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Magazine Chapter 6: Problem 5
Find \(\int_{c}(2 z-1)\left(z^{2}-z\right)^{-1} d z\) for the (a) circle \(C=C_{2}^{+}(0)=\\{z:|z|=2\\}\) having positive orientation. (b) circle \(C=C_{j}^{+}(0)=\left\\{z:|z|=\frac{1}{2}\right\\}\) having positive orientation.
Short Answer
Expert verified
(a) 0
(b) \(-2\pi i\)
Step by step solution
01
Analyze the Integral
We need to evaluate the integral \(\int_{C}(2z-1)(z^{2}-z)^{-1} dz\) over given circles. Notice that the integrand is \(\frac{2z-1}{z(z-1)}\). The singularities are at \(z = 0\) and \(z = 1\). These are the poles of the function.
02
Determine Singularities Location
For part (a), we analyze the circle \(|z| = 2\), which contains both singularities \(z = 0\) and \(z = 1\) inside. For part (b), we analyze the circle \(|z| = \frac{1}{2}\), which contains only the singularity \(z = 0\) since \(|1| > \frac{1}{2}\).
03
Apply Residue Theorem for Circle (a)
By the Residue Theorem, for a function with isolated singularities, the integral of \(f(z)\) around a closed curve is \(2\pi i\) times the sum of the residues inside the curve. For \(|z| = 2\), both 0 and 1 are inside, so we find residues at both.
04
Calculate Residue at z=0
The residue at \(z = 0\) is found by multiplying \((2z -1)\) by \(z\), so \(\lim_{{z \to 0}} \frac{2z-1}{z-1} = -1\).
05
Calculate Residue at z=1
Similarly, the residue at \(z = 1\) is determined by \((2z-1)\), so \(\lim_{{z \to 1}} \frac{2z-1}{z} = 1\).
06
Compute Integral for Circle (a)
The sum of residues for \(|z| = 2\) is \(-1 + 1 = 0\). Hence, \(\int_{C}(2z-1)(z^{2}-z)^{-1} dz = 2\pi i \times 0 = 0\).
07
Apply Residue Theorem for Circle (b)
For the circle \(|z| = \frac{1}{2}\), only \(z = 0\) is inside. Thus, we only consider this residue.
08
Compute Integral for Circle (b)
The residue at \(z = 0\) is \(-1\), so the integral is \(2\pi i \times (-1) = -2\pi i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Residue Theorem
The Residue Theorem is a powerful tool in complex analysis that helps evaluate complex integrals of functions with isolated singularities. Specifically, it states that the integral of a function around a closed contour is equal to \(2\pi i\) times the sum of the residues of the function inside that contour.
Here's a quick way to understand it:
Here's a quick way to understand it:
- Find the singular points of your function, also known as poles, that lie inside the contour of integration.
- Determine the residues at these poles.
- Sum up these residues.
- Multiply the total residue by \(2\pi i\) to get the value of the integral.
Contour Integration
Contour integration involves integrating a complex function over a path, or contour, in the complex plane. When you work with contour integration:
- Select the contour over which the integration is to be performed. Common choices include circles, rectangles, or other simple curves.
- Ensure the function is analytic (differentiable) everywhere except possibly at a set of isolated points within or on the contour.
- Apply theorems from complex analysis, like the Residue Theorem, to evaluate the integral along the specified path.
Singularities
In complex analysis, singularities refer to points at which a function ceases to be analytic. These can be classified mainly into three types:
- Removable Singularities: A point where a function is undefined but can be redefined such that the function becomes analytic.
- Poles: Points where a function goes to infinity. The order of the pole describes how the function behaves near that point (for example, a simple pole at \(z = a\) is of order one).
- Essential Singularities: Points at which the function's behavior is chaotic and can't be simplified to either of the previous types.
