If $a\operatorname{Re}(f(z))+b\operatorname{Im}(f(z))=c$ then f is constant

Let $A$ be an open connected subset of $\mathbb C$ and $f:A\to B$ analytic. Suppose there exist reals constants $a$, $b$ and $c$ such that $a^2+b^2\neq 0$ and $a\operatorname{Re}\bigl(f(z)\bigr)+b\operatorname{Im}\bigl(f(z)\bigr)=c$ for all $z \in \mathbb C$. Prove $f$ is constant.

Proof: as f is analytic then cauchy-riemann equations holds.

I computed the partial derivatives of $a\operatorname{Re}\bigl(f(z)\bigr)+b\operatorname{Im}\bigl(f(z)\bigr)=c$ with respect to $x$ and $y$ and tried to make a relation with the cauchy equations but things become complicated, because I got this:

$a\frac{\partial u}{\partial x}=a\frac{\partial v}{\partial y}=-b\frac{\partial v}{\partial x}=b\frac{\partial u}{\partial y}$ and I don't know how to get the result.

Solution 1:

Let $$Re(f)=u(x,y),\quad Im(f)=v(x,y),$$ Then we have $$au(x,y)+bv(x,y)=c.$$ Now differentiating both sides yields $$au_x+bv_x=0,\quad au_y+bv_y=0,$$ use Cauchy-Riemann equations we get $$au_x-bu_y=0,\quad au_y+bu_x=0.$$ Rewrite the above equations in matrix form: $$\begin{bmatrix}a &-b\\ b&a\end{bmatrix}\begin{bmatrix}u_x\\ u_y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}.$$ Note that $$det\begin{bmatrix}a &-b\\ b&a\end{bmatrix}=a^2+b^2\neq0,$$ therefore the system has a unique solution $$\begin{bmatrix}u_x\\ u_y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix},$$ and by Cauchy-Riemann equations again we get $v_x=v_y=0.$ Hence $f$ is constant.

Solution 2:

Note that $a\operatorname{Re}\bigl(f(z)\bigr)+b\operatorname{Im}\bigl(f(z)\bigr)=\operatorname{Re}\bigl((a-bi)f(z)\bigr)$. So, let $g(z)=(a-bi)f(z)$. Then the real part of $g$ is constant. From the Cauchy-Riemann equations you will be able to deduce that $g$ itself is constant. Therefore, $f$ is constant.

Solution 3:

If you know the open mapping theorem, you can argue this way: $f(A)$ is contained in the line $ax + by = c.$ Therefore $f(A)$ is not open in $\mathbb C.$ By the OMT this implies $f$ is constant.

Solution 4:

Here's an argument with a geometrical flavor:

Observe that if $u(x, y)$ and $v(x, y)$ are the real and imaginary parts of the holomorphic function $f(z)$, then

$\nabla u \cdot \nabla v = 0; \tag{1}$

indeed,

$\nabla u \cdot \nabla v = u_x v_x + u_y v_y = -u_x u_y + u_y u_x = 0, \tag{2}$

where we have used the Cauchy-Riemann equations

$u_x = v_y, \tag{3}$

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

in establishing (2). Now if

$au(x, y) + bv(x, y) = c, \tag{5}$

with $a, b, c$ constants, then taking the gradient yields

$a \nabla u + b\nabla v = 0; \tag{6}$

If we then take the scalar products of (6) with each of $\nabla u$, $\nabla v$, we obtain

$a \nabla u \cdot \nabla u + b \nabla v \cdot \nabla u = 0, \tag{7}$

and

$a \nabla u \cdot \nabla v + b \nabla v \cdot \nabla v = 0; \tag{8}$

using (1), we obtain

$a \nabla u \cdot \nabla u = 0; \tag{9}$

$b \nabla v \cdot \nabla v = 0. \tag{10}$

Since $a^2 + b^2 \ne 0$, at least one of $a, b$ does not vanish; suppose then that $a \ne 0$; by (9),

$\nabla u \cdot \nabla u = 0, \tag{11}$

whence

$\nabla u = 0, \tag{12}$

and hence by Cauchy-Riemann,

$\nabla v = 0 \tag{13}$

as well. From (12) and (13) we conclode that both $u(x, y)$ and $v(x, y)$, and hence $f(z)$, are constant. In the event $b \ne 0$, a similar argument produces the same result.

The essential idea here is that two vectors, in this case $\nabla u$ and $\nabla v$, cannot be both orthogonal (1) and collinear (6) without vanishing.

The preceding argument has the advantage that it is very basic, even elementary. A somewhat more sophisticated demonsteation may be based on the observation that the equation

$ax + by = c \tag{14}$

is that of a line $L$ in the $x-y$ plane; as such, $L$ is closed with empty interior; that is, it contains no open set; but the Open Mapping Theorem (https://en.m.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)), affirms that for non-constant $f(z)$, the image of an open set is open. But our hypotheses affirm that the image of the open set $A$ is contained in $L$, the interior of which is empty. So $f(z)$ must be a constant.

We may also generalize the present result to the case of an implicit curve $\gamma$ defined in the $u-v$ plane by a function $F(u, v)$; we take $\gamma$ to be the set $F(u, v) = c$. Then if $f = u + iv$ maps $A$ into $\gamma$, we must have

$F(u(x, y), v(x, y)) = c, \tag{15}$

whence by the chain rule

$F_x = F_u u_x + F_v v_x = 0, \tag{16}$

$F_y = F_u u_y + F_v v_y = 0; \tag{17}$

(16) and (17) may be re-written as

$F_u \nabla u + F_v \nabla v = 0, \tag{18}$

in parallel with (6); and, here as above so below we have

$F_u \nabla u \cdot \nabla u = 0, \tag{19}$

$F_v \nabla v\cdot \nabla v = 0, \tag{20}$

so now if we assume

$F_u^2 + F_v^2 \ne 0, \tag{21}$

or what is the same,

$\nabla_{(u, v)} F \ne 0, \tag{22}$

analogously to affirming $a^2 + b^2 \ne 0$, then we may conclude once again that

$\nabla u = \nabla v = 0, \tag{23}$

and thus $f(z) = u(x, y) + iv(x, y)$ is constant.

I think a similar result binds iff $\gamma$ is parametrically presented as $\gamma(t)$. One could always in principle, I believe, convert $\gamma(t)$ to implicit form. But I'll leave such researches to my myriad readers.

Of course, the Open Mapping technique applies to this latrer, more general case as well.