Consider a pair of holomorphic functions $f,g \in \mathcal{O}(\Delta)$ on the complex unit disk $\Delta = \{|z| < 1\}$ that both satisfy $f(0) = g(0) = 0$ and $f'(0) = g'(0) = 1$. *Does the domain
$$
f(\Delta) \cdot g(\Delta) := \{f(z)g(w) \mid z, w \in \Delta \} \subset \mathbb{C}
$$
contain the unit disk $\Delta$, with equality if and only if $f(\Delta) = g(\Delta) = \Delta$?*


In this setting, it may be useful to recall Bloch's theorem, which states that $f(\Delta)$ and $g(\Delta)$ contain disks of a fixed (absolute) radius; however those disks need not of course be centered at the origin. 



The largest disk contained by a 'product' of two simply connected plane regions with unit conformal radii
Let's start with the simple reduction. Notice that $f(\Delta)$ and $g(\Delta)$ are connected open sets containing small disks near the origin, so if one of them is unbounded, $f(\Delta)g(\Delta)=\mathbb C$. 


Let $a\in\Delta\setminus\{0\}$ (we certainly have $0=f(0)g(0)$, so the origin is never problematic) be not in $f(\Delta)g(\Delta)$. Consider $f(\Delta)\ni 0$ and $(a/g)(\Delta)\ni\infty$. Those are connected but not necessarily simply connected disjoint open sets. Let $\Omega$ be the union of $f(\Delta)$ and all bounded connected components of $\mathbb C\setminus f(\Delta)$ (or, which is the same, the complement of the connected component of $\mathbb C\setminus f(\Delta)$ containing $\infty$). If $(a/g)(\Delta)$ intersects $\Omega$, it must intersect $f(\Delta)$ as well (you cannot reach a bounded connected component of a complement of an open set by a path (actually even an open sausage, if you want) from $\infty$ without crossing the set itself. Now replace $f$ with the conformal mapping $\varphi$ from $\Delta$ to $\Omega$ with $\varphi(0)=f(0)$, $\Phi=\varphi'(0)>0$. Then, by the Schwarz lemma, $\Phi\ge f'(0)=1$, so if we consider $\widetilde f=\Phi^{-1}\varphi$, we will have $\Phi^{-1}a\notin \widetilde f(\Delta)g(\Delta)$ and $\widetilde f$ is now univalent. Similarly we can make $g$ univalent.


Now comes the main 


**Lemma:** Let $f$ be a (bounded and, if you want, analytic up to the boundary) univalent function such that $f(0)=0, f'(0)=1$. Let $A$ be the area on $\mathbb C\setminus\{0\}$ given by $dA(z)=|z|^{-2}dm\_2(z)$, which is invariant under $z\mapsto az (a\ne 0)$ and $z\mapsto z^{-1}$. Then
$$
A(f(\Delta\setminus r\Delta))\ge 2\pi\log\frac{1}{r}+o(1)\text{ as }r\to 0^+\,.
$$
**Proof:**
Let $S=\Delta\setminus r\Delta$. We have (since $f$ is univalent)
$$
A(f(S))\times 2\pi\log\frac 1r=\left[\int\_S\frac{|f'|^2}{|f|^2}\,dm\_2\right]\left[\int\_S\frac{1}{|z|^2}\,dm\_2\right]
\\
\ge
\left[\int\_S\frac{|f'|}{|f|}\,\frac{dm\_2(z)}{|z|}\right]^2=I^2\,.
$$
Note that 
$$
I=\int\_{[0,2\pi]}d\theta\int\_r^1d\rho \frac{|f'(\rho e^{i\theta})|}{|f(\rho e^{i\theta})|}\ge \int\_{[0,2\pi]}d\theta\int\_r^1 d\rho \frac d{d\rho}(\log|f(\rho e^{i\theta})|)
\\
=
\int\_{[0,2\pi]}d\theta\log|f(e^{i\theta})|-\int\_{[0,2\pi]}d\theta\log|f(re^{i\theta})|=I\_1-I\_2\,.
$$
However, under our assumptions we have $I\_1=0$ while $I\_2=2\pi\log r+O(r)$ as $r\to 0^+$ whence the lemma.


Now life gets easy. Notice that for sufficiently small $r>0$ both $f(S)$ and $(a/g)(S)$ lie in the annulus $\{w\in\mathbb C\,:\,(1-o(1))r\le|w|\le (1+o(1))|a|r^{-1}$. (we use both the boundedness and the univalence properties here). But the invariant area of this annulus is only $4\pi\log\frac 1r+\log|a|+o(1)$, so the images must overlap somewhere, thus finishing the story.


Cute question, by the way :-)


