Hi guys just wondering if anyone can help simplifying Cauchy Integral Formula for me with the following question:
Evaluate:
$\ \int_C fdz $ where $\ f(z) = \frac {1} {z-3} $ and C is:
i) any simple closed curve in a counterclockwise direction with $\ z = 3 $ inside C;
ii) any simple closed curve in a counterclockwise direction with $\ z = 3 $ outside C.
What I would do is using $\ f = 1 $ and $\ z_0=3$ but where I'm confused is if I sub $\ z = 3$ in is this not undefined?
Seeing as the Cauchy Integral Formula is :
$$\ \int_C \frac {f(z)}{z-z_0}dz = 2\pi i f(z_0) $$
does this mean my answer is just $\ 6\pi i? $ (I'm just thinking outload here)
and any help with the second part would be brilliant also! Thanks
Best Answer
Let us first consider Part (ii).
As the only singularity of the function $f(z)$ is at $z = 3$, any integral over a closed curve that does not contain the point $z = 3$ is just $0$, as there are no singularities contained within the curve.
Therefore, $\oint_C f(z)dz = 0$ for any closed curve $C$ that doesn't contain the point $z = 3$.
Now we consider Part (i), where $z = 3$ is contained within the curve. Here, one must evaluate the Residue of $f(z)$ the point $z = 3$.
The Residue is easily evaluated, as we only have to deal with a simple pole. It is given by: $$\text{Res}_3 f(z) = 2\pi i\lim\limits_{z \rightarrow 3}{(z - 3)f(z)} = 2\pi i$$
Thus, in this case, $\oint_C f(z)dz = 2\pi i$, if I haven't made any mistakes (no guarantees there, I'm pretty new to Residue Calculus myself).