I am trying to solve the following problem, "Show that if $h(z)$ is a complex-valued harmonic function such that $zh(z)$ is also harmonic, then $h(z)$ is analytic."

My approach was to calculate first $f = zh(z)$ the partial $f_{xx}$ and $f_{yy}$, add them and then I get the equation $h_x +h_y = 0$. However, from here I tried substituting $h = u +iv$ and derive Cauchy Riemann equations to show $h(z)$ is analytic. However, when I substitute this $h$, I get an equation of the form $u_x +iv_x +u_y + iv_y = 0$ and I don't know how to derive the Cauchy Riemann Equations from here. Any suggestions on another approach or how I could finish this argument? Thanks!

  • $\begingroup$ It's easier if you use the Wirtinger derivatives and $\Delta = 4\dfrac{\partial^2}{\partial \overline{z}\partial z}$. $\endgroup$ – Daniel Fischer Feb 20 '16 at 21:25

$h = u+iv$ is a complex-valued harmonic function if both real and imaginary part are harmonic: $$ u_{xx} + u_{yy} = 0 \\ v_{xx} + v_{yy} = 0 \\ $$

In the same manner, $$ f(z) = zh(z) = (x + iy)(u(z) + iv(z)) = (xu(z)-yv(z)) + i (xv(z)+yu(z)) $$ is a complex-valued harmonic function if both real and imaginary part are harmonic functions: $$ (xu(z)-yv(z))_{xx} + (xu(z)-yv(z))_{yy} = 0 \\ (xv(z)+yu(z))_{xx} + (xv(z)+yu(z))_{yy} = 0 $$

If you expand the last two equations and simplify the result by using the fact that $h$ is harmonic (the first two equations), then you can conclude that $h = u+iv$ satisfies the Cauchy-Riemann differential equations.

Alternatively (as suggested by Daniel Fischer), using the Laplace operator $\Delta = \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2}$ and the Wirtinger derivatives: $$ \Delta (zh(z))= 4\dfrac{\partial^2}{\partial \overline{z}\partial z}(zh(z)) = 4\dfrac{\partial}{\partial \overline{z}} \bigl( h(z) + z \dfrac{\partial}{\partial z} h(z) \bigr) \\ = 4\dfrac{\partial}{\partial \overline{z}} h(z) + 4\dfrac{\partial^2}{\partial \overline{z}\partial z} h(z) = 4\dfrac{\partial}{\partial \overline{z}} h(z) + \Delta h(z) $$ because $\dfrac{\partial z}{\partial z} = 1$ and $\dfrac{\partial z}{\partial \overline{z}} = 0$.

If $h(z)$ and $zh(z)$ are harmonic then $\Delta h(z) = \Delta (zh(z)) = 0$ and therefore $\dfrac{\partial}{\partial \overline{z}} h(z) = 0$, which is equivalent to $h$ satisfying the Cauchy-Riemann differential equations.


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