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.

  • $\begingroup$ Gave NET, today? $\endgroup$ – Sahiba Arora Jun 18 '17 at 17:50

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 $r$: $$f(z)=rz+b \quad (r\in\mathbb R, b\in \mathbb C)$$
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!


Break out the Cauchy-Riemann equations. What does the symmetry say about them?

  • $\begingroup$ $u_{y}=-v_{x}$ so $u_{y}=0?$ $\endgroup$ – neelkanth Jun 18 '17 at 17:53
  • 1
    $\begingroup$ can u ex plane little more? $\endgroup$ – neelkanth Jun 18 '17 at 17:56
  • $\begingroup$ This gives $f'(z)$ constant and hence polynomial of degree at most 1? am i right ? $\endgroup$ – neelkanth Jun 18 '17 at 17:59
  • $\begingroup$ Being of degree at most one is necessary but not sufficient. $\endgroup$ – Georges Elencwajg Jun 18 '17 at 19:34

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:


For what choices of $a_{i's}$ do you always have Re($f$) free of variable $y$ such that $u_y=0$?

  • $\begingroup$ so 1st and 2nd option is write ? $\endgroup$ – neelkanth Jun 18 '17 at 18:09
  • $\begingroup$ Yes....@neelkanth $\endgroup$ – Nitin Uniyal Jun 18 '17 at 18:13
  • $\begingroup$ Thanks a lot.... $\endgroup$ – neelkanth Jun 18 '17 at 18:13
  • $\begingroup$ @neelkanth..my pleasure.. $\endgroup$ – Nitin Uniyal Jun 18 '17 at 18:14

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

$u_x = v_y \tag{3}$


$u_y = - v_x; \tag{4}$

combining (2) and (4) we find

$v_x = u_y = -v_x, \tag{5}$


$v_x = 0, \tag{6}$

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),

$u_{xx} = v_{yx} = 0 \tag{10}$


$v_{yy} = u_{xy} = 0. \tag{11}$

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

$u_1 = u_x = v_y = v_1 = \alpha \in \Bbb R; \tag{14}$


$f(z) = u + iv = u_0 + iv_0 + \alpha x + i \alpha y = u_0 + i v_0 + \alpha (x + iy) = u_0 + i v_0 + \alpha z, \tag{14}$

in agreement with Georges Elencwajg's result. Assertion (2) of the question binds.

  • $\begingroup$ I didn't want to bust this entirely open for the OP. $\endgroup$ – ncmathsadist Jun 18 '17 at 23:50

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