Let $f$ be an analytic function. If $\Im(f) \Re(f) = 1$, prove that $f$ is constant. A function $f:\mathbb{C} \rightarrow \mathbb{C}$ is everywhere differentiable, and $f(x+iy) = u(x,y) + iv(x,y)$, where $u,v$ are real-valued functions. Suppose that $u(x,y)v(x,y) = 1$ for all real $x,y$. Show that $f$ is constant.  
I'm not sure how to approach this. This is what I tried.
Since $f$ is analytic, then the C-R equations hold:
$$u_x = v_y \ \ \ \ u_y = -v_x.$$  
Also, using $uv = 1$, differentiating w.r.t $x$,
$$u_x v + v_x u = 0$$
and w.r.t $y$,
$$u_y v + v_y u = 0.$$
Hence, the C-R equations will simplify to
$$u_x = v_y = -\frac{u}{v}v_x$$
and
$$u_v = -v_x = \frac{u}{v}v_y$$
Not very sure what to do now.
 A: Probably the easiest way to obtain the result is by looking at $f^2$. The assumption says $f^2$ has constant imaginary part, and that implies $f^2$ is constant (if that's not yet officially known, it's straightforward to deduce it from the Cauchy-Riemann equations). Say $f^2(z) \equiv c$. If $a^2 = c$, then $f(z) \in \{a, -a\}$ for all $z$, and by continuity both, $f^{-1}(a)$ and $f^{-1}(-a)$ are open. Since the domain of $f$ is connected, one of the two preimages must be empty, i.e. $f$ must be constant.

We can write the relations you found as
$$\begin{pmatrix} u_x & v_x \\ u_y & v_y \end{pmatrix} \begin{pmatrix} v \\ u\end{pmatrix} = 0.$$
Since $(u,v) \neq 0$, it follows that
$$0 = \det \begin{pmatrix} u_x & v_x \\ u_y & v_y \end{pmatrix} = u_x v_y - u_y v_x,$$
and by the Cauchy-Riemann equations, that determinant is $u_x^2 + u_y^2 = v_x^2 + v_y^2$. A sum of squares of real numbers is only zero if all the numbers are zero, so that implies $u_x = u_y = v_x = v_y = 0$, hence $f$ is constant.
