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Suppose $u(x,y)=\ln(x^2+y^2)$

i) Show that $u$ is harmonic on $\mathbb{C} \backslash \lbrace 0 \rbrace$

ii) Show that $u$ is not the real part of a function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$

I manage to show the first part. For the second part, note that $u(x,y)=\ln(x^2+y^2)=\ln(|z|^2)=2 \Re \log(z)$

But this only show that $u$ is not a real part of $\log(z)$. I don know how to show $u$ cannot be the real part of a function which analytic on $\mathbb{C} \backslash \lbrace 0 \rbrace$

Can anyone guide me?

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  • $\begingroup$ If two holomorphic functions have the same real part, their difference is a purely imaginary constant. $\endgroup$ – Daniel Fischer Oct 6 '13 at 17:54
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$z=0$ is a branch point of the complex logarithm, so it cannot be defined continuously (so not analytically) on $\Bbb C\setminus\{0\}.$ Since the real part of an analytic function determines the imaginary part up to imaginary constant, we're done.

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If $u$ were the real part of a holomophic function $f$ in $D = \Bbb C\setminus\{0\} $ then $$ f'(z) = u_x(z) - i u_y(z) = \frac{2x - 2iy}{x^2+y^2} = \frac{2}{z} $$ in $D$, and integration along a circle $\gamma$ surrounding $z=0$ gives a contradiction: $$ 0 = \int_\gamma f'(z) \, dz = \int_\gamma \frac{2}{z} \, dz = 4 \pi i \, . $$

(The idea of the proof is to use that $\frac 1z$ has no anti-derivative in $\Bbb C\setminus\{0\} $, without working with “holomorphic branches of the logarithm” explicitly.)

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