MHB Complex Contour Keyhole Integration Methods

[Amad27]

This is an interesting complex analysis problem; **The figure on the bottom left is what is being referred to,Fig7-10.**

View attachment 3736**Firstly: (1)** How is the branch point $z=0$ at $z=0$?? We have $f(0) = 0$ that is not a discontinuity is it?

**Secondly:(2)** It says that: $AB$ and $GH$ are coincident with the $x-$axis. Then *This isn't really a keyhole but rather a circle isn't it??*

**Thirdly: (3)** They say that:

$$\int_{BDEFG} f(z) dz = \int_{0}^{2\pi} \frac{ (Re^{i\theta})^{p-1}iRe^{i\theta} d\theta }{1 + Re^{i\theta}}$$

This **Third: (3)** is the most important question. How are they doing this transformation? Where does this transformation come from. Why choose $z = Re^{i\theta}$??

--*Same* for the other transformation and other substitutions. Where are what are these substitutions?

Thanks

 

[Euge]

**Firstly: (1)** How is the branch point $z=0$ at $z=0$?? We have $f(0) = 0$ that is not a discontinuity is it?

The reason is because the function $z \mapsto z^{p-1}$ has a branch point at $z = 0$. Since $0 < p < 1$, $p - 1$ is not an integer. So if $f(z) = z^{p-1}$, then $f(e^{0i}) = e^0 = 1$ and $f(e^{2\pi i}) = e^{2\pi i(p-1)} \neq 1$ (since $p - 1$ is not an integer). However, $e^{0i}$ and $e^{2\pi i}$ are both equal to $1$. So $f$ is multi-valued at $z = 0$ and $0$ is a branch point of $f$. More generally, $0$ is a branch point of $g(z) = z^c$ for every non-integer complex number $c$.

**Secondly:(2)** It says that: $AB$ and $GH$ are coincident with the $x-$axis. Then *This isn't really a keyhole but rather a circle isn't it??*

No, it's not a circle. Their contour is the annulus of inner radius $\varepsilon$ and outer radius $R$.

**Thirdly: (3)** They say that:

$$\int_{BDEFG} f(z) dz = \int_{0}^{2\pi} \frac{ (Re^{i\theta})^{p-1}iRe^{i\theta} d\theta }{1 + Re^{i\theta}}$$

This **Third: (3)** is the most important question. How are they doing this transformation? Where does this transformation come from. Why choose $z = Re^{i\theta}$??

Again, they're using the polar representation for complex numbers. Every point on $BDEFG$ is a distance $R$ from the origin and has argument between $0$ and $2\pi$. So every point on $BDEFG$ is of the form $Re^{i\theta}$, where $\theta$ is between $0$ and $2\pi$. If you don't see it this way, then think in terms of real variables. A circle of radius $R$ centered at the origin is parametrized by parametric equations $x = R\cos(\theta)$, $y = R\sin(\theta)$, where $0 \le \theta \le 2\pi$. So every point $(x,y)$ on a circle of radius $R$ is of the form $(R\cos(\theta), R\sin(\theta))$. The point $(R\cos(\theta), R\sin(\theta))$ corresponds to the complex number $Re^{i\theta}$.

--*Same* for the other transformation and other substitutions. Where are what are these substitutions?

You should review polar representation for complex numbers before proceeding.

 

[Amad27]

You should review polar representation for complex numbers before proceeding.

Hello Euge,

Here is a mini review, tell me if this "review," is enough to move on:
(1) $Re^{i\theta} = R[\cos(\theta) + i\sin(\theta)]$ Euler's general formula.
(2) in $z = a + bi$, we have $a = R\cos(\theta)$ and $b = R\sin(\theta)$.
(3) $|z| = (R)\cdot\sqrt{\cos^2(\theta) + \sin^2(\theta)} = R$

So these are the polar representations.

