I'm working now on Complex Integration, and I'm on Cauchy's Theorem for Simply Connected Domains.
I have issue about a simple question. It's about a function's integral over a rectangle-like piecewise smooth path in the Complex Plane.
The question is as follows:
Given a rectangle $R$ with vertices: $(-2-i),(2-i),(2+i),(-2+i)$, and $z_0=5i$. What is the value of $$\int _{ \partial R }^{ }{ \frac { 1 }{ z-{ z }_{ 0 } } dz } $$
The answer provided is as follows:
As $z_0$ is not contained in $R$, we can find a disk $D$ (for ex: $\left| z \right| =\sqrt { 5 } $) which contains $R$ but not $z_0$, and we verify that $f(z) = \frac { 1 }{ z-{ z }_{ 0 } }$ is analytic in $D$. Furthermore, as $\partial R$ is a piecewise smooth and closed curve in $D$, by Cauchy's Theorem for Simply Connected Domains, $$\int _{ \partial R }^{ }{ \frac { 1 }{ z-{ z }_{ 0 } } dz } = 0$$
I have no trouble with the answer to the question, but when I tried to proceed straightforwardly (ie. by direct definition's method) evaluating it on my own, I reached to an integral expression which I couldn't resolve.
What I did is as follows:
I defined $f: \mathbb{C}\setminus \{ { z }_{ 0 }\} \rightarrow \mathbb{C} \setminus \{ { z }_{ 0 }\} $ with $z\mapsto f(z)=\frac { 1 }{ z-{ z }_{ 0 } } $,
I defined parametrization for my path over which to integrate, ie. circle $\left| z \right| =\sqrt { 5 } $, which gave: $\gamma (t)=\sqrt { 5 } { e }^{ it }$ with $0\le t\le 2\pi $ (also verifying indeed that $z_0$ is not a point of this curve),
So that I have:
$$\int _{ \partial R }^{ }{ \frac { 1 }{ z-{ z }_{ 0 } } dz } = \int _{ \gamma }^{ }{ \frac { 1 }{ z-{ z }_{ 0 } } dz } = \int _{ 0 }^{ 2\pi }{ [f\circ \gamma (t)]{ \gamma }^{ \prime }(t)dt }\quad (1)$$
Then,
$$ f\circ \gamma (t)=\frac { 1 }{ \sqrt { 5 } { e }^{ it }-5i } =(...)=\frac { \sqrt { 5 } }{ { e }^{ it }-\sqrt { 5 } i } $$
By inserting the previous expression in the integral, we have,
$$ (1)\Leftrightarrow \int _{ 0 }^{ 2\pi }{ \frac { \sqrt { 5 } }{ { e }^{ it }-\sqrt { 5 } i } \sqrt { 5 } i{ e }^{ it }dt } $$
$$ (...) $$
$$ (1)\Leftrightarrow 5\int _{ 0 }^{ 2\pi }{ \frac { i-{ e }^{ it } }{ 5+{ e }^{ 2it } } dt } $$
I skipped arithmetic calculations because we can proceed by many other ways between steps I mentioned.
I kindly want you to show me how I can resolve -if possible- this last integral expression.
Your integral in $(1)$ can be written as $$ i \int \limits_0^{2\pi} \frac{e^{it}}{e^{it} - \sqrt{5}i} dt $$ Substitute $u = e^{it}$, then $dt = \frac{du}{iu}$, yielding $$ i \int \limits_0^{2\pi} \frac{e^{it}}{e^{it} - \sqrt{5}i} dt = i \int \limits_{e^0}^{e^{2 i\pi}} \frac{u}{u - \sqrt{5}i} \frac{du}{iu} = \int \limits_1^1 \frac{du}{u - \sqrt{5}i} = 0 $$ by the definition of the (Riemann) integral, since the integral bounds coincide. The integral bounds originate from the applied substitution due to $e^{0} = e^{2 i \pi} = 1$.