# Harmonic function with vanishing partial derivative

Let $u:D(0,1)\to \mathbf{R}$ be harmonic on the unit disc, and suppose there exists a $z_0\in D(0,1)$ such that all partial derivatives of $u$ vanish. Show that $u$ is constant.

I found this problem on a complex analysis qual, so it shouldn't require much more than the harmonic mean value property, but I have no idea how to proceed.

Let's use the fact that $$u = \text {Re} \,f$$ for some $$f$$ holomorphic in $$D(0,1).$$ Write $$f=u+iv.$$ Then for any $$z\in D(0,1),$$

$$f'(z) = u_x(z) + iv_x(z) = u_x(z) - iu_y(z),$$

where have used the Cauchy-Riemann equations to get the second equality. This implies $$f''(z) = u_{xx}(z) - iu_{yx}(z),$$ $$f'''(z) = u_{xxx}(z) - iu_{yxx}(z),$$ etc. Hence all of $$f$$'s derivatives can be written in terms of the partial derivatives of $$u.$$ This tells us that at $$z_0,$$ the Taylor series of $$f$$ is just the one term $$f(z_0).$$ Thus $$f=f(z_0)$$ in some $$D(z_0,r).$$ By the identity principle, $$f = f(z_0)$$ in $$D(0,1),$$ which implies $$u=u(z_0)$$ in $$D(0,1).$$

• how do we know that there exists a function $v$ such that $f=u+iv$ is holomorphic? – alpastor Apr 20 '19 at 2:57
• @alpastor This is a standard result in complex analysis: Any real harmonic function in a simply connected domain is the real part of a holomorphic function there. – zhw. Apr 20 '19 at 3:27