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