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
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).