MHB Properties of Contour Integrals - Palka Lemma 2.1 (vi) .... ....

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 4: Complex Integration, Section 2.2 Properties of Contour Integrals ...

I need help with some aspects of the proof of Lemma 2.1, part (vi), Section 2.2, Chapter 4 ...

Lemma 2.1, Chapter 4 reads as follows:View attachment 7429
View attachment 7430In the above text from Palka, at the start of the proof of (vi), we read the following:

" ... ... We suppose that $$\int_{ \gamma } f(z)dz \ne 0$$ - (vi) holds trivially otherwise - and set $$u = e^{ -i \theta }$$ , where $$\theta$$ is any argument of $$\int_{ \gamma } f(z)dz$$. Thus $$\lvert u \rvert = 1$$ and $$\left\lvert \int_{ \gamma } f(z)dz \right\rvert = u \int_{ \gamma } f(z)dz$$ ... ... "
My questions are as follows:
Question 1

Is $$u = e^{ -i \theta }$$ a simple change of variable process?
Question 2

What is meant by " ... $$\theta$$ is any argument of $$\int_{ \gamma } f(z)dz$$" ... ?
Question 3

Can someone please explain why/how $$\left\lvert \int_{ \gamma } f(z)dz \right\rvert = u \int_{ \gamma } f(z)dz$$ ... ...?
Help will be much appreciated ..

Peter

 

[Opalg]

Question 1

Is $$u = e^{ -i \theta }$$ a simple change of variable process?

Question 2

What is meant by " ... $$\theta$$ is any argument of $$\int_{ \gamma } f(z)dz$$" ... ?

Question 3

Can someone please explain why/how $$\left\lvert \int_{ \gamma } f(z)dz \right\rvert = u \int_{ \gamma } f(z)dz$$ ... ...?

1. No, it's not a change of variable. $$u = e^{ -i \theta }$$ is a complex number, defined as in the answer to your Question 2.

2. $$\int_{ \gamma } f(z)\,dz$$ is a complex number, so it has a modulus-argument form that we can call $re^{i\theta}$. Here, $r$ is the modulus (or absolute value) of $$\int_{ \gamma } f(z)\,dz$$, and $\theta$ is an argument of $$\int_{ \gamma } f(z)\,dz$$. (The reason for calling it "an" or "any" argument is that the argument is only defined modulo $2\pi$.)

3. So $$\int_{ \gamma } f(z)\,dz = re^{i\theta} = ru^{-1}$$, and therefore $$u\int_{ \gamma } f(z)\,dz = r = \left|\int_{ \gamma } f(z)\,dz \right|.$$

 

[Math Amateur]

1. No, it's not a change of variable. $$u = e^{ -i \theta }$$ is a complex number, defined as in the answer to your Question 2.

2. $$\int_{ \gamma } f(z)\,dz$$ is a complex number, so it has a modulus-argument form that we can call $re^{i\theta}$. Here, $r$ is the modulus (or absolute value) of $$\int_{ \gamma } f(z)\,dz$$, and $\theta$ is an argument of $$\int_{ \gamma } f(z)\,dz$$. (The reason for calling it "an" or "any" argument is that the argument is only defined modulo $2\pi$.)

3. So $$\int_{ \gamma } f(z)\,dz = re^{i\theta} = ru^{-1}$$, and therefore $$u\int_{ \gamma } f(z)\,dz = r = \left|\int_{ \gamma } f(z)\,dz \right|.$$

Thanks for the help, Opalg ...

Peter