# Show that $f'(z_0)=0$

Let $f:\mathbb{C}\to\mathbb{R}$ be a real valued function defined on a domain $A\subset \mathbb{C}$. Given that $f$ is complex differentiable at $z_0 \in A$, show that $f'(z_0)=0$.

My attempt so far:

Since $f$ is a real valued function it is of the form $f(z)=u(x,y)$. So the partial derivatives satisfy $v_x=v_y=0$. Since $f$ is differentiable at $z_0$, it satisfies the Cauchy-Riemann equations and so $u_x=v_y=0$ and $-u_y=v_x=0$. This means that $u(x,y)=C$ for some constant $C$. Thus $f$ is constant and so it is such that $f'(z_0)=0$.

Can someone verify whether this is correct or not?

Second attempt.

Since $f$ is a real valued function, it is of the form $f(z)=u(x,y)$. Therefore, $v_x(x,y)=v_y(x,y)=0$.

Since $f$ is differentiable at $z_0$ it must satisfy the Cauchy-Riemann equations at this point. i.e., $u_x(x_0,y_0)=v_y(x_0,y_0)=0$ and $-u_y(x_0,y_0)=v_x(x_0,y_0)=0$. Therefore, $f'(z_0)=0$.

• You have already shown that $f'(z_0)=0$ using the Cauchy-Riemann equations. Your argument about $f$ being constant is wonky. As $f$ needs not to be differentiable on a neighborhood of $z_0$. For example $f(z) = z\bar z = |z|^2$ is real, complex differentiable at $z_0 = 0$, and not constant on any neighbor of $z_0$. – user251257 Dec 14 '16 at 21:16
• @user251257 So showing that $u_x=u_y=0$ at $z_0$ is enough to show that $f'(z_0)=0$? – user374859 Dec 14 '16 at 21:23
• @user251257 I've added a second attempt to my post. Is it correct now? – user374859 Dec 14 '16 at 21:30
• you might want to say what $v$ is. If $f'(z_0)$ exists then by Cauchy-Riemann, we have $f'(z_0) = u_x(x_0, y_0) - u_y(x_0,y_0) i = v_y(x_0, y_0) + v_x(x_0,y_0) i$. – user251257 Dec 14 '16 at 21:37

Yes your proof is correct. The main step is to prove that every real-valued holomorphic function $f$ is necessarily constant, as you show. Then $f'(z)\equiv 0$.
• Only differentiability at a point $z_0$ is assumed. $f$ needs not to be entire. – user251257 Dec 14 '16 at 21:08