Question

Show that $$ \int_{C} \frac{d z}{(z-1-i)^{n+1}}=\left\\{\begin{array}{cl} 0, & n \neq 0 \\ 2 \pi i, & n=0 \end{array}\right. $$ for \(C\) the boundary of the square \(0 \leq x \leq 2,0 \leq y \leq 2\) taken counterclockwise. [Hint: Use the fact that contours can be deformed into simpler shapes (like a circle) as long as the integrand is analytic in the region between them. After picking a simpler contour, integrate using parametrization.].

Short answer

For the given complex integral, when \(n = 0\), the integral evaluates to \(2 \pi i\), and for \(n ≠ 0\), the integral evaluates to 0.

Step-by-step solution

Deformation of Contour

First, change the contour from the square to a circle centered at \(z=1+i\). This is possible since between the square and the circle, the integrand function is analytic (has no singularities or poles). The radius of the circle should be such that it lies within the square.

Parametrization of Circle

Parametrize the complex variable \(z\) on the circle with center at \(1+i\) and radius \(r>0\) such that \(z = 1+i+re^{it}\), where \(t\) is the parameter (angle in polar coordinates) varying from \(0\) to \(2\pi\). This describes a counterclockwise circle around the complex number \(1+i\). Compute \(dz = ire^{it}dt\). Substitute these into the integral expression.

Evaluate the Integral

Now evaluate the integral for the two cases \(n = 0\) and \(n ≠ 0\). For \(n = 0\), the integral is direct, yielding \(2 \pi i\). For \(n ≠ 0\), use the power series expansion \(1/(1+x) = \sum_{k=0}^{\infty}(-x)^k\), replace \(x\) by \(re^{it}\), and integrate term-by-term. Here, all terms except the zeroth vanish, leading to an total integral of 0.

Key concepts

Contour Integration

Contour integration is a method in complex analysis used to evaluate integrals of complex functions over a specific path, or "contour," in the complex plane. The main advantage of contour integration is that it provides a way to integrate functions that might be difficult to handle using real analysis alone.
By considering the path along which the integration is performed, we can often simplify complicated integrals.

• In the given exercise, we start with a square contour in the complex plane.
• We are asked to evaluate the integral of a function over this contour.
• Contour integration allows us to deform the path into a simpler shape, like a circle, without changing the integral's value.

As long as the function is analytic between the original and deformed contours - which means it has no singularities or discontinuities - the integral remains unchanged. This highlights the power and elegance of contour integration in solving complex integrals.

Analytic Functions

An analytic function in complex analysis is a function that is locally given by a convergent power series. These functions exhibit a number of interesting and useful properties, including being infinitely differentiable and having derivatives expressed as power series themselves.
In our exercise, the function \(\frac{1}{(z-1-i)^{n+1}}\) is considered, where its analytic nature plays an essential role.

• An important feature of analytic functions is that they don't have poles or singularities within the region of interest.
• For contour deformation, this characteristic ensures the integral’s value remains unchanged when the path of integration is altered.

Understanding whether a function is analytic or not is crucial for applying various techniques in contour integration, especially the Residue Theorem.

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis for evaluating contour integrals. It connects the integral of a function around a closed contour to the sum of residues of the function’s singularities inside the contour.
The residues represent the coefficients of the \(\frac{1}{z}\) term when the function is expanded into a Laurent series around a singularity.

• If a function has no singularities inside the contour, like in the case where \(n eq 0\), the integral yields zero, as per our problem's result.
• When \(n = 0\), the function \[\frac{1}{(z-1-i)}\] has a simple pole at \(z = 1+i\) inside the contour, contributing a residue that results in an integral of \(2 \pi i\).

This theorem simplifies the evaluation of complex integrals, making it an essential concept in solving contour integration problems.

Complex Contours

Complex contours are essentially the paths in the complex plane over which integrals are evaluated in complex analysis. These paths can be lines, circles, or any other irregular shape. In our exercise, the contour started as a square and was deformed into a circle to facilitate the integration.
Some key concepts about complex contours include:

• The ability to deform contours helps manage the complexity of integration, as simpler shapes often make computation more straightforward.
• As long as the function is analytic, you can switch between these contours without altering the result of the integral.
• This flexibility is particularly useful for circumventing singularities and focusing on regions where the function behaves nicely.

Hence, understanding and utilizing complex contours is fundamental for effectively applying complex analysis techniques, such as contour integration and the Residue Theorem.