Complex Analytic functions that are lines are constant 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? 
 A: 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]. 
A: 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.
