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Can someone help me understand this passage in a student-written wiki article? The question is whether $u(x+iy) = \log \sqrt{x^2+y^2}$ has a conjugate harmonic function on $\mathbb{C}\setminus \{0\}$. First it establishes that if there is such a conjugate, then $f'(z) = u_x - iu_y$, where $f=u+iv$. So far, so good. Then they write $$f'(z) = \left( \sqrt{x^2+y^2}\right)_x - i \left( \sqrt{x^2+y^2}\right)_y \tag{1}$$

where they seem to have left out the logs. OK, maybe a typo, but then all the computations follow from it. It would seem we would need $f'(z) = \frac{-x+iy}{x^2+y^2}$, but all the rest of their computations follow from (1). Am I missing something, or is the whole answer wrong?

From these differences, I get that the integral of $f'$ about the unit circle should be $-2\pi i$, not $2\pi i$. A minor difference, perhaps, and the idea of the proof still works, but I want to know if I am screwing up somewhere, or if the article is screwed up.

Edit: As Daniel points out, there is a mistake which cancels out in the end. My calculation above mistakenly added a minus sign.

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  • $\begingroup$ @julien Does that show that there cannot be a harmonic conjugate in $\mathbb{C} \setminus \{0\}$? Certainly $\arg(z)$ is not a single-valued harmonic conjugate, but does that show that one doesn't exist? $\endgroup$
    – Eric Auld
    Commented Dec 2, 2013 at 16:09
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    $\begingroup$ Yes: if $v$ is a harmonic conjugate there, then $u+iv$ must coincide with $\log z$ on its domain since they have the same real part. But if you don't want to use that there is no holomorphic log there, it is neither easier nor faster: just the same thing. $\endgroup$
    – Julien
    Commented Dec 2, 2013 at 16:17
  • $\begingroup$ @julien Very good, that's much simpler $\endgroup$
    – Eric Auld
    Commented Dec 2, 2013 at 16:24
  • $\begingroup$ @julien Would it be more precise to say that the harmonic conjugate must agree with $\arg z$ up to a constant? E.g. the constant functions $1$ and $1+i$ have the same real part but do not coincide. $\endgroup$
    – Eric Auld
    Commented Dec 2, 2013 at 17:32
  • $\begingroup$ Right, I should have said up to a constant. If $f$ is holomorphic with null real part, then $f$ is constant on every connected component. $\endgroup$
    – Julien
    Commented Dec 2, 2013 at 18:08

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Yes, the computation is wrong, the logarithm is missing, the correct derivative would be

$$f'(z) = \frac{x-iy}{x^2+y^2} \quad \left(\; = \frac1z\right).$$

However, in the integral, as written, the denominator was replaced by $1 = \sqrt{x^2+y^2}$, and thus the mistake was cancelled (the author would probably have noticed, if (s)he made a mistake that didn't cancel), and the further computation is correct.

$$\int_0^{2\pi} \frac{\cos\theta - i\sin\theta}{\cos^2\theta + \sin^2\theta} ie^{i\theta}\,d\theta = \int_0^{2\pi} \frac{\cos\theta - i\sin\theta}{\sqrt{\cos^2\theta + \sin^2\theta}} ie^{i\theta}\,d\theta$$

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  • $\begingroup$ Right, thank you. I was doing the chain rule wrong, and I added a minus sign. $\endgroup$
    – Eric Auld
    Commented Dec 2, 2013 at 16:07

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