# contour integration, complex triangle inequality, residue theorem

Source: https://www.maths.ed.ac.uk/~jmf/Teaching/MT3.html by Jose Figueroa-O'Farrill

Background:

p.v. is Cauchy's principal value

which can be "complexified" to the form,

To make use Cauchy's residue theorem, we close the contour with the following curve,

My confusion:

How the integral over the $$C^+_\rho$$ part of the contour vanishes...

Most of my confusion here seems to lie in understanding the referenced equation 2.28, which reads,

Any pointers are appreciated :(

• Don't you understand why 2.28 holds? Or don't you understand how it is applied here? Commented Dec 1, 2019 at 22:48
• Ah of course, thank you for the question. I do not understand how 2.28 holds. I find the step from the 2nd line to the final line to be unclear. Using 2.24 is straight forward, absolute value of each factor of the integrand, followed by dt. I find the pulling of the f(z(t)) out of the integral and this "max" function confusing. Moreover, I am not sure how the integral in the 3rd line is clearly the arc length. And finally, I am confused about why dz appears in the middle term of the boxed inequality line. Commented Dec 2, 2019 at 10:27
• There is no need to wrote "p.v." since the integral converges absolutely. Commented Dec 2, 2019 at 11:36

The inequality (2.28) states that the absolute value of the integral $$\int_\gamma f(z)\,\mathrm dz$$ of a function $$f$$ along a path $$\gamma\colon[a,b]\longrightarrow\mathbb C$$ is smaller than or equal to the product of two numbers:
• the maximum of $$\lvert f\rvert$$ restricted to $$\gamma\bigl([a,b]\bigr)$$;
• the length of $$\gamma$$.
Let us now apply this to your problem. The length of $$C_\rho^+$$ is $$\pi\rho$$. Furthermore, if $$z$$ belongs to the semicircle, then $$\lvert z^2+4\rvert\geqslant\rho^2-4$$ and therefore (if $$\rho>1$$), $$\max\left\lvert\frac1{z^2+4}\right\rvert\leqslant\frac1{\rho^2-4}$$. So,$$\left\lvert\int_{C_\rho^+}\frac{\mathrm dz}{z^2+4}\right\rvert\leqslant\frac{\pi\rho}{\rho^2-4}.$$Since $$\lim_{\rho\to\infty}\frac{\pi\rho}{\rho^2-4}=0$$,$$\lim_{\rho\to\infty}\left\lvert\int_{C_\rho^+}\frac{\mathrm dz}{z^2+4}\right\rvert=0.$$