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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?

Idonknow
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2 Answers2

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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.

Cameron Buie
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  • Could you tell me the details ' we're done' ? log is not analytic on C {0}, and real part determine the imaginary part up to imaginary constant, from that, why can we deduce that there are no holomorphic function which has log(x^2＋y^2) on real part? – Pont Jun 12 '21 at 07:49
  • @Nekojiru Let's suppose that $f$ were an analytic function on $\Bbb C\setminus{0},$ such that $\Re f=\ln(x^2+y^2).$ Then the real parts of $f$ and $2\log z$ would agree on the disk $|z-1|<1,$ and so there would be some real $\alpha$ such that $f(z)=2\log(z)+i\alpha$ on the disk $|z-1|<1.$ Put another way, $2\log(z)=f(z)-i\alpha$ on this disk, and so $2\log(z)$ would have an analytic continuation to $\Bbb C\setminus{0}.$ Contradiction. – Cameron Buie Jun 12 '21 at 11:29
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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.)

Martin R
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  • Could you tell me why ∮r f(z)dz is $0$? – Pont Jun 12 '21 at 02:20
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    @Nekojiru: The integral of a derivative over a closed curve is always zero: $\int_\gamma f'(z) , dz = f(\gamma(1)) - f(\gamma(0)) = 0$. – Martin R Jun 12 '21 at 05:50
  • I'm worried about {0} inside the closed curve. There are no need to worry about nonholomorphic point inside the curve? – Pont Jun 12 '21 at 06:12
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    @Nekojiru: No need to worry, it is just the definition and the fundamental theorem of calculus: $\int_\gamma f'(z) , dz = \int_0^1 f'(\gamma(t)) \gamma'(t) , dt = \int_0^1 (f \circ \gamma)'(t) , dt = f(\gamma(1)) - f(\gamma(0)) = 0$. – Martin R Jun 12 '21 at 06:14
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    Complete explanation. Thank you so much! – Pont Jun 12 '21 at 06:29