Complex Analytic functions that are lines are constant

Question

If a complex function is analytic on D and its point lie on a line, it is constant with respect to x and y. How does one prove this?

Answer 1

Here is an elementary proof: There exist $m,c \in \mathbb R$ such that $v=mu+c$ where $u$ and $v$ are the real and imaginary parts of $f$. This gives $v_x=mu_x$ and $v_y=mc_y$. Use Cauchy Riemann equations to get $(1+m^{2})u_x=0$. Hence $u_x=0$. Similarly $u_y=0$ so $u$ is a constant. Similarly $v$ is also a constant and so is $f$. [Notation: $u_x=\frac {\partial } {\partial x} u$ etc].

Answer 2

A line in the plane $\Bbb R^2$ may always be represented as

$ax + by = d, \tag 1$

where

$a, b, d \in \Bbb R \tag 2$

and

$a^2 + b^2 \ne 0, \tag 3$

that is, not both $a$ and $b$ vanish.

If the range of the holomorphic function

$f(z) = f(x, y) = u(x, y) + iv(x, y) \tag 4$

lies in this line, then

$au(x, y) + bv(x, y) = d; \tag 5$

thus,

$au_x + bv_x = 0, \tag 6$

$au_y + bv_y = 0; \tag 7$

now using the Cauchy-Riemann equations

$u_x = v_y, \; u_y = -v_x \tag 8$

in (6)-(7) we find

$au_x - bu_y = 0, \tag 9$

$au_y + bu_x = 0; \tag{10}$

we may write these last two equations in matrix-vector form as

$\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \begin{pmatrix} u_x \\ u_y \end{pmatrix} = 0; \tag{11}$

also, we have

$\det \left (\begin{bmatrix} a & -b \\ b & a \end{bmatrix} \right ) = a^2 + b^2 \ne 0; \tag{12}$

thus

$\begin{pmatrix} u_x \\ u_y \end{pmatrix} = 0; \tag{13}$

again with the aid of Cauchy-Riemann we obtain

$-v_x = u_y = u_x = v_y = 0, \tag{14}$

from which we conclude that $u(x, y)$ and $v(x, y)$, and hence

$f(z) = f(x, y) = u(x, y) + iv(x, y), \tag{15}$

are constant.

The above presents a solution which is cast in terms of the real functions $u(x, y)$, $v(x, y)$, and relies strictly on elementary complex analysis; a somewhat more purely complex-analytic version of what we have just done, though a little more advanced, may be had by writing the line (1) in terms of $z$ and $\bar z$, as follows:

$x = \dfrac{z + \bar z}{2}, \tag{16}$

$y = \dfrac{z - \bar z}{2i}; \tag{17}$

$a\dfrac{z + \bar z}{2} + b\dfrac{z - \bar z}{2i} = d, \tag{18}$

that is,

$\left (\dfrac{a}{2} + \dfrac{b}{2i} \right) z + \left ( \dfrac{a}{2} - \dfrac{b}{2i} \right ) \bar z = d; \tag{19}$

setting

$c = \dfrac{a}{2} + \dfrac{b}{2i} = \dfrac{a}{2} - i\dfrac{b}{2}\ne 0, \tag{20}$

the equation of the line takes the form

$cz + \bar c \bar z = d; \tag{21}$

now if $f(z)$ lies in such a line, then

$cf(z) + \bar c \overline {f(z)} = d; \tag{21}$

now just take the Wirtinger derivative $\partial/\partial z$, exploiting the facts that

$\overline{f(z)} = \bar f(\bar z) \tag{22}$

and

$\dfrac{\partial {\bar f(\bar z)}}{\partial z} = 0; \tag{23}$

we find

$c f'(z) = 0, \tag{24}$

which since $c \ne 0$ yields

$f'(z) = 0, \tag{25}$

that is, $f(z)$ is constant.