Complex Contour Integral

Yes, that is correct. So in summary, the line integrals along the given paths are all equal to zero because the function |z| is not holomorphic, and therefore the value of the integral depends on the path chosen. This is in contrast to holomorphic functions, where the value of the integral is independent of the path chosen.
  • #1
kreil
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Homework Statement



Calculate the following line integrals from point z'=(0,-1) to z"=(0,1) along three different contours, [itex]C_j=(0,1,2)[/itex].

[tex]\int_{C_j}|z|dz[/tex]

where [itex]C_0[/itex] is the straight line along the y-axis, [itex]C_1[/itex] is the right semi-circular contour of radius 1, and [itex]C_2[/itex] is the left semi-circular contour of radius 1.

The Attempt at a Solution



(i) Along [itex]C_0[/itex], [tex]z=iy \implies dz = idy[/tex] and the integral is

[tex]\int_{C_0}|z|dz=i^2 \int_{-1}^1ydy=-\frac{y^2}{2}|_{-1}^1=-\frac{1}{2}+\frac{1}{2}=0[/tex](ii) Along [itex]C_1[/itex], [itex]z=re^{i \theta} \implies dz = ire^{i \theta}d \theta[/tex] with [itex]\theta:\frac{3 \pi}{2} \rightarrow \frac{\pi}{2}[/itex]. Note that r=1.

So, [tex]\int_{C_1}|z|dz = ir^2\int_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}e^{2i \theta}d \theta=\frac{1}{2}e^{2 i \theta}|_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}=\frac{1}{2} ( e^{i \pi}-e^{3i \pi})=0[/tex](iii) Along [itex]C_2[/itex], [tex]z=re^{i \theta} \implies dz = ire^{i \theta}d \theta[/tex] with [itex]\theta:-\frac{\pi}{2} \rightarrow \frac{\pi}{2}[/itex].

The integral is similar to (ii), and one obtains:

[tex]\frac{1}{2} ( e^{i \pi}-e^{-i \pi})=0[/tex]

Did I do these integrals correctly (correct limits in ii and iii)? If so then geometrically, why are these integrals equal to zero?

Thanks for your comments.
 
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  • #2
For (ii) and (iii), don't forget that you are integrating |z|, not z.

The point of the problem is to show you that the path integral does depend on the path you choose. You will see later that the value of a path integral is independent of the path if a function is holomorphic. That is because holomorphic functions have antiderivatives. It's the same theorem as in multivariable calculus, when you learned that the value of a path integral over a vector field depends only on the start and end points if the vector field is the gradient of a function.

In this example, the value of the integrals does depend on the path because |z| is not holomorphic.
 
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  • #3
oops forgot i was integrating |z|..i fixed (ii) quickly and got an answer of 2i, is that correct?

[tex]
\int_{C_1}|z|dz = ir^2\int_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}e^{i \theta}d \theta=i [sin(\theta)-icos(\theta)]|_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}=2i
[/tex]
 
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1. What is a complex contour integral?

A complex contour integral is a mathematical concept used in complex analysis to calculate the value of a function along a certain path in the complex plane. It involves integrating a complex-valued function over a closed curve or contour, which can be a line, circle, or any other shape. The result of the integral is a complex number.

2. Why are complex contour integrals important?

Complex contour integrals are important because they allow us to evaluate complex functions by breaking them down into simpler components. They also have many applications in physics, engineering, and other fields, such as calculating electric field and fluid flow in complex systems.

3. How do you calculate a complex contour integral?

To calculate a complex contour integral, you first need to parameterize the contour using a complex-valued function. Then, you integrate the function along the contour, which can be done using techniques such as the Cauchy Integral Theorem or Cauchy's Residue Theorem. Finally, you evaluate the integral to obtain a complex number.

4. What are some common types of contours used in complex contour integrals?

Some common types of contours used in complex contour integrals include circles, straight lines, and closed curves such as rectangles or ellipses. Other more complex contours can also be used, depending on the function being integrated.

5. Are there any tips for solving complex contour integrals?

One tip for solving complex contour integrals is to carefully choose the contour based on the function being integrated. It can also be helpful to break down the contour into simpler segments and evaluate each segment separately. Additionally, understanding the properties of the function, such as poles and branch cuts, can aid in solving the integral.

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