Proposition

Consider $u(x, y) = \ln(x^2 + y^2)$.

  1. Show that $u$ is harmonic on $\mathbb{C} \setminus \{ 0 \}$.
  2. Prove that $u$ is not the real part of a function that is holomorphic in $\mathbb{C} \setminus \{ 0 \}$.

Solution

1

$u_x = 2x / (x^2 + y^2)$ and $u_y = 2y / (x^2 + y^2)$. This gives us

\[\begin{align*} u_{xx} &= \frac{2}{x^2 + y^2} - \frac{4x^2}{(x^2 + y^2)^2}, \\ u_{yy} &= \frac{2}{x^2 + y^2} - \frac{4y^2}{(x^2 + y^2)^2}. \end{align*}\]

Then $u_{xx} + u_{yy} = 0$, so $u$ is harmonic.

2

Suppose that there exists a holomorphic function $f$ defined on $\mathbb{C} \setminus \{ 0 \}$ such that $\Re(f) = u$. Then

\[\begin{align*} f' &= f_x \\ &= u_x + iv_x \\ &= u_x - iu_y & \text{(Cauchy-Riemann)} \\ &= \frac{2x}{x^2 + y^2} - i\frac{2y}{x^2 + y^2} \\ &= \frac{2x - iy}{x^2 + y^2} \\ &= \frac{2}{x + iy} \\ &= \frac{2}{z}. \end{align*}\]

Therefore,

\[\begin{align*} \int_{C[0, 1]} f'dz &= \int_{C[0, 1]} \frac{2}{z} dz \\ &= 2\pi i \cdot 2 = 4\pi i. \end{align*}\]

$f$ is clearly an antiderivative of $f’$. By Corollary 4.13[A first course in complex analysis], $\int_{C[0, 1]} f’ = 0$. This is a contradiction, so such an $f$ does not exist.