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Define $$f(z)=\ln r + i\theta$$ on the domain $\{z:r\gt , 0\lt \theta \lt 2\pi\}$.

This domain is just a punctured disk of radius $\ln r$, correct?

How does one determine whether this is analytic, I can't see how I would take the CREs

$$u(r,\theta) = \ln r$$ $$v(r,\theta) = \theta.$$

Should I convert this back to $x+iy$ form and proceed? How can I do such a thing with what appears to be a punctured open neighborhood?

How do I show that the function is analytic and find its derivatives?

Travis Willse
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coal
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  • @Ilham Well they are equivalent, but you are right, that is probably better formatting(although in my post above is how it was written) – coal Apr 27 '15 at 04:01
  • I have already answer this question using the polar form of the Cauchy Riemann equations. Look here http://math.stackexchange.com/questions/1245754/cauchy-riemann-equations-in-polar-form/1245803#1245803 – Alonso Delfín Apr 27 '15 at 04:05
  • @AlonsoDelfín I have not enough rep to comment on your answer, but you should have $\frac{\partial v}{\partial \theta} = \frac{\partial u}{\partial r}\implies 1=\frac1r$ right, so $r=1$? – coal Apr 27 '15 at 04:15
  • @AlonsoDelfín Oh I see, the Jacobian, thanks – coal Apr 27 '15 at 04:19
  • @AlonsoDelfín And what is the derivative once this is known? – coal Apr 27 '15 at 04:21
  • Well that is not in my linked answer but it must be $$ f'(z)=e^{-i\theta}\left( \frac{\partial u}{\partial r} + i \frac{\partial v}{\partial r} \right) =\frac{-i}{z}\left( \frac{\partial u}{\partial \theta} - i \frac{\partial v}{\partial \theta} \right) $$ – Alonso Delfín Apr 27 '15 at 04:26

2 Answers2

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The domain given is $U=\mathbb {C}\setminus [0,\infty).$ I'm not sure why people are talking about a punctured disc.

Assuming the basic properties of $e^z,$ we can do the following. First, $f(re^{it}) = \ln r +it$ is continuous and 1-1 in $U.$ Second, $e^{f(z)} = z$ for all $z\in U.$ So fix a point $a\in U.$ Then for small $h$ we have

$$1 = \frac{e^{f(a+h)} - e^{f(a)}}{h} = \frac{e^{f(a+h)} - e^{f(a)}}{f(a+h)-f(a)}\cdot \frac{f(a+h)-f(a)}{h}.$$

Thus

$$\left (\frac{e^{f(a+h)} - e^{f(a)}}{f(a+h)-f(a)}\right )^{-1} = \frac{f(a+h)-f(a)}{h}.$$

As $h\to 0,$ $f(a+h) \to f(a),$ hence the left side tends one over the derivative of $e^z$ at $f(a).$ Because $(e^z)' = e^z,$ this is just $1/e^{f(a)} = 1/a.$ Therefore $f'(a) = 1/a$ as expected. This shows $f$ is analytic on $U.$

zhw.
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The domain is the disk of radius $r$ with the closed positive real half-axis removed; in particular this is not a punctured disk.

A good way to check analyticity of $f$ is to write the C.-R. equations in polar coordinates. (The alternative is to write $f$ in rectangular coordinates, but it is slightly tricky to handle taking derivatives of the imaginary part of $f$: At least for $\theta \in (0, \frac{\pi}{2})$, the imaginary part of $f$ coincides with $\arctan\left(\frac{y}{x}\right)$, but this latter function is not defined on the $y$-axis, and it does not agree with the imaginary part of $f$ in the other three quadrants.)

Here, $$f(z) = \ln z,$$ where $\ln$ is the branch of the (natural) logarithm function whose argument takes values in $(0, 2 \pi)$, or in rectangular coordinates, $$f(x + iy) = \frac{1}{2} \ln(x^2 + y^2) + i \arg(x, y),$$ where $\arg$ here denotes the function that returns the anticlockwise angle in $(0, 2\pi)$ from the positive $x$-axis to the ray from the origin through $(x, y)$.

Travis Willse
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  • Thank you for the detailed answer, it is much appreciated – coal Apr 27 '15 at 04:07
  • @coal You're welcome, I hope you found it useful. (To be clear, by writing the "C.-R. equations in polar coordinates" I do not mean writing $\frac{\partial u}{\partial r} = \frac{\partial v}{\partial \theta}$, etc. If it is not clear how to do this, it could be worthwhile to ask how to do this as a separate question.) – Travis Willse Apr 27 '15 at 04:11
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    As a comment in the question , In my answer you will find a proof of the Cauchy Riemann equations in polar form an later the answer to your question, here is the link again http://math.stackexchange.com/questions/1245754/cauchy-riemann-equations-in-polar-form/1245803#1245803 – Alonso Delfín Apr 27 '15 at 04:15
  • @Ilham Roughly speaking, $\log$ is holomorphic on both of these "Pacman with closed mouth" regions; this is roughly because even the rules for $\log$ are different on these different sets. For example, in the latter case (on the domain with the negative half-axis removed), we can define an analytic log function by the same formula as in my answer, but insisting that $\arg$ takes values in $(-\pi, \pi)$ instead of $(0, 2 \pi)$. – Travis Willse Apr 27 '15 at 04:18
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    @AlonsoDelfín Thanks for linking this, this is exactly what I had in mind. – Travis Willse Apr 27 '15 at 04:20
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    (@AlonsoDelfín In fact, it is even better than what I had in mind, as it treats $\log$, exactly the question Ilham is interested in.) – Travis Willse Apr 27 '15 at 04:21
  • @Travis Oh yes, so here the complex logarithm function described gives to each $z\in \mathbb{C}$, $ln|z| + i Arg(z)$ but $Arg(z)$ is in $[0,2\pi)$ rather than $[-\pi, \pi)$? Edit: It's okay I think I'll read up on that rather than hijacking someone else's question. – Ilham Apr 27 '15 at 04:23
  • How do I find the derivative after showing this? – coal Apr 27 '15 at 04:27
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    Yes, that's exactly right, and other choices are possible too. (Strictly speaking, not all branches are given by insisting that $\arg$ take values in a particular interval, but those are the branches that occur in practice.) – Travis Willse Apr 27 '15 at 04:28
  • Hint What is the derivative of the real function $\ln$? – Travis Willse Apr 27 '15 at 04:28
  • @Travis $\frac1x$ so my function is $\ln r + i\theta$ and I can termwise derive( with the chain rule?)? – coal Apr 27 '15 at 04:29