Determining whether the Cauchy-Goursat theorem applies

[Bashyboy]

Homework Statement

Let $C$ be the negatively oriented contour $2e^{-i \pi t}$ $0 \le t \le 1$ For which of
the following functions does the Cauchy-Goursat theorem apply so that the integral of the
function along $C$ is zero?

Homework Equations

The Attempt at a Solution

The first function is $f(z) = e^{z^7}$, where $z^7 = e^{7 \log z}$.

Obviously, the contour over which we wish to integrate $f(z)$ is the unit circle. For $f(z)$ to even be integrable, it must first be a function, which can be accomplished by removing a branch so that $\log z$ becomes a function. By removing a branch, we are actually removing a ray which makes some angle $\theta$ with the positive real-axis that begins at the origin and extends away from it. For instance, I could remove the ray which corresponds to the negative real-axis, all those points of the form $(a,0)$, where $a \in \mathbb{R} \setminus \{0\}$; consequently, there can be no points whose angle (argument) is $\theta = \pi$.

Here is why I do not believe one can apply the Cauchy-Goursat theorem: there are singularities interior to the contour (infinitely many, actually), and one singularity on the contour--namely, $(-1,0)$.

Despite this, I believe we can integrate the function over the contour because there is only one singularity point. This would be analogous to an improper integral because we are integrating up to a singularity point, right?

My question is, how would I set up the integral? When $t = 1/2$, we get $C(1/2) = (-1,0)$. So, would I do an improper integral on the variable $t$?

Also, would the integral be the same independent of which branch I choose to remove?

 

[Ray Vickson]

Homework Statement

Let $C$ be the negatively oriented contour $2e^{-i \pi t}$ $0 \le t \le 1$ For which of
the following functions does the Cauchy-Goursat theorem apply so that the integral of the
function along $C$ is zero?

Homework Equations

The Attempt at a Solution

The first function is $f(z) = e^{z^7}$, where $z^7 = e^{7 \log z}$.

Obviously, the contour over which we wish to integrate $f(z)$ is the unit circle. For $f(z)$ to even be integrable, it must first be a function, which can be accomplished by removing a branch so that $\log z$ becomes a function. By removing a branch, we are actually removing a ray which makes some angle $\theta$ with the positive real-axis that begins at the origin and extends away from it. For instance, I could remove the ray which corresponds to the negative real-axis, all those points of the form $(a,0)$, where $a \in \mathbb{R} \setminus \{0\}$; consequently, there can be no points whose angle (argument) is $\theta = \pi$.

Here is why I do not believe one can apply the Cauchy-Goursat theorem: there are singularities interior to the contour (infinitely many, actually), and one singularity on the contour--namely, $(-1,0)$.

Despite this, I believe we can integrate the function over the contour because there is only one singularity point. This would be analogous to an improper integral because we are integrating up to a singularity point, right?

My question is, how would I set up the integral? When $t = 1/2$, we get $C(1/2) = (-1,0)$. So, would I do an improper integral on the variable $t$?

Also, would the integral be the same independent of which branch I choose to remove?

Why would you define $z^7$ as $e^{7 \log z}$? Did YOU do this, or did the person setting the problem force you to use that weird definition?

 

[Bashyboy]

No, my textbook defines in such a way. Here is the book: http://www.uwyo.edu/selden_homepage/4230/complex-jiblm.pdf page 28 definition 183

 

[pasmith]

$z^n$ is single-valued for integer $n$: $\exp(n(\mathrm{Log}(z) + 2im\pi)) = \exp(n \mathrm{Log}(z))$ for any $m \in \mathbb{Z}$, since $\exp(2imn\pi) = 1$.

 

[Ray Vickson]

No, my textbook defines in such a way. Here is the book: http://www.uwyo.edu/selden_homepage/4230/complex-jiblm.pdf page 28 definition 183

But, BY DEFINITION, $z^7 = z \cdot z \cdot z \cdot z \cdot z \cdot z \cdot z$. You only need $z^a = e^{a \log z}$ for non-integer values of $a$.

No, my textbook defines in such a way. Here is the book: http://www.uwyo.edu/selden_homepage/4230/complex-jiblm.pdf page 28 definition 183

But, BY DEFINITION, $z^7 = z \cdot z \cdot z \cdot z \cdot z \cdot z \cdot z$.

You only need $z^a = e^{a \log z}$ for non-integer values of $a$. While the book does not specifically say so, I think the intention is that definition 183 applies to non-integer values of the power---at least, that is how it is explained in all complex variable books I have ever seen before. One hint that this might really be the author's intention is the fact that in various places much before page 28 he writes things like $z^2$, $z^3$, etc, as though he expects the reader to understand what these mean without having to reach page 28.

 

[Bashyboy]

$z^n$ is single-valued for integer $n$: $\exp(n(\mathrm{Log}(z) + 2im\pi)) = \exp(n \mathrm{Log}(z))$ for any $m \in \mathbb{Z}$, since $\exp(2imn\pi) = 1$.

Ah, of course. I should have been able to easily deduce that from the definition. So, then $f(z) = e^{z^7}$ would be analytic in and on the contour, thereby allowing us to apply the Cauchy-Goursat theorem.

Okay, suppose we didn't have such a function, an analytic one, but one for which we had to remove a branch. Would all I said in my first post be valid? For instance, $f(z) = \log z$, where $\log z = \ln|z| + i \arg z$ would be one such function, right?