-z0)n+1dz=2πif(n)(z0)n!here r is the radius of a circle center at orogin .  and z0 lies inside the circle .

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