Visualizing and solving Complex Integrals.pptx

This MATLAB sample assignment simplifies complex mathematical concepts through practical programming. It covers Cauchy's integral formula, numerical integration, contour visualization, and symbolic computation. This example demonstrates how MATLAB efficiently addresses complex analysis problems, offering a comprehensive understanding of both theory and practice.

1. Title: Visualizing and Solving Complex Integrals with MATLAB Tools Visit: www.matlabhomeworkhelper.com Email: [email protected] Phone: [phone]

2. Introduction This sample assignment from MATLABHomeworkHelper.com simplifies advanced mathematical concepts through practical programming applications. It delves into complex analysis techniques, focusing on Cauchy's integral formula to evaluate integrals over specific contours in the complex plane. Utilizing MATLAB, you’ll numerically integrate complex functions, visualize contours, and apply symbolic computation to handle singularities and validate formulas. This example enhances your understanding of how MATLAB can efficiently solve complex analysis problems, ensuring a well-rounded grasp of both theoretical and practical aspects.

3. Problem 1

4. Solution a)

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6. Solution b )

7. Solution c)

8. Solution c)

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11. Solution MATLAB code for part (a): syms z f = sin(pi*z^2) + cos (pi*z^2); integrand = f / ((z - 1) * (z - 2)); residue_at_1 = subs(f, z, 1) / (1 - 2); residue_at_2 = subs(f, z, 2) / (2 - 1); integral_value = 2 * pi * 1i * (residue_at_1 + residue_at_2); disp ( vpa ( integral_value ));

12. Solution MATLAB code for part (b): syms z f = (z^2 + 1) / z^2; % Calculate residue at z = 0 (pole of order 2) residue = limit(diff(f * z^2, z, 1) / factorial(1), z, 0); integral_value = 2 * pi * 1i * residue; disp ( vpa ( integral_value ));

13. Solution MATLAB code for part (c): syms z f = z^2; integrand = f / (z - 1); residue_at_1 = subs(f, z, 1); integral_value = 2 * pi * 1i * residue_at_1; disp ( vpa ( integral_value ));

14. Solution MATLAB code for part (d): % Define the function z(t) along the contour z_contour = @(theta) (1 - 0.1 * cos (100 * theta)) .* exp (1i * theta); % Define the integrand z^2 dz / dtheta ( dz / dtheta is needed for the line integral) integrand = @(theta) z_contour (theta).^2 .* (1i * (1 - 0.1 * cos (100 * theta)) .* exp (1i * theta) ... - 10 * sin(100 * theta) .* exp (1i * theta)); % Perform the numerical integration over the interval [0, pi] integral_value = integral(@(theta) integrand(theta), 0, pi, ' ArrayValued ', true); disp ('Numerical value of the integral:'); disp ( vpa ( integral_value , 8));

15. Problem 2 Let 𝑓(𝑧) = 𝑧𝑛, where 𝑛 is a positive integer. By directly computing the integral, show that Cauchy’s integral formula holds for 𝑓(𝑧0) and Cauchy’s formula for derivatives holds for 𝑓′ (𝑧0). You may need the binomial formula for expanding (𝑎 + 𝑏)𝑛. As a hint: you may want to make a short argument, based on Cauchy’s theorem, reducing the integrals to circles centered on the point of interest.

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19. MATLAB Code to Demonstrate These Formulas % Define symbolic variables syms z z0 n % Define the function f(z) = z^n f = z^n ; % Define the integrand for Cauchy's integral formula integrand_cauchy = f / (z - z0); % Simplify the integral over a contour % Assume the contour C is a circle centered at z0 with radius R R = 1; % Can be any positive value contour_param = z0 + R* exp (1i*theta); % Parametrize the contour with theta dz = diff( contour_param , theta); % Derivative dz / dtheta

20. % Substitute the parametrization into the integrand integrand_param = subs( integrand_cauchy , z, contour_param ) * dz ; % Integrate with respect to theta over [0, 2*pi] cauchy_integral = int ( integrand_param , theta, 0, 2*pi); % Simplify the result and divide by (2*pi*i) f_z0 = simplify( cauchy_integral / (2*pi*1i)); % Display the result disp ('Cauchy Integral Formula Result for f(z) = z^n :'); disp (f_z0); % Now, calculate the derivative of f(z) = z^n f_derivative = diff(f, z);

21. % Define the integrand for the derivative formula integrand_derivative = f_derivative / (z - z0); % Substitute the parametrization into the integrand integrand_param_derivative = subs( integrand_derivative , z, contour_param ) * dz ; % Integrate with respect to theta over [0, 2*pi] cauchy_derivative_integral = int ( integrand_param_derivative , theta, 0, 2*pi); % Simplify the result and divide by (2*pi*i) f_prime_z0 = simplify( cauchy_derivative_integral / (2*pi*1i)); % Display the result disp ('Cauchy Integral Formula Result for f''(z) = n*z^{n-1}:'); disp (f_prime_z0);

22. C onclusion In conclusion, this sample assignment from MATLABHomeworkHelper.com highlights the effectiveness of MATLAB in simplifying complex analysis. By focusing on Cauchy's integral formula and exploring numerical integration, contour visualization, and symbolic computation, you gain practical experience in addressing intricate mathematical problems. This exercise not only reinforces your theoretical knowledge but also enhances your ability to solve real-world challenges using MATLAB. Whether handling singularities or validating formulas, you'll develop a well-rounded skill set for efficiently managing complex integrals and other advanced mathematical tasks.