The reason is because the function $z \mapsto z^{p-1}$ has a branch point at $z = 0$. Since $0 < p < 1$, $p - 1$ is not an integer. So if $f(z) = z^{p-1}$, then $f(e^{0i}) = e^0 = 1$ and $f(e^{2\pi i}) = e^{2\pi i(p-1)} \neq 1$ (since $p - 1$ is not an integer). However, $e^{0i}$ and $e^{2\pi i}$ are both equal to $1$. So $f$ is multi-valued at $z = 0$ and $0$ is a branch point of $f$. More generally, $0$ is a branch point of $g(z) = z^c$ for every non-integer complex number $c$.

So a branch point is where a function has multiple values? I am not understanding this so easily. Can you explain a branch point? Online research hasnt helped with this too much, it goes too indepth with disks, and what not. Please?

No, it's not a circle. Their contour is the annulus of inner radius $\varepsilon$ and outer radius $R$.

So it is still not a keyhole? Its kind of like a washer isn't it?

Again, they're using the polar representation for complex numbers. Every point on $BDEFG$ is a distance $R$ from the origin and has argument between $0$ and $2\pi$. So every point on $BDEFG$ is of the form $Re^{i\theta}$, where $\theta$ is between $0$ and $2\pi$. If you don't see it this way, then think in terms of real variables. A circle of radius $R$ centered at the origin is parametrized by parametric equations $x = R\cos(\theta)$, $y = R\sin(\theta)$, where $0 \le \theta \le 2\pi$. So every point $(x,y)$ on a circle of radius $R$ is of the form $(R\cos(\theta), R\sin(\theta))$. The point $(R\cos(\theta), R\sin(\theta))$ corresponds to the complex number $Re^{i\theta}$.

Ok. I can't understand one thing. You are integrating, along a contour correct?
What does this mean? To integrate along a contour?

So when they integrate from 0 to $2\pi$, so

How is : $\displaystyle \int_{BDEFG} = \int_{0}^{2\pi} \frac{(Re^{i\theta})iRe^{i\theta} d\theta}{1 + Re^{i\theta}}$

How are these equal?

I know they use the substitution $z = Re^{i\theta}$

But how do the limits of integration become $0 \to 2\pi$?

Thanks Euge, I will wait for your reply.

 

[Euge]

Hello Euge,

Here is a mini review, tell me if this "review," is enough to move on:
(1) $Re^{i\theta} = R[\cos(\theta) + i\sin(\theta)]$ Euler's general formula.
(2) in $z = a + bi$, we have $a = R\cos(\theta)$ and $b = R\sin(\theta)$.
(3) $|z| = (R)\cdot\sqrt{\cos^2(\theta) + \sin^2(\theta)} = R$

So these are the polar representations.

You have the basic idea, so let's move on. :D

So a branch point is where a function has multiple values? I am not understanding this so easily. Can you explain a branch point? Online research hasnt helped with this too much, it goes too indepth with disks, and what not. Please?

A branch point $z_0$ of a function $f(z)$ in the complex plane is a point whose general argument $\text{arg}(z_0)$ gives more than one value of $f(z_0)$. That may seem strange because a function is supposed to be single-valued. But branch points of a function $f$ are not in the domain of $f$; they belong to a class of singularities known as non-isolated singularities. In more advanced courses in complex variables, branch points are classified in carefully studied, for they are important for understanding the algebraic structure and topology of ramified Riemann surfaces. Don't worry about all this, though. :)

So it is still not a keyhole? Its kind of like a washer isn't it?

Think about the path of the contour. The path traces out a "keyhole".

Ok. I can't understand one thing. You are integrating, along a contour correct?
What does this mean? To integrate along a contour?

