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Find all entire functions $f(z)$ such that $Re(f(z))=1+2x^2+x^3-2y^2-3xy^2$ where $z=x+iy$.

My attempt to solve it follows.

Let $f(x,y)=u(x,y)+iv(x,y)$ where $u(x,y)=1+2x^2+x^3-2y^2-3xy^2$.

Because $f(z)$ is entire, it respects Cauchy - Riemann equations, so:

$$\begin{cases}u_x(x,y)=v_y(x,y)\\ u_y(x,y)=-v_x(x,y)\end{cases} \Longrightarrow \begin{cases}v_y(x,y)=4x+3x^2-3y^2\\ v_x(x,y)=4y+6xy\end{cases}$$

Integrating the two equations, I obtain that:

$$\begin{cases}v(x,y)=4xy+3x^2y-y^3+h(x)\\ v(x,y)=4xy+3x^2y+k(y)\end{cases}$$

These conditions hold if and only if $k(y)=-y^3$ and $h(x)$ is constant. Therefore, $v(x,y)=4xy+3x^2y-y^3+c$ where $c\in \mathbb{R}$.

So, I have that:

$$f(x,y)=1+2x^2+x^3-2y^2-3xy^2+i\left(4xy+3x^2y-y^3+c \right)=$$

$$=1+\left[2x^2-2y^2+4ixy\right]+\left[x^3-3xy^2+3ix^2y-iy^3\right]+ic=$$ $$=1+2(x+iy)^2+(x+iy)^3+ic$$

Therefore, if $z=x+iy$, the functions requested are:

$$f(z)=z^3+2z^2+ic+1, \quad c\in \mathbb{R}$$

(clearly entire for all $c$ because it is a polynomial of the variable $z$)

Is this correct? Is the step in which I've found $v(x,y)$ formally correct or it can be written better? Thanks!

Sigma Algebra
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    Looks right in approach. I haven't checked the details, but there is always one solution, up to a constant, for $v,$ and this is the right approach. – Thomas Andrews Jan 16 '25 at 22:28
  • Thank you very much! – Sigma Algebra Jan 16 '25 at 22:29
  • seems so but i'm not good in calculation tho. one easy to check is that if you expect the answer to be in polynomial form, enough to consider real line. once the value at real line is fixed, the uniqueness gives the answer. this case the form of function restricts to have real values at real line except for constant imaginary – klerys Jan 16 '25 at 22:35
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    Some point of vocabulary : you are looking for a harmonic conjugate of the given polynomial : see for example here – Jean Marie Jan 16 '25 at 22:42

1 Answers1

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I think this is more precise:

Since $v_y(x,y)=4x+3x^2-3y^2$ we have $v(x,y)=4xy+3x^2y-y^3+h(x)$. Derivating with respect to $x$ we get $v_x(x,y)=4y+6xy+h'(x)$, so comparing with $v_x(x,y)=4y+6xy$ we have $h'(x)=0$, so $h(x)=c \in \Bbb R$

jjagmath
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