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6. Calculate the following integral \[ \int_{\mathcal{C}} \frac{1}{(z-1)\left(z...
May 14, 2024
Solution by Steps
step 1
To solve the integral ∫C1(z−1)(z3−1) dz\int_{\mathcal{C}} \frac{1}{(z-1)(z^3-1)} \, dz, we need to identify the singularities of the integrand. The singularities occur at the points where the denominator is zero
step 2
The denominator (z−1)(z3−1)(z-1)(z^3-1) has singularities at z=1z=1 and the roots of z3−1=0z^3-1=0. The roots of z3−1=0z^3-1=0 are z=1z=1, z=ωz=\omega, and z=ω2z=\omega^2, where ω=e2πi/3\omega = e^{2\pi i / 3} and ω2=e4πi/3\omega^2 = e^{4\pi i / 3}
step 3
For part (a), the contour C\mathcal{C} is the anticlockwise circle of radius one about the point 3. This contour does not enclose any of the singularities z=1z=1, z=ωz=\omega, or z=ω2z=\omega^2
step 4
Since the contour does not enclose any singularities, by the Cauchy-Goursat theorem, the integral is zero
step 5
For part (b), the contour C\mathcal{C} is the square contour with edges −i,i,2+i-i, i, 2+i, and 2−i2-i, traversed in the anticlockwise direction. This contour encloses the singularity at z=1z=1 but not the singularities at z=ωz=\omega or z=ω2z=\omega^2
step 6
We can use the residue theorem to evaluate the integral. The residue at z=1z=1 is given by Res(1(z−1)(z3−1),z=1)\text{Res}\left(\frac{1}{(z-1)(z^3-1)}, z=1\right)
step 7
To find the residue at z=1z=1, we factor the integrand as 1(z−1)(z−1)(z−ω)(z−ω2)=1(z−1)2(z−ω)(z−ω2)\frac{1}{(z-1)(z-1)(z-\omega)(z-\omega^2)} = \frac{1}{(z-1)^2(z-\omega)(z-\omega^2)}
step 8
The residue at z=1z=1 is the coefficient of 1z−1\frac{1}{z-1} in the Laurent series expansion of the integrand around z=1z=1. This is given by lim⁡z→1(z−1)1(z−1)2(z−ω)(z−ω2)=1(1−ω)(1−ω2)\lim_{z \to 1} (z-1) \frac{1}{(z-1)^2(z-\omega)(z-\omega^2)} = \frac{1}{(1-\omega)(1-\omega^2)}
step 9
Simplifying, we get 1(1−ω)(1−ω2)=1(1−e2πi/3)(1−e4πi/3)\frac{1}{(1-\omega)(1-\omega^2)} = \frac{1}{(1-e^{2\pi i / 3})(1-e^{4\pi i / 3})}
step 10
Using the fact that 1−ω=1−e2πi/31-\omega = 1 - e^{2\pi i / 3} and 1−ω2=1−e4πi/31-\omega^2 = 1 - e^{4\pi i / 3}, we can simplify further
step 11
The integral ∫C1(z−1)(z3−1) dz\int_{\mathcal{C}} \frac{1}{(z-1)(z^3-1)} \, dz for the square contour is 2πi2\pi i times the residue at z=1z=1
step 12
Therefore, the integral is ∫C1(z−1)(z3−1) dz=2πi⋅1(1−ω)(1−ω2)\int_{\mathcal{C}} \frac{1}{(z-1)(z^3-1)} \, dz = 2\pi i \cdot \frac{1}{(1-\omega)(1-\omega^2)}
Answer
(a) The integral is 00.
(b) The integral is 2πi⋅1(1−ω)(1−ω2)2\pi i \cdot \frac{1}{(1-\omega)(1-\omega^2)}.
Key Concept
Residue Theorem
Explanation
The residue theorem allows us to evaluate complex integrals by summing the residues of the integrand's singularities enclosed by the contour.
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