You're integrating over a parametric curve. Let $\gamma$ be a continuously differentiable curve in the complex plane, and suppose $\gamma(t)$, $a \le t \le b$ is a parametrization of $\gamma$. Let $f$ be a continuous function on $\gamma$. The integral of $f$ along $\gamma$ is given by

$$\int_{\gamma} f(z)\, dz := \int_a^b f(\gamma(t))\, \dot{\gamma}(t)\, dt$$

So when they integrate from 0 to $2\pi$, so

How is : $\displaystyle \int_{BDEFG} = \int_{0}^{2\pi} \frac{(Re^{i\theta})iRe^{i\theta} d\theta}{1 + Re^{i\theta}}$

How are these equal?

I know they use the substitution $z = Re^{i\theta}$

But how do the limits of integration become $0 \to 2\pi$?

Look back at my previous point. You have $\gamma(\theta) := Re^{i\theta}, 0 \le \theta \le 2\pi$ is a parametrization of $BDEFG$. So

$$\int_{BDEFG} f(z)\, dz = \int_0^{2\pi} f(\gamma(\theta)) \dot{\gamma}(\theta)\, d\theta = \int_0^{2\pi} f(Re^{i\theta}) iRe^{i\theta}\, d\theta$$

where $f(z) = z^{p-1}/(1 + z)$.

 

[Amad27]

You have the basic idea, so let's move on. :DA branch point $z_0$ of a function $f(z)$ in the complex plane is a point whose general argument $\text{arg}(z_0)$ gives more than one value of $f(z_0)$. That may seem strange because a function is supposed to be single-valued. But branch points of a function $f$ are not in the domain of $f$; they belong to a class of singularities known as non-isolated singularities. In more advanced courses in complex variables, branch points are classified in carefully studied, for they are important for understanding the algebraic structure and topology of ramified Riemann surfaces. Don't worry about all this, though. :)Think about the path of the contour. The path traces out a "keyhole".You're integrating over a parametric curve. Let $\gamma$ be a continuously differentiable curve in the complex plane, and suppose $\gamma(t)$, $a \le t \le b$ is a parametrization of $\gamma$. Let $f$ be a continuous function on $\gamma$. The integral of $f$ along $\gamma$ is given by

$$\int_{\gamma} f(z)\, dz := \int_a^b f(\gamma(t))\, \dot{\gamma}(t)\, dt$$
Look back at my previous point. You have $\gamma(\theta) := Re^{i\theta}, 0 \le \theta \le 2\pi$ is a parametrization of $BDEFG$. So

$$\int_{BDEFG} f(z)\, dz = \int_0^{2\pi} f(\gamma(\theta)) \dot{\gamma}(\theta)\, d\theta = \int_0^{2\pi} f(Re^{i\theta}) iRe^{i\theta}\, d\theta$$

where $f(z) = z^{p-1}/(1 + z)$.

Hello Euge, thanks for answering.

Its the problem of my knowledge of "multivalued" function which is ultimately related to sine and cosine and the angle revolutions.

also by the way, even though it is laid out as a keyhole, they say that the AB and GH coincide with the x-axis.

One integral they have is:

$$\oint_{GH} = \int_{R}^{\epsilon} \frac{(xe^{2\pi i})^{p-1} dx}{1 + xe^{2\pi i}} $$

I have two questions:

(1:) How is it possible integral from a length to another length/ $R$ and $\epsilon$ are both lengths.

I have an idea on this. Because this is on the x-axis, they are FIXED values to that endpoint right. On the x-axis, that endpoint doesn't change does it? Same with $\epsilon$ right?

Also,

(2:) they say the revolution has gone $2\pi$ but is there some reason why the $e^{2\pi i}$ is there? $e^{2\pi i} = 1$ so why does that need to be there?

Also,

I am working on a new integral, which I completed:

$$\int_{0}^{\infty} \frac{\log^2(x)}{x^2 + 1} dx$$

I just need someone to see if the proofs for when I prove the contours are $=0$

Here is the link: http://mathhelpboards.com/analysis-50/evaluating-logarithmic-integral-using-complex-analysis-13807.html#post65588

It is quite lengthy, hopefully you can peer review it.

 

[Euge]

Hi Olok,

Your questions are unclear -- could you explain more clearly what you're having problems with?