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

Expert verified
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

01

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.
02

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.
03

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.

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