We have an open set $U$ in $\mathbb{R}^2$ and two functions $u, v \in C^1 (U)$. $u$ and $v$ satisfy the Cauchy-Riemann equations in $U$, i.e., $u_x = v_y$ and $u_y = -v_x$. Also, $u^2 + v^2 \neq 0$ for all $(x,y) \in U$. Then, we have to show that $$\frac{uu_x + vv_x}{u^2 + v^2}$$ is harmonic in $U$, i.e., the Laplacian of this function is zero.

On the face of it, it seems there is nothing to do but calculate the second partial derivatives and verify the result. But that would be very messy and complicated. Is there something easier that I am missing?

Note: $u_x$ means the partial derivative of $u$ with respect to $x$, etc.


1 Answer 1


Hint: If $u$ and $v$ are continuously differentiable and satisfy the Cauchy-Riemann equations then $f = u + iv$ is holomorphic. $f$ is not zero, therefore $g = f'/f$ is holomorphic in $U$ as well. What can you say about the real and imaginary part of $g$?


$ \frac{f'}{f} = \frac{f'\overline f}{|f|^2} = \frac{(u_x+iv_x)(u - iv)}{u^2 + v^2} \Longrightarrow \Re \frac{f'}{f} = \frac{uu_x + vv_x}{u^2 + v^2}$ is harmonic.

  • $\begingroup$ Why are you write $f'$ as derivative of $x$ only? what about $u_y $ and $v_y$? $\endgroup$
    – Sagigever
    Jun 6, 2021 at 20:26
  • $\begingroup$ @Sagigever: If $f$ is complex differentiable then $f' = f_x = u_x + i v_x$. $\endgroup$
    – Martin R
    Jun 6, 2021 at 20:31
  • $\begingroup$ Is that a deffinition? because $f(z) = f(x+iy)$ so it doesn't make sense to me that $f' = f_x$ and not $f' = f_x + f_y$ $\endgroup$
    – Sagigever
    Jun 6, 2021 at 20:44
  • $\begingroup$ Ok, I will wait till tomorrow. thank you $\endgroup$
    – Sagigever
    Jun 6, 2021 at 20:56
  • 1
    $\begingroup$ @Sagigever: See Complex differentiability . $\endgroup$
    – Martin R
    Jun 7, 2021 at 4:25

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