Prove: f(z) = log z cannot be analytic

[Tsunoyukami]

I'm having difficulty completing the last problem of an assignment due tomorrow evening. I feel as if I'm missing something; every time I attempt the problem I get stuck or confused.

"22. Use the conclusion of Exercise 21 and Example 13 of Section 6, Chapter 1, to prove that f(z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve $\gamma$ that surrounds the origin. (Hint: What is the value of $\int_{\gamma} f'(z) dz$?) (from Complex Variables, 2nd. edition by Stephen D. Fisher)


As this problem refers to both Exercise 21 and Example 13 I will summarize them here as well:

"21. Let $\gamma$ be a piecewise smooth simple closed curve, and suppose that F is analytic on some domain containing $\gamma$. [Then] $\int_{\gamma} F'(z) dz = 0$"

"Example 13 Suppose that $\gamma$ is a piecewise smooth positively oriented simple closed curve. The value of the integral

$\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p}$ , p not in $\gamma$ is

$\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p} = 0$, p is outside $\gamma$, or

$\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p} = 1$, p is inside $\gamma$"



I attempted to show that f(z) = log z is analytic by applying the Cauchy-Riemann equations.

$f(z) = log z = ln|z| + iarg(z) = ln|(x^2 + y^2)^\frac{1}{2}| + i arctan(y/x)$
$f(z) = u + iv$ with
$u = ln|(x^2 + y^2)^\frac{1}{2}|$ and
$v = arctan(y/x)$

I then computed the partial derivatives of both u and v with respect to x and y and showed that u and v satisfy the Cauchy-Riemann equations. As a result, I expect f(z) = log z to be analytic.

However, the question asks me to show that f(z) is not analytic...

If I follow the hint given in the question:

$\int_{\gamma} f'(z) dz$
$\int_{\gamma} (log z)' dz$
$\int_{\gamma} \frac{1}{z} dz$

I'm not too sure where I should go from here...if I simply integrate this I "just" get f(z) back (I'm not sure how to account for integrating over the path...should I write:

$\int_{\gamma} \frac{1}{z} dz = \int_{a}^{b} \frac{1}{\gamma(t)} \gamma'(t) dt$

But then how should I proceed? I need to show that such an integral does not equal 0, because by exercise 21 if the integral equals 0 the function is analytic...


Any help will be greatly appreciated! Thanks a lot in advance! :)

 

[Fightfish]

prove that f(z) = log z cannot be analytic on any domain D that contains a piecewise smooth simple closed curve γ that surrounds the origin.

If I follow the hint given in the question:
$\int_{\gamma} f'(z) dz = \int_{\gamma} \frac{1}{z} dz$

"Suppose that $\gamma$ is a piecewise smooth positively oriented simple closed curve. The value of the integral
$\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - p} = 1$, p is inside $\gamma$"

"21. Let $\gamma$ be a piecewise smooth simple closed curve, and suppose that F is analytic on some domain containing $\gamma$. Then $\int_{\gamma} F'(z) dz = 0$"

The answers are there. What is the value of p for the function in your question?

 

[Tsunoyukami]

Oh! The value of p is 0. Then, since the origin (ie. 0) is interior to $\gamma$ by example 13 the value of the integral is non-zero and therefore f(z) = log z is not analytic by exercise 21!

 

[Fightfish]

Yup, you've got it :)

 

[diggory]

I kinda get what yall are sayin but...

where did this example 13 business come from? I have to answer the same question but I don't think we ever learned that relation.

 

[Dick]

where did this example 13 business come from? I have to answer the same question but I don't think we ever learned that relation.

It's the Cauchy Integral Theorem. You've got to have seen it.

 

[diggory]

Aw dang well okay. Thanks for the help though

 

[Bacle2]

Some other ideas, just to illustrate:

1)Notice that the argument is not even continuous in the plane, let alone analytic.

2)If you accept that Logz and e^z are" inverses " , then notice that e^z is not 1-1 in the plane ( it is actually oo->1), so that it cannot have a global inverse in the plane.