Show that $f(z)$ is constant 
If $f(z)=u(x,y) + iv(x,y)$ is an entire function such that $u\cdot v$ is constant then $f(z)$ is constant. 

I know that I need to use the Cauchy-Riemann equations, but I don't know how to start. Should I differentiate $u\cdot v$ with respect to $x$ or $y$ then substitute with Cauchy-Riemann?
 A: I think this might work:
Consider the entire function $(f(z))^2$; with
$f(z) = u(x, y) + iv(x, y) \tag{1}$
we have
$(f(z))^2 = (u(x, y) + iv(x, y))^2 = (u(x, y))^2 - (v(x, y))^2 + 2i u(x, y) v(x, y).  \tag{2}$
A very simple application of the Cauchy-Riemann equations to (2), using the fact that $u(x, y) v(x, y)$ is constant, shows that $u(x, y))^2 - (v(x, y))^2$ is constant is well; i. e., we have, by hypothesis,
$(u(x, y) v(x, y))_x = (u(x, y) v(x, y))_y = 0,  \tag{3}$
or
$\nabla(u(x, y) v(x, y)) = 0; \tag{4}$
thus CR yields
$\nabla((u(x, y))^2 - (v(x, y))^2) = 0; \tag{5}$
$(u(x, y))^2 - (v(x, y))^2$ is also constant.  We see then that $(f(z))^2$ is itself constant, and so $f(z)$ must be . . . you guessed it, constant.
Those feeling the need for yet more verbiage can note that 
$f(z)f'(z) = \dfrac{1}{2}((f(z))^2)' = 0; \tag{6}$
if $f(z) = 0$, then it is . . .  well, constant; if $f(z) \ne 0$, then $f'(z) = 0$, and again, $f(z)$ is . . . constant!  QED!
Note added Sunday 31 May 2015 10:10 AM PST: This in response to Samar Hayek's comment below:  for general holomorphic $g(z) = r(x, y) + is(x, y)$, Cauchy-Riemann asserts that $r_x = s_y$ and $r_y = - s_x$; if now $\nabla s = 0$, $s_x = s_y = 0$, so $r_x = r_y = 0$, i.e. $\nabla r = 0$.  Applying this to the above shows (4) implies (5), since $u^2 - v^2$ and $2uv$ and $u^2 - v^2$ are the real and imaginary parts of $(u + iv)^2$.  End of Note.
A: More easiest way:
Consider the function $$g(z)=e^{if^2(z)}.$$
Given , $uv=\text{ constant }=c$(say).
Now $|g(z)|=e^{-2c}$
Then $g(z)$ is bounded entire function and hence it is constant and consequently $f(z)$ is constant.
A: The condition given implies that the range of $f(z)$ is a subset of a hyperbola $xy = c$. By the open mapping theorem this is only possible if $f(z)$ is constant.
If you wish to use the Cauchy-Riemann equations, it would be better to use them on $g(z) = (f(z))^2$, since Im$(g(z)) = 2uv$ is constant and therefore the Cauchy-Riemann equations immediately give that all partials of $g(z)$ are zero. This implies $g(z)$ is constant, and it's not hard to go from there to show that $f(z)$ is constant.
A: Hint: Given u.v=constant.....(1)
Differentiate partially (1) w.r.t. x to obtain,

$uv_x+vu_x=0$ ......(2)

and  partial differentiation of (1) w.r.t. y gives,

$uv_y+vu_y=0 $

Use C-R equations i.e. v_y=u_x and v_x=-u_y we get,           

$uu_x-vv_x=0$  ...(3)

Solve (2) and (3) to get u=constant. Similarly, obtain v=constant. Hence proved.
