integral of $z^{-1}$

Question

I know something similar might been asked, but I learn by example. The assignment is to find the integral of $z^{-1}$ in the square with points $(1+i) , (1-i), (-1-i), (-1+i)$

Do we rewrite $z$ as $a+bi$? How is this done?

Answer 1

Let $\gamma$ be the curve traveling along the perimeter of the square with vertices $(1+i),(-1+i),(-1-i),(1-i)$ traveling around the perimeter once in a counter-clockwise direction. Let us label the edges of the square counterclockwise with edge $1$ being the edge from $(1+i)$ to $(-1+i)$

$\oint_\gamma z^{-1}dz = \int_{\gamma_1} z^{-1}dz + \int_{\gamma_2} z^{-1}dz+\int_{\gamma_3} z^{-1}dz+\int_{\gamma_4} z^{-1}dz$ where $\gamma_n$ is the $n^{th}$ part of the trip traveling along the $n^{th}$ edge.

Let us look at each in more detail. $\int_{\gamma_1} \frac{dz}{z} = \int_1^{-1} \frac{dx}{x+i}$. This comes from realizing that we can think of a complex number $z$ instead as the form $x+iy$ where both $x$ and $y$ are real, and that all along the first edge, $y=1$.

We have $\int_{\gamma_1}\frac{dz}{z}=\int_1^{-1}\frac{dx}{x+i} = \ln(-1+i)-\ln(1+i) = \ln(\frac{-1+i}{1+i}) = \ln(\frac{2i}{2})=\ln(i)=\frac{i\pi}{2}$

Continuing in this fashion, find the rest of the integrals $\int_{\gamma_2}\frac{dz}{z}, \int_{\gamma_3}\frac{dz}{z}, \int_{\gamma_4}\frac{dz}{z}$ and then add the results.

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Alternatively, using Cauchy's Integral Formula and the calculus of residues, we know that for analytic $f$ over a domain $\Omega$, and the boundary of omega $\partial \Omega$, for each $a \in \Omega^\circ$ we have:

$$f(a) = \frac{1}{2\pi i} \oint_{\partial \Omega} \frac{f(z)}{z-a} dz$$

What this means to us is that if we want to find $\oint \frac{1}{z}dz$ we could write it instead using the above with $f(z)=1$ for all $z$, and $a=0$ as:

$$f(0)=1 = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-0}dz = \frac{1}{2\pi i}\oint_\gamma \frac{1}{z}dz$$

Multiplying both sides by $2\pi i$ we get $\oint_\gamma \frac{1}{z}dz = 2\pi i$ (which should agree with your calculations using the first method).