Show that $\displaystyle u=\frac{1}{2}\log(x^2+y^2)$ is harmonic and find its harmonic conjugate function.

I did the first part to show that $\displaystyle \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$

Now, to find v:

$\displaystyle\frac{\partial u}{\partial x}=\frac{x}{x^2+y^2}=\frac{\partial v}{\partial y}$

I get $\displaystyle v=\tan^{-1}(\frac{y}{x})$

$\displaystyle\frac{\partial u}{\partial y}=\frac{y}{x^2+y^2}=-\frac{\partial v}{\partial x}$

Here I get $\displaystyle v=-\tan^{-1}(\frac{x}{y})$

What did I do wrong here ?

  • $\begingroup$ The arguments $\theta$ of $x+iy$ and of $i(x+iy)=-y+ix$ are related by the constant $\pi/2$. $\endgroup$ Jul 6, 2014 at 15:46

2 Answers 2


You've already got a complete answer (+1), but I'd like to add the complex analysis way to do it. If you set $z = x^2 + y^2$, then notice that

$$f(z) = \ln\Big(\sqrt{x^2 + y^2}\Big) = \ln |z|$$

Now this is exactly the real part of (a branch of) the logarithm, for if we write $z = re^{i\theta}$ and ignore some technicalities about branches,

$$\ln(z) = \ln(re^{i\theta}) = \ln r + \ln e^{i\theta} = \ln |z| + i \theta = \ln |z| + i \operatorname{Arg}(z)$$

where $\operatorname{Arg}(z)$ is the argument. Thus we've identified $\ln |z|$ as the real part of a complex analytic function, and so its harmonic conjugate is the imaginary part - in particular, $\operatorname{Arg}(z)$. Geometrically, one can check that the argument $\theta$ of a number satisfies

$$\tan \theta = \frac{y}{x}$$

so this is consistent with the answer you've already got.


What you got in the first step is correct. That is, when we integrate $$\frac{\partial v}{\partial y}=\frac{x}{x^2+y^2}=\frac{\frac{1}{x}}{1+(\frac{y}{x})^2} =\frac{\frac{\partial}{\partial y}(\frac{y}{x})}{1+(\frac{y}{x})^2}=\frac{\partial}{\partial y}\left(\tan^{-1}(\frac{y}{x})\right),$$ we obtain $v=\tan^{-1}(\frac{y}{x})$. Note that in your second step: $$\frac{\partial v}{\partial x}=-\frac{y}{x^2+y^2}=\frac{-\frac{y}{x^2}}{1+(\frac{y}{x})^2} =\frac{\frac{\partial}{\partial x}(\frac{y}{x})}{1+(\frac{y}{x})^2}=\frac{\partial}{\partial x}\left(\tan^{-1}(\frac{y}{x})\right).$$ Integrate it, again we get $v=\tan^{-1}(\frac{y}{x})$.

  • 1
    $\begingroup$ u r in Antarctica and u r 94 ?? I can't decide which one is more awesome.. $\endgroup$
    – square_one
    Jul 6, 2014 at 7:57

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