# Cauchy–Goursat theorem for absolute value of $\log$

Suppose $D$ is a simple closed curve, and $z_0$ is away from $\partial D$. Can I apply the Cauchy–Goursat theorem and say that

$$\oint_{\partial D} \log|z_0 - z| \, dz = 0 \ ?$$

I think that I can. $\log$ is analytic away from its singularity, so I believe the criteria for Cauchy–Goursat should be satisfied. I suspect there may be some issue with choice of a branch cut, but I think that that can be solved just by choosing an appropriate cut.

• Is $z_0$ inside or outside $D$? – Chappers Jun 12 '17 at 15:38
• @Chappers outside – Alex Jun 12 '17 at 15:43

No, this is true for a circle by direct calculation, but not for a more general shape: for example, take $z_0=0$ (so the integrand is $\log{\lvert z \rvert}$), and $D$ as the interior of the region $0<\arg{z}<\theta$, $a<\lvert z \rvert <b$ (i.e. a sector of an annulus centred at $z=0$). Then if $z=re^{it}$, the boundary is composed of the sections $$a<r<b, t=0, \\ r=b, 0<t<\theta, \\ a<r<b, t=\theta, \\ r=a, 0<t<\theta,$$ with anticlockwise orientation. Thus, substituting into the integral gives \begin{align} I &= \int_a^b (\log{r}) \, dr + \int_0^{\theta} (\log{b}) \, ibe^{it} \, dt + \int_b^a (\log{r}) \, e^{i\theta} \, dr + \int_{\theta}^0 (\log{a}) \, iae^{it} \, dt \\ &= (1-e^{i\theta})(b\log{b}-a\log{a}-b+a) + (b\log{b}-a\log{a})(e^{i\theta}-1) \\ &= (a-b)(1-e^{i\theta}), \end{align} which is not zero unless $\theta=0$ or $a=b$.

The problem is that $\log{\vert z_0-z \rvert}$ is the real part of a holomorphic function, namely $\log{(z_0-z)}$, and is therefore not holomorphic itself (it is real-valued, so cannot preserve angles, for example). It is true that $$0=\int_{\partial D} \log{(z_0-z)} \, dz=\int_{\partial D} \big( \log{\vert z_0-z \rvert} + i\arg{(z_0-z)} \big)\, dz,$$ but because, to put it crudely, $dz$ is complex, the real and imaginary parts of the integrand cannot be separated so that their individual contributions give zero: the complex path mixes them up. All you can have from Cauchy's theorem is $$\int_{\partial D} \log{\vert z_0-z \rvert} \, dz = -i\int_{\partial D} \arg{(z_0-z)} \, dz.$$

No. $\log z$ is analytic (outside a branch cut), but $\log|z|$ is most certainly not.

(It is possible to apply Cauchy's integral formula, but you need to do some legwork first, it's not immediate.)

• Why is $\log|z|$ not analytic away from its singularity? $\frac{d}{dz} \log|z| = \frac{1}{z}$, no? – Alex Jun 12 '17 at 16:23
• No, for example $\log |z|$ is real valued. – mrf Jun 12 '17 at 17:04

You can. You only have to worry about branch cuts if you are dealing with the log of a complex argument (note that $|z - z_0|$ is real). In the former case we have $\log z = \log |z| + \arg z$ from the polar form of the complex number and the only multivalued part comes from the argument function.

Fun fact: you can actually use the integral of $1/z$, defined as $\log z$ as the basis for proving the Cauchy and residue theorems and formulating the usual theory of complex integration. See Ahlfors' classic book for the details.