Why Do These Complex Contour Integrals Equal Zero?

[kreil]

Homework Statement

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

$$\int_{C_j}|z|dz$$

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

The Attempt at a Solution

(i) Along $C_0$, $$z=iy \implies dz = idy$$ and the integral is

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

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

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

$$\frac{1}{2} ( e^{i \pi}-e^{-i \pi})=0$$

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.

 

[owlpride]

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.

 

[kreil]

oops forgot i was integrating |z|..i fixed (ii) quickly and got an answer of 2i, is that correct?

$$\int_{C_1}|z|dz = ir^2\int_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}e^{i \theta}d \theta=i [sin(\theta)-icos(\theta)]|_{\frac{3 \pi}{2}}^{\frac{\pi}{2}}=2i$$