Let $f=u+iv$ be an entire function where $u,v$ are the real and imaginary parts of $f$ respectively. If the Jacobian $$ J_a= \left[ {\begin{array}{cc} u_x(a) & u_y(a) \\ v_x(a) & v_y(a) \\ \end{array} } \right] $$ is symmetric $\forall a\in \mathbb C$, then

(1) $f$ is a polynomial.

(2)$f$ is a polynomial of $\deg\le 1$.

(3)$f$ is necessarily a constant function.

(4)$f$ is a polynomial of degree strictly greater than 1.

$f$ is analytic. Using C-R equation, we have $\det J_a=u_x(a)v_y(a)-v_x(a)u_y(a)=u_x(a)u_x(a)+u_y(a)u_y(a)$. We know $J_a$ is symmetric $\implies u_y(a)=v_x(a)$. How do I proceed further. How to judge the option? Please help me.


1 Answer 1


The second Cauchy-Riemann equation says that the off-diagonal elements of the Jacobian differ by a sign, $v_x=-u_y$, so that to be symmetric, $v_x=u_y$, they have to be zero.

Which means that $u(x+iy)=p(x)$ is a function of $x$ alone and $v(x+iy)=q(y)$ of $y$ alone. Then the first CR-equation gives $p'(x)=u_x=v_y=q'(y)$ for all $x,y$ independent of each other which implies $$p'(x)=q'(0)=a=p'(0)=q'(y).$$ This gives $$f(z)=az+b$$ with real $a$ and complex $b$ as the only possible solutions and indeed these functions are all solutions.

  • $\begingroup$ If $a$ and $b$ are both zero then which of the option/s is/are correct. Since we always assume that zero polynomial has infinite degree. According to the answer you gave it seems that $1$ and $2$ are the only correct options. But it may so happen that both of $a$ and $b$ are zero. Then there is a slight problem. Isn't it so? $\endgroup$
    – D_C
    Commented May 31, 2018 at 16:04
  • 1
    $\begingroup$ Yes, it is one sensible convention to set $\deg 0=-\infty$. As $-\infty\le 1$ there is no problem. $\endgroup$ Commented May 31, 2018 at 17:22

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