Let $\Omega\subsetneqq\mathbb{C}$ be a simply connected domain. Given some $a\in\Omega$, by the Riemann mapping theorem, the family
$$\mathcal{A}=\{g:\Omega\to\mathbb{C},\ g\text{ is analytic},\ g(a)=0,\ g'(a)=1\}$$ is nonempty. Let $g_0\in\mathcal A$ be the one given by the Riemann mapping theorem, that is, $g_0$ is the conformal equivalence from $\Omega$ to some disk $D_0$ centered at $0$. For any $g\in\mathcal A$, let $$R(g)=\sup_{z\in\Omega}|g(z)|$$ Then $D_0$ has radius $R(g_0)$. Prove that
(1) $R(g)\geq R(g_0)$ for any $g\in\mathcal A$
(2) If $g(\Omega)$ is Jordan measurable, then $V(g(\Omega))\geq\pi(R(g_0))^2$, where $V$ denotes the area of a set.
My progress: I have shown (1) by contradiction and Schwarz's lemma. As to (2), I have $$V(g(\Omega))=\iint_{g(\Omega)}dudv=\iint_\Omega|g'(\omega)|^2dxdy$$ (the second equality follows from $u=u(x,y), v=v(x,y)$ where $g(z)=u(x,y)+iv(x,y)$) But how can I show $|g'|\geq|g_0'|$, or is this a dead end?
Instead of looking at $g$ directly, look at $g\circ g_0^{-1}\colon\mathbb{D}\to\mathbb{C}$. We have $V(g(\Omega))=V((g\circ g_0^{-1})(\mathbb{D}))$ and $(g\circ g_0)(0)=0$, $(g\circ g_0)'(0)=1$. Maximum modulus principle on $(g\circ g_0)'$ and integrating over $\mathbb{D}$ gives the result.