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Residue Theory Calculator

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Example
Created on 2024-06-20Asked by Olivia Green (Solvelet student)
Evaluate the integral ∫∣z∣=3z2+1z2−4z+4 dz \int_{|z|=3} \frac{z^2 + 1}{z^2 - 4z + 4} \, dz using the residue theorem.

Solution

To evaluate the integral ∫∣z∣=3z2+1z2−4z+4 dz\int_{|z|=3} \frac{z^2 + 1}{z^2 - 4z + 4} \, dz using the residue theorem: 1. **Identify the singularities inside the contour ∣z∣=3|z| = 3:** The singularities are at the roots of z2−4z+4=(z−2)2=0z^2 - 4z + 4 = (z-2)^2 = 0, which gives z=2z = 2. 2. **Calculate the residue at z=2z = 2:** Since z=2z = 2 is a double pole, the residue is found using: Res(z2+1(z−2)2,2)=lim⁡z→2ddz[(z−2)2⋅z2+1(z−2)2]=lim⁡z→2ddz(z2+1)=2z∣z=2=2⋅2=4. \text{Res}\left(\frac{z^2 + 1}{(z-2)^2}, 2\right) = \lim_{z \to 2} \frac{d}{dz}\left[ (z-2)^2 \cdot \frac{z^2 + 1}{(z-2)^2} \right] = \lim_{z \to 2} \frac{d}{dz} (z^2 + 1) = 2z \big|_{z=2} = 2 \cdot 2 = 4. 3. **Apply the residue theorem:** ∫∣z∣=3z2+1z2−4z+4 dz=2πi⋅Res(z2+1z2−4z+4,2)=2πi⋅4=8πi. \int_{|z|=3} \frac{z^2 + 1}{z^2 - 4z + 4} \, dz = 2\pi i \cdot \text{Res}\left(\frac{z^2 + 1}{z^2 - 4z + 4}, 2\right) = 2\pi i \cdot 4 = 8\pi i. Therefore, the integral evaluates to 8πi8\pi i. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Jack Williams on Solvelet
1. Calculate the residue of the function f(z)=z2+1z4+4 f(z) = \frac{z^2 + 1}{z^4 + 4} at the pole z=2i z = 2i .2. Use residue theory to evaluate the integral ∮Czz3+1 dz \oint_C \frac{z}{z^3 + 1} \, dz , where C C is the unit circle centered at the origin.,
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