Let $f(x+iy) = u(x,y) + iv(x,y)$ be a holomorphic function on $\Omega \subset \mathbb{C}$, where $u\cdot v = 1$. Prove that $f$ must be constant. 
Let $f(x+iy) = u(x,y) + iv(x,y)$ be a holomorphic function on $\Omega
 \subset \mathbb{C}$, where $u\cdot v = 1$. Prove that $f$ must be
  a  constant function.

I take for granted that $ u'_x = u'_y = v'_x = v'_y = 0$ means the function is constant (I'm not entirely sure about this). 
Since $f$ is holomorphic, it's true that
$$u'_x = v'_y$$
$$u'_y = -v'_x$$
Since $u(x,y) = \frac{1}{v(x,y)}$ we get 
$$u'_x = -\frac{v'_x}{v(x,y)^2},$$ 
and similarily $$v'_y = -\frac{u'_y}{u(x,y)^2}.$$
Substituting $u'_x, v'_y$ in the the Cauchy-Riemann equation nets
$$-\frac{v'_x}{v(x,y)^2} = -\frac{u'_y}{u(x,y)^2}$$
$$u(x,y)^2v'_x=u'_yv(x,y)^2$$
$$u(x,y)^2v'_x=-v'_xv(x,y)^2$$
Dividing by $v'_x$ gives us the impossible equation
$$u(x,y)^2=-v(x,y)^2$$
($u^2$ cannot be negative), which must mean we performed an illegal operation and the only possibility is that $v'_x = 0$. This directly implies that $u'_y = 0$ from the Cauchy-Riemann equations. And setting these to zero in the first equations gives $u'_x = v'_y = 0$
Thus, $u'_x = u'_y = v'_x = v'_y = 0$ which means f is constant.
Is there anything wrong with my proof, and is there an easier way to do it?
 A: From $u\cdot v={1\over2}{\rm  Im}(f^2)$ it follows that the holomorphic function $f^2$ has constant imaginary part. This implies that $f$, being continuous, is locally constant, hence constant.
A: I think that here we need that $\Omega$ is an open connected subset of $\mathbb{C}$. In that case your proof is fine. 
Alternative way. Consider the function $\exp(if^2(x))$ then
$$|\exp(if^2(x))|=\exp(-\mbox{Im}(f^2))=\exp(-(2uv))=e^{-2}$$
then, by Why: A holomorphic function with constant magnitude must be constant. , $\exp(if^2(x))$ is constant and therefore also $f$ is constant.
A: I am assuming $\Omega$ is an open, connected subset of $\Bbb C$.  More remarks pertinent to this assumption are made near the end of his answer.
With holomorphic 
$f(x + iy) = u(x, y) + iv(x, y), \tag 1$
we have 
$f^2(x + iy) = (u(x, y) + iv(x, y))^2 = u^2(x, y) - v^2(x, y) + 2i u(x, y) v(x, y) \tag 2$
is also holomorphic.  Thus we may apply the Cauchy-Riemann equations to $ u^2(x, y) - v^2(x, y)$ and $2u(x, y) v(x, y)$ and see that
$(u^2(x, y) - v^2(x, y))_x = (2u(x, y) v(x, y))_y, \tag 3$
$(u^2(x, y) - v^2(x, y))_y = -(2u(x, y) v(x, y))_x. \tag 4$
If, then, 
$u(x, y) v(x, y) = \beta, \tag 5$
a constant, we see from (3)-(4) that
$(u^2(x, y) - v^2(x, y))_x = 0, \tag 6$
$(u^2(x, y) - v^2(x, y))_y = 0, \tag 7$
i.e., that
$\nabla(u^2(x, y) - v^2(x, y)) = 0. \tag 8$
If now $(x_0, y_0)$ and $(x, y)$ are any two points in a given connected, open set $\Omega \subset \Bbb C$, we may join them by a differentiable path
$\gamma:[0, 1] \to \Omega, \tag 9$
such that
$\gamma(t) = (\gamma_x(t),\gamma_y(t)), \; \gamma(0) = (x_0, y_0), \; \gamma(1) = (x, y); \tag{10}$
then
$(u^2(x, y) - v^2(x, y)) - (u^2(x_0, y_0) - v^2(x_0, y_0)) = \displaystyle \int_0^1 \dfrac{d(u^2(\gamma_x(t), \gamma_y(t)) - v^2(\gamma_x(t), \gamma_y(t))}{dt} dt$
$= \displaystyle \int_0^1 \nabla(u^2(\gamma_x(t), \gamma_y(t) - v^2(\gamma_x(t), \gamma_y(t)) \cdot \gamma'(t) dt = \int_0^1 0 \cdot \gamma'(t) dt = 0;\, \tag{11}$
or
$u^2(x, y) - v^2(x, y) = u^2(x_0, y_0) - v^2(x_0, y_0) \tag{12}$
for any $(x, y)$ in the same component of $\Omega$ as $(x_0, y_0)$.  It follows that $u^2(x, y) - v^2(x, y)$ is constant on components of $\Omega$, hence $f^2(x + iy) = u^2(x, y) - v^2(x, y) + 2iu(x, y) v(x, y)$ is also, hence $f(x + iy)$ as well.
We see upon scrutiny of the preceding argument that it requires $\Omega$ to be path connected to be effecitve; thus we have in actuality shown that $f(x + iy)$ is constant on the path components of $\Omega$; but these are the same as the components of $\Omega$ for open $\Omega \subset \Bbb R^2$. It is possiible for $f(x + iy)$ to take on different values on different components of $\Omega$.
As far as our OP Heuristics' proof, the only feature which bothers me is the division by $v'_x$, which may vanish on some subset of $\Omega$.  For example, taking $f(x + iy) = (x + iy)^2 = (x^2 - y^2) + 2ixy$, we see that $v = 2xy$ hence $v'_x = 2y = 0$ on the entire real axis.  I think there may be a way to work around this, but I'm not sure what it is at the moment.
