A quick question about analytic functions on the unit disk with $Re(z) \leq 0$ Suppose f is analytic in the domain $$D(0; 1) = \{z ∈ \Bbb C : |z| < 1\}$$ 
and $\operatorname{Re} f(z) ≤ 0$ for all $z \in D(0; 1)$. If $\operatorname{Re} f(0) = 0$, show that $f$ is constant on $D(0; 1)$.
I have a feeling this can be solved quickly with the Cauchy-Riemann equations and/or the complex difference quotient and forcing a zero derivative everywhere on a radius from zero, but I can't get them to work out. 
My other attempt was via assuming the contrary and trying to show the existence of a point that would have to have a positive real part in its image if the function was not constant, maybe through some application of the Maximum Modulus Theorem, but just because $f(0$ has an upper bound on the image of the real components of $f(D)$ doesn't mean it is the maximum modulus of course. But geometrically, I envision analytic complex functions to stretch/compress and turn its preimage, like making a design from pizza dough (not folding an edge into the interior), but I am not sure how to express this mathematically or even if this is the correct topological intuition in general for analytic functions of a single variable in the complex plane.  
Thanks in advance - sorry if this is too chatty, but I have the feeling getting a hint for this question may help me root out neglected or misunderstood parts of Complex in general in my thinking. 
 A: Hint: Apply the Maximum Modulus Theorem to $g(z)=e^{f(z)}$.
A: One way to say Cauchy's Integral Formula is that the average value of an analytic function on the boundary of a disk in its domain is the value of the function at the center. The same is then true of the real and imaginary components of the function; this is expressed by saying they are harmonic functions. If ${\rm Re}f(z_0) < 0$ for some $z_0 \in D(0, 1)$, the average value on the circle $|z|=|z_0|$ is negative since ${\rm Re}f \leq 0$ is continuous, but then ${\rm Re}f(0) < 0$. Hence ${\rm Re}f(z) = 0$ everywhere. Now $f$ can't be an open map, so from the open mapping theorem, it must be constant.
I'm not sure how to use just the maximum modulus principle for analytic functions or the Cauchy-Riemann equations to solve it. (Edit: Ted Shifrin gave a maximum modulus proof.)
A: Since $f$ is analytic, hence $\Re{ f(z)}$ is harmonic which means it follows the maximum principle. But $\Re{f(z)} \leq 0 \text{ for } z \in \mathbb{D}$ and $\Re{f(0)} = 0 $  implies that $\Re{f(z)} \equiv 0$ in $\mathbb{D}$. Now using the Cauchy- Riemann equations you can show that $\Im{f(z)} = $ constant.  
A: Many thanks - I just verified one of my intuitions by checking the details of the Open Mapping Theorem. 
Choose $r<1$ and consider the connected open set $B(0,r) \subset D$. Then by the Open Mapping Theorem, since $f$ is holomorphic on $D$ and thus holomorphic on $B$, $f(B)$ is an open set. But that means there exists an $s>0$ such that, for some $y_0 \in \mathbb R, \ B(y_0i,s) \subset f(D)$. But this means there exists a $z_0 \in B(y_0i,s)$ with $Re(z_0)>0$ and so there exists a $\omega \in B(0,r) \subset D$ with $Re(f(\omega))>0$, a contradiction. So $f$ must be constant. 
