Determine all holomorphic functions $f :\mathbb{C} → \mathbb{C}$ with $Re(f(z))) = x^2 + axy + by^2$ for $x = Re(z), y = Im(z)$.

Question

I am having problems solving this exercise:

Let $a, b ∈ \mathbb{R}$ be given. Determine all holomorphic functions $f :\mathbb{C} → \mathbb{C}$ with $Re(f(z))) = x^2 + axy + by^2$ for $x = Re(z), y = Im(z)$.

My idea was to solve it using the Cauchy–Riemann equations.

Let $u(x,y):=Re(f(x+iy))=x^2 + axy + by^2$ and $v(x,y):=Im(f(x+iy))$ so that $f(x,y)=x^2+axy+by^2+i(v(x,y))$

We have with CRE $u_x=v_y,u_y=-v_x$

So we have that $v_y=2x+ay \Rightarrow v(x,y)=2xy+\frac{1}{2}ay^2+h(x)$ where $h(x)$ is a differentiable function.

We also have that $v_x=-(ax+2by) \Rightarrow v(x,y)=-ax^2-2bxy+g(y)$ so I find the $g(y)=\frac{1}{2}ay^2$ and $h(x)=-ax^2$. So we have that $v(x,y)=-ax^2-2bxy+\frac{1}{2}ay^2 \neq 2xy+\frac{1}{2}ay^2-ax^2$

I think I wrote a lot of nonsense. Surely there is at least one mistake somewhere but I can't understand what I'm doing wrong in this exercise. can you help me?

Answer 1

$u_x=2x+ay$ and $u_{xx}=2$. Also, $u_y=ax+2by$ and $u_{yy}=2b$. We get $u_{xx}+u_{yy}=2+2b$. So no such function $f$ can exist unless $2+2b=0$ or $b=-1$. So no contradiction is seen from what you have done.