Let $f=u+iv$ be an entire function. If the Jacobian
$$J_a=\Bigg[ \begin{matrix} u_x(a) & u_y(a) \\ v_x(a) & v_y(a) \end{matrix} \Bigg]$$ symmetric for all $a\in \mathbb C,$ then
(A) $f$ is a polynomial
(B) $f$ is a polynomial of degree $\leq 1$
(C) $f$ is necessarily a constant function
(D)$f$ is a polynomial of degree $> 1$
My Try
If the Jacobian
$$J_a=\Bigg[ \begin{matrix} u_x(a) & u_y(a) \\ v_x(a) & v_y(a) \end{matrix} \Bigg]$$ symmetric for all $a\in \mathbb C,$ then $u_y=v_x$. So, by Cauchy-Riemann equation $v_x=-u_y$. So, $2v_x=0$. $\mathbb C$ is connected. So, $v_x=0$
$v= g(y)$. By cauchy Riemann equation, $u_y=0 \implies u=h(x)$ Hence, $f(x,y)=g(y)+ih(x)$
Applying Cauchy Rieman equation $u_y=-v_x\implies g'(y)=-h'(x).$ I am not able to conclude from this. Can you please help me?
Best Answer
$g'(y)=-h'(x)$ for all real numbers $x$ and $y$ implies that there is a constant $c$ such that $g'(y)=-h'(x)=c$ for all $x$ and $y$. [This is because LHS does not depend on $x$ and RHS does not depend on $y$]. Hence $g(y)=cy+d$ and $h(x)=-cx+e$ fo some constants $d$ and $e$. Finally $f(x+iy)=cy+d+i(-cx+e)$. Can you complete the answer now?
The answer is B).