If \(C\) is a circle of radius \(\rho\) about \(z_{0},\) show that $$\oint_{C} \frac{d z}{\left(z-z_{0}\right)^{n}}=2 \pi i \quad \text { if } \quad n=1$$ but for any other integral value of \(n\), positive or negative, the integral is zero. Hint: Use the fact that \(z=z_{0}+\rho e^{i \theta}\) on \(C\).

Contour integration is a fundamental technique in complex analysis used to evaluate integrals along a path or contour in the complex plane. In the given exercise, the contour is a circle centered at a point \(z_0\) with radius \(\rho\). This specialized integral is written as \(\oint_C \frac{dz}{(z - z_0)^n}\), where \(C\) denotes the contour path.
This method involves parameterizing the contour (using a function that maps a parameter to the contour), converting the integral into a more manageable form. Understanding when and how to apply contour integration is crucial because it simplifies complex integrals and reveals properties that are not easily seen through direct computation.
The integral around a closed contour is particularly interesting because of its deep connections with other key results in complex analysis, such as the residue theorem.

The residue theorem is a powerful tool in complex analysis that simplifies the evaluation of complex integrals. It states that for a function with isolated singularities inside a contour, the integral around the contour is related to the sum of the residues at these singularities.
In the given exercise, the residue theorem is implicitly used. For \(n = 1\), the function \(\frac{1}{(z - z_0)}\) has a singularity at \(z_0\). The residue at this point is 1, making the contour integral \(2\pi i\) according to the residue theorem.
When \(n eq 1\), the function \(\frac{1}{(z - z_0)^n}\) doesn't contribute a residue in the same way, leading to an integral result of zero for all other values of \(n\). This highlights the significance of the residue theorem in solving problems with complex integration.

The complex exponential function \(e^{i\theta}\) plays a vital role in parameterizing contours, particularly circles in the complex plane. In the exercise, the circle is parameterized by \(z = z_0 + \rho e^{i\theta}\), allowing us to describe any point on the circle using the angle \(\theta\).
This representation simplifies the integral by converting it into an integral over the parameter \(\theta\) from 0 to \(2\pi\), making the function more manageable. The properties of the exponential function, such as periodicity and its behavior under complex conjugation, are crucial to manipulating and simplifying complex integrals.
Understanding the behavior of the complex exponential function is key to solving many integrals in complex analysis. It allows for elegant simplifications that result in exact solutions.

Parameterization is the process of expressing a curve or surface using a parameter. In complex integration, this simplifies computation by converting a complex path integral into a real integral over a parameter, often \(\theta\).
In the exercise, the circle of radius \(\rho\) centered at \(z_0\) is parameterized by \(z = z_0 + \rho e^{i\theta}\). By differentiating, we find \(dz = \rho i e^{i\theta} d\theta\). This parameterization is then substituted into the integral, transforming it into a form that can be evaluated straightforwardly.
The parameterization approach is widely applicable and essential for dealing with contours that have specific geometrical shapes, providing a universal method for transforming and solving complex integrals.