# How do you integrate this complex function

The integration is a closed integral/circulation (I couldn't find out the mathjax for closed integral sign !)

$$\int_c \frac{-3z+4}{z^2+4z+5}\, dz, \space \space \space \space \space \text {where \Bbb c is the circle } |z| = 1$$

My question is
- In order to evaluate this integral, what/which concept(s) do I need to go through?
- Is there a general approach to solve this kind of questions like

• Do a partial fraction
• Do Substitution/ Integrate by parts/ apply formula ...

I have done some elementary calculus and complex analysis but this is something I don't find familiar.

• Residue theorem. Yes, partial fraction decomposition is often part of it, substitutions/integration by parts very rarely. The typical way to evaluate such integrals is the residue theorem. Find the singularities, compute the residues, sum, done. – Daniel Fischer Jan 2 '14 at 17:17
• That is very helpful and a complete answer. I'm going through the residue theorem right now. – vvy Jan 2 '14 at 17:48

You're forgetting to take into consideration the curve $C$! Let's parametrize the curve. We are looking at $|z|=1$. This is the unit circle. So we will parametrize in the 'typical' fashion, $z=e^{it}$ for $t \in [0,2\pi]$. Then $dz=ie^{it}dt$ then what we have is the integral $$\int_C \frac{-3z+4}{z^2+4z+5}\;dz=\int_0^{2\pi} \frac{-3e^{it}+4}{e^{2it}+4e^{it}+5}\;dt$$ Of course, this really isn't much different than the integral we started with. Rewrite the integral as $$\int_0^{2\pi} \frac{10}{e^{2it}+4e^{it}+5}-\frac{3(2e^{it}+4)}{2(e^{2it}+4e^{it}+5)}\;dt$$ The first term can be integrated by completing the square, making a substitution, then using inverse functions (a standard technique). The second term can be solved using a simple $u$-substitution.

This is the long way of doing this problem. Since you hinted at only an elementary approach to the problem, this method only requires basic Calculus techniques after the initial parametrization. However, the fast way of doing this problem is to use the Residue Theorem. The denominator is only zero when $z^2+4z+5=0$, that is when $z=-2\pm i$. These "correspond" to the points $(-2,\pm 1)$ in $\mathbb{R}^2$. However, neither of these are inside our closed curve $C$-the unit circle. Therefore, the Residue Theorem says.........

The integral is $0$. All that work for nothing!

• This reminds me of "differentiation using the first principle". The transformation $|z| \text{ to } e^{i\theta}$ is very helpful. Thanks for the insight! – vvy Jan 2 '14 at 17:55
• It's not so much a transformation as how we have to do contour integration. Luckily, in complex analysis there are many powerful theorems-such as the Residue Theorem-to avoid having to do these long and complicated integrations. – mathematics2x2life Jan 2 '14 at 17:57

The only poles of this rational function are at $-2\pm i$. Those are OUTSIDE the circle. So what's the integral around a curve of any function that's holomorphic on and inside the curve?

• On third glance I get what you mean. Thanks ! And you have given me more food(holomorphic) to digest! – vvy Jan 2 '14 at 17:49
• . . . in other words, you see at once that the integral is $0$ without doing any more than observing that those poles are outside the circle. – Michael Hardy Jan 2 '14 at 21:45
• Hi, I saw this video and I think now I fully understand your answer. The Residue Theorem is awesome! – vvy Jan 3 '14 at 18:15