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Let $u(x,y)= \ln(x^2+y^2)$. We have that $u$ is harmonic and is defined on $\mathbb{R}^2 - \{ (0,0)\}$ (not simply connected).

How can I show that there is no function $v(x,y)$ such that, for every $z = z=x+iy \in \mathbb{C}-\{(0,0)\}$, $$f(z) = u(x,y)+iv(x,y)$$

is holomorphic?

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    Do you know that _locally_ there exist harmonic conjugates of $u$? What is the derivative of the locally existing holomorphic functions with real part $u$? Why does that imply that there is no global harmonic conjugate? – 2017-01-17
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    @DanielFischer The *local* holomorphic function should be $$f(z) = \ln(x^2+y^2) -i2\arctan\left(\frac{x}{y}\right),$$ which is the primitive of $g(z) = u_x(x,y)-u_y(x,y)$. But why does this imply that there is no global harmonic conjugate? – 2017-01-17
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    How do you get $f'(z) = 0$ there? $f$ is clearly not constant. – 2017-01-17
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    @DanielFischer Pardon: $$\frac{2x}{x^2+y^2} - i\frac{2y}{x^2+y^2}.$$ – 2017-01-17
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    Okay. Can you simplify that? (Express it in terms of $z$ rather than in terms of $x$ and $y$.) Since the derivative does not depend on the chosen $v$ (and its domain), it would be the derivative of the global holomorphic function too, if one existed. Do you know a property that the derivatives of holomorphic functions have that this function doesn't have? – 2017-01-17
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    @DanielFischer Why would it be the derivative of the global holomorphic function too, if one existed? We can simplify it as $$\frac{z+\overline{z}}{|z|^2} - i \frac{z-\overline{z}}{|z|^2}$$ What property does it lack? – 2017-01-17
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    Because a global harmonic conjugate is in particular a local harmonic conjugate. In your simplification, you made an error, $z - \overline{z} = 2i y$, so it becomes $$\frac{z + \overline{z}}{\lvert z\rvert^2} - \frac{z - \overline{z}}{\lvert z\rvert^2},$$ and that is easy to simplify further. If $g$ is holomorphic, then there is something interesting you can say about $\int_{\gamma} g'(z)\,dz$. What? – 2017-01-17
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    @DanielFischer So $$2\frac{\overline{z}}{|z|^2},$$ whose integral around the unitary circle is $4\pi i$, which is non-zero. Therefore we can conclude the proof. Right? – 2017-01-17
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    Or more simply $\frac{2}{z}$. Yes, that has a nonzero integral along some closed curves, hence it's not globally a derivative. – 2017-01-17
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    @DanielFischer Thanks a lot. – 2017-01-17
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    *Locally* $v$ must be $2\arg z$. But it's impossible define a continuous $\arg$ function in $\Bbb C\setminus\{0\}$ (nontrivial continuous homomorphism $\Bbb C\setminus\{0\}\longrightarrow\Bbb R$). – 2017-03-01

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