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.
Best Answer
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.