If $f$ is a holomorphic function in a rectangle in the first quadrant, and $|f(z)| \leq Re(z)$, prove that $f = 0$ for all the rectangle. If $f$ is a holomorphic function in a rectangle in the first quadrant, and $|f(z)| \leq Re(z)$, prove that $f = 0$ for all the rectangle.
The rectangle is all $z \in \mathbb{C} = x + iy$ s.t. $0<x<1$ and $ 0<y<1$.
An idea that I have to solve the problem is to create new holomorphic functions on the rectangle by "flipping" the image of f 90 degrees (multiplying it by a natural power of $i$). Then I can multiply these 4 functions together to get a new function $g$, holomorphic on the rectangle, and by the maximus modulus principle, $g$ has to be constant (0). 
I'm having a little bit of trouble formalizing these thoughts, and would appreciate any help! Thanks.
 A: The idea you have should work, except that I wouldn't flip the image of $f$ by 90 degrees, I would pre-compose $f$ by 90 degree rotations about the center of the square.
Let $\phi(z) = iz + 1$. This is the holomorphic map which rotates the complex plane 90 degrees around the point $\frac{1}{2} + \frac{1}{2}i$, the center of your square. 
Let $$g(z) = f(z)f(\phi(z))f(\phi^{\circ 2}(z))f(\phi^{\circ 3}(z)).$$ Then $g(z)$ is holomorphic, and you want to show that $|g(z)|\to 0$ as $z$ approaches the boundary of the square. As $f(z)$ is bounded by $1$ on the square, it suffices to show that one of the factors $f(\phi^{\circ k}(z))$ of $g$ tends to $0$ as $z$ approaches a boundary edge. If $z$ is approaching the left edge, $f(z)\to 0$. If $z$ approaches the right edge, then $f(\phi(z))$ approaches $0$, etc. Thus $|g(z)|\to 0$ as $z$ approaches the boundary of the rectangle, and by the maximum modulus principle, it follows that $g = 0$.
Now you want to conclude from this that $f$ is constant $ = 0$. I'm sure there are lots of ways to do this, but one is the following: if $f$ were not constant, then the set of $z$ where $f(z) = 0$ cannot have a cluster point in the rectangle, and hence is at most countable. Thus the set of $z$ where $g(z) = 0$ is at most countable, contradicting that $g\equiv 0$.
