Understanding the Cauchy Integration Formula for Analytic Functions

[Peter Alexander]

Hello everyone! I'm having a bit of a problem with comprehension of the Cauchy integration formula. I might be missing some key know-how, so I'm asking for any sort of help and/or guideline on how to tackle similar problems. I thank anyone willing to take a look at my post!

Homework Statement

Compute the given integral $$ f\left(z\right)=\frac{e^{z^{2}}}{z^{2}-6z} $$ using Cauchy integration formula for

1. $\left|z-2\right|=1$
2. $\left|z-2\right|=3$

Homework Equations

My way of tackling problems in mathematics is to write down concise formulas and their explanations. In this particular case, I believe the relevant formulas are

1. According to Cauchy theorem, if $f(z)$ is a complex analytical function and its derivative $f'(z)$ is continuous everywhere within the given region bounded by enclosed curve $C$, then $$
   \oint_{C} f(z)dz=0
   $$
2. Cauchy integration formula states that for $f=f(z)$ being an analytical function defined on a region bounded by enclosed curve $C$, we can write $$f\left(z_{0}\right)=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-z_{0}}dz$$ provided that $z_0$ lays within such region.
3. Lastly, we know that $$\oint_{C}\frac{1}{z-z_{0}}dz=2\pi i$$

The Attempt at a Solution

To begin with, we test analyticity of $f(z)$ using Cauchy-Riemann equations. As it turns out, both equations are satisfied.

Now, this is where issues begin. To clear any ambiguities, CT refers to Cauchy Theorem (check "relevant equations" under point 1) and CIF to Cauchy integration formula (check "relevant equations" under point 2).

For both cases, I looked at singularities (singularity points). As we can see, for the given function $f(z)$ two poles exist; one at $z_{1}=0$ and the other at $z_{2}=6$. We know that in order to use CT, the enclosed path of integration must not include any of the poles/singularities. If it does, we might be able to solve the integral using CIF.

The first case, where $|z-2|=1$ traces a unit circle with center at $z=2$, we can see that no singularities are embraced. This way we state that according to CT, $$\int_{C:\;|z-2|=1}f(z)dz=0$$ For the second case, however, we see that a traced circle with radius 3 and center at $z=2$ does include a singularity point $z_{1}=0$. We initially begin by writing $$f(a)=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-a}dz$$ $$\oint\frac{f(z)}{z-a}dz=2\pi if(a)$$ All we need to do now is determine what $f(z)$ and $z-a$ is. By splitting the denominator into $$\frac{e^{z^{2}}}{z(z-6)}$$ question arises; if $f(z)=e^{z^{2}}$, then what is "$z-a$'' term?

 

[FactChecker]

Consider z-a, where a=0 and f(z) = e^z2/(z-6)

 

[Peter Alexander]

Consider z-a, where a=0 and f(z) = e^z2/(z-6)

It actually helped! Now what I've gathered thus far, is to seek singularities and compute the integral at the point of singularity? Why would you do that/why that works?

Because the same procedure has produced the correct answer in some other tasks I've attempted solving, yet I still don't understand why. Would you be kind enough to elaborate a bit more on that or at least share sources which explain why that works?

 

[FactChecker]

That is a good question. IMO, the answer is profound and beautiful mathematics. It turns out that when a function is analytic in a region, its values inside the entire region are completely determined by its values around the border (or really just inside, where it is analytic). And when taking an integral on a path, the path can be deformed continuously within the region without changing the integral. So in this example, a path integral can be shrunk to a tiny circle around a singularity without changing the answer. The integral around the tiny circle is easy to approximate ( like ≈ 2πi f(a) ) .

 

[Peter Alexander]

This is actually interesting and explains a lot! It is only now that problems I've solved actually make sense. Thank you so much for your time and energy.