Prove that the entire function $f$ is linear.

Question

Suppose $f=u+iv$ be an entire function such that $u(x,y)=\phi(x)$ and $v(x,y)=\psi(y)$ for all $x,y\in\mathbb{R}$. Prove that $f(x)=az+b$ for some $a\in\mathbb{C},b\in\mathbb{C}$.

My approach was: $u(x,y)=\phi(x)\Rightarrow\frac{\partial u}{\partial x}=\phi'(x),\frac{\partial u}{\partial y}=0$. Now $\frac{\partial u}{\partial x}=\frac{1}{i}\frac{\partial u}{\partial y}$ and $\frac{\partial u}{\partial z}=\frac{1}{2}\left(\frac{\partial u}{\partial x}+\frac{1}{i}\frac{\partial u}{\partial y}\right)=0$. Now $\frac{\partial f}{\partial z}=2\frac{\partial u}{\partial z}=0$, implying $f$ is constant.

Where did I go wrong?

Answer 1

Cauchy-Riemann equations ($u_x=v_y$ and $u_y=v_x$) provide: $$ u_x=\varphi'(x)=\psi'(y)=v_y $$ Thus $\varphi'(x)=c=\psi'(y)$, and hence $$ \varphi(x)=cx+a,\quad \psi(y)=cy+b, $$ and finally $$ f(z)=cz+a+bi. $$

Answer 2

One doesn't need to find $\dfrac{du}{dz}$ in this case. Instead, let's look at $\dfrac{df}{dz} := \frac{\partial f}{\partial x} = \phi' (x)$, a real-valued holomorphic function as derivatives of holomorphic functions are holomorphic too. The only real-valued holomorphic functions are the constant functions. Thus, we have that $f'(z) \equiv a$ for some $a \in \mathbb{R}$. In other words, $f(z) = az + b$ for some $a \in \Bbb{R}, b \in \Bbb{C}$.