Complex analysis objective type question. Let $f(z)=u+iv$ be an entire function. If the Jacobian matrix $$\begin{bmatrix}
    u_{x}(a)       & u_{y}(a)  \\
    v_{x}(a)       & v_{y}(a)   \\ 
\end{bmatrix}$$ is symmetric for all complex $a$. Then
$1.$ $f$ is a polynomial.
$2.$ $f$ is a polynomial of degree at most $1.$
$3.$ $f$ is constant.
$4.$ $f$ is a polynomial of degree at least $2.$
I have no idea how to solve this problem. It is clear that $4$th option is wrong as constant function satisfied given condition. From given symmetric matrix i just have that $u_{y}(a)=v_{x}(a) $. How to proceed further. Please help. Thanks a lot. 
 A: Since $u_y=v_x$ by the symmetry condition and $u_y=-v_x$ by Cauchy-Riemann we deduce that $u_y=v_x=0$, so that $u$ and $v$ are functions depending only on $x$ and $y$ respectively, i.e. $$f(z)=f(x+iy)=u(x)+iv(y)$$ 
Now $f'(z)=\frac {\partial f}{\partial x}(z)=u'(x)$ is a real-valued holomorphic function, which implies that it is a real constant: $f'(z)=r\in \mathbb R$.
This immediately forces  the conclusion that the required entire functions are the polynomials of degree at most one with real linear coefficient $r$: $$f(z)=rz+b \quad (r\in\mathbb R, b\in \mathbb C)$$
Warning
Note carefully that the class  of entire functions with symmetric Jacobians is none of the suggested four classes  and note also that $iz$ is a polynomial of degree one whose Jacobian is not symmetric!
A: Break out the Cauchy-Riemann equations.  What does the symmetry say about them?
A: For symmetry $u_y=v_x$ for all $z\in\mathbb C$. As $f$ is entire, use C-R equations in it to get $u_y=-u_y$ which gives $u_y=0$. 
Now think of an entire $f=u+iv$ having expansion like this:
$f(z)=a_0+a_1.z+a_2.z^2+....$.
For what choices of $a_{i's}$ do you always have Re($f$) free of variable $y$ such that $u_y=0$?
A: Here's a way of doing it which is, I think, similar in spirit to the solution of Georges Elencwajg, although the view presented here is a little different:
If the Jacobean matrix
$\begin{bmatrix} u_{x} & u_{y} \\ v_{x} & v_{y} \end{bmatrix} \tag{1}$
is in fact symmetric, we have
$u_y = v_x; \tag{2}$
we also have the Cauchy-Riemann equations, which tell us that
$u_x = v_y \tag{3}$
and
$u_y = - v_x; \tag{4}$
combining (2) and (4) we find
$v_x = u_y = -v_x, \tag{5}$
forcing
$v_x = 0, \tag{6}$
and hence
$u_y = 0 \tag{7}$
as well, also by virtue of (2)-(4).  It of course follows from (6) and (7) that $u$ depends only on $x$ and $v$ depends only on $y$.
We compute the second derivatives of $u$ and $v$, $u_{xx}$, $u_{xy} = u_{yx}$, $u_{yy}$, $v_{xx}$, $v_{xy} = v_{yx}$, and $v_{yy}$.  Since $u_y = 0$, (7), we have
$u_{yy} = u_{yx} = u_{xy} = 0; \tag{8}$
likewise, $v_x = 0$ yields
$v_{xx} = v_{xy} = v_{yx}  = 0; \tag{9}$
also, from (3), (8) and (9),
$u_{xx} = v_{yx} = 0 \tag{10}$
and
$v_{yy} = u_{xy} = 0. \tag{11}$
We see that all the second partials of $u$ and $v$ with respect to $x$ and $y$ are $0$.  It follows that $u$ is a linear function of $x$, and that $v$ is a linear function of $y$, since the only possible non-vanishing derivatives are $u_x$ and $v_y$.  We may thus write
$u = u_0 + u_1x, \tag{12}$
$v = v_0 + v_1y, \tag{13}$
and it follows from (2) that
$u_1 = u_x = v_y = v_1 = \alpha \in \Bbb R; \tag{14}$
thus
$f(z) = u + iv = u_0 + iv_0 + \alpha x + i \alpha y = u_0 + i v_0 + \alpha (x + iy) = u_0 + i v_0 + \alpha z, \tag{14}$
in agreement with Georges Elencwajg's result.  Assertion (2) of the question binds.
