Let $w$ be a complex number such that $|w| < 1$. Evaluate the integral

$$\int_0^{2\pi} \log |1-w e^{i\theta}| \, d\theta.$$

I am having a hard time moving forward on this question.

I tried substituting $z=e^{i\theta}$, and to get $$\oint_\gamma \frac{\log|1-wz|}{iz} \, dz,$$ where $\gamma$ is the unit circle with positive orientation.

Attempt 1: With this substitution, the integrand is not holomorphic, because multiplying by $iz$ (a holomorphic function), we get a purely real valued function (at least in some open set in the right half plane), which can only be holomorphic if it is constant, which this function is not. The strategies for contour integrals will probably not help me.

Attempt 2: I tried finding some symmetry (either antipodal point or point with equal imaginary part; horizontally across) to cancel out the integral, but couldn't.

Attempt 3: After searching, I found this post on Math SE. The suggestion there is focus on the function $\log(1-wz)$ recognizing that the real part of this is $\log|1-wz|$ and either (1) "think mean value" or (2) differentiate this function. I don't know how it helps to observe that $\oint_\gamma \frac{1}{1-wz} dz = 0$. I appreciate any suggestion that will help me understand the hints (or another way to see that the integral is 0).

  • $\begingroup$ I have answered my question, but will wait for a day before accepting my own answer. If you have a more insightful answer, please feel free to add it. $\endgroup$
    – Snow
    Jan 3, 2016 at 19:46
  • $\begingroup$ Do you know the mean value property for harmonic functions? $\endgroup$
    – zhw.
    Jan 3, 2016 at 20:33
  • $\begingroup$ Ah, I didn't know that, but thank you. It looks like there is a wealth of related information there. $\endgroup$
    – Snow
    Jan 3, 2016 at 20:37

2 Answers 2


It's not hard to see that the real part of $\log(1-wz)$ is $\log|1-wz|$. It takes one more step to see that the real part of the integral (not just the integrand) $\oint \frac{\log(1-wz)}{iz} dz$ is $\oint \frac{|\log(1-wz)|}{iz} dz$.

In the standard parameterization (i.e., $z = e^{i\theta}$) of the unit circle $\gamma$ with a real parameter, we have $\arg(dz) = \arg(z)+\frac{\pi}{2}$ and hence $\frac{dz}{iz}$ is real. Therefore, the real part of the integral $$\oint_\gamma \frac{\log(1-wz)}{iz} dz = \oint_\gamma \log(1-wz)\frac{dz}{iz}$$ is $$\oint_\gamma \frac{|\log(1-wz)|}{iz} dz.$$

There is an open set $\Omega$ containing the unit disk (i.e., simply connected open set containing $\gamma$) on which the integrand $\frac{|\log(1-wz)|}{iz}$ is holomorphic (the singularity is removable) and hence the integral is 0.

Therefore $$\oint \frac{|\log(1-wz)|}{iz} dz = 0.$$


If you set $w = a+\mathrm{i}b$, and write $e^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta$, you have $$ w e^{\mathrm{i}\theta}=(a+\mathrm{i}b)(\cos\theta+\mathrm{i}\sin\theta)= a\cos\theta-b\sin\theta+\mathrm{i}(b\cos\theta+a\sin\theta)\ . $$ Therefore $$ |1-w e^{\mathrm{i}\theta}|=\sqrt{(1-a\cos\theta+b\sin\theta)^2+(b \cos\theta+a\sin\theta)^2}\ . $$ Taking the log, we get $$ \log|1-w e^{\mathrm{i}\theta}|=\frac{1}{2}\log\left(1+a^2+b^2-2 a \cos \theta +2 b \sin \theta \right)\ . $$ Applying now the integral formula (see Gradshteyn and Ryzhik Table of Integrals, Series, and Products, ed. 1996, formula 4.225.4) $$ \int_0^{2\pi}dx\ln (1+a^2+b^2+2 a \sin x+2 b \cos x)=0\mbox{ if }a^2+b^2\leq 1\mbox{ and }2\pi\ln (a^2+b^2)\mbox{ if }a^2+b^2\geq 1 $$ one has immediately that the requested integral is zero (for $|w|\leq 1$) and equal to $$ \int_0^{2\pi}\log|1-w e^{\mathrm{i}\theta}|=2\pi\log |w|\ , $$ for $|w|>1$.

  • $\begingroup$ Sorry, this is not a proof. I want to prove my answer so no Mathematica. $\endgroup$
    – Snow
    Jan 3, 2016 at 18:52
  • $\begingroup$ If I am not mistaken, you did not post your answer. To me, the question was 'Evaluate the integral'... $\endgroup$ Jan 3, 2016 at 18:57
  • $\begingroup$ It seems (here and here) that the answer is 0. For future reference, in the context of higher level math, you typically have to prove (or cite theorems) for everything you say. If someone posts a question under the tag "(real/complex)-analysis" they usually want a rigorous argument for evaluating an integral. $\endgroup$
    – Snow
    Jan 3, 2016 at 19:40
  • $\begingroup$ Fair enough. Just edited my answer. The integral (involving only real parameters) is known in closed form (not sure if this qualifies as 'rigorous argument'). The advantage of this elementary argument is that you get for free the value of your original integral for any $w$ (not just $|w|\leq 1$). $\endgroup$ Jan 3, 2016 at 20:18
  • $\begingroup$ Unfortunately, I can't call it elementary because I don't know how such an integral is proved (apart from complex analysis) and I won't be able to prove it on a homework or exam by citing the table. $\endgroup$
    – Snow
    Jan 3, 2016 at 22:36

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