Statement:
Complex function $f=u+\textbf i v$ is analytic in domain $D\subset \mathbb C$ and everywhere in $D$ $u=\phi(v)$, where $\phi=\phi(t)\in C^{1}(\mathbb R\to\mathbb R)$ and $\phi$ is strictly monotone. Prove that $f$ is constant in $D$.
My question is about sufficiency of strict monotonicity condition. I came up with the solution that does not use it.
Solution:
Let's write the Cauchy-Riemann conditions:
$$ \left\{ \begin{aligned} &u_x=v_y\\ &u_y=-v_x \end{aligned} \right. $$
Now use chain rule:
$$ \left\{ \begin{aligned} &\phi_tv_x=v_y\\ &\phi_tv_y=-v_x \end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned} &\phi_tv_x-v_y=0\\ &v_x+\phi_tv_y=0 \end{aligned} \right. \Leftrightarrow \begin{bmatrix} \phi_t & - 1\\ 1 & \phi_t \end{bmatrix} \begin{bmatrix} v_x\\ v_y \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Now we can find the determinant: $$ \det \begin{bmatrix} \phi_t & -1\\ 1 & \phi_t \end{bmatrix}= \phi^2_t+1\ne 0 $$ So we have the homogenous linear system with non-zero determinant, that implies the variables vector to be zero, so $v_x=v_y=0\big|_D$. It implies that $\Im f$ is constant in $D$ and it is an easy task to prove that it implies $f$ being constant in $D$.
As I mentioned before, the strict monotonicity is not used anywhere, but it is in the statement. Where am I wrong? Or may it be is a trap from the task writers?