 Solution

The uniformization theorem states that every simply connected Riemann surface is biholomorphic to exactly one of the Riemann sphere $\hat{\mathbb{C}}$, the complex plane $\mathbb{C}$, or the unit disc $\mathbb{D}$. A surface is uniformized by one of these when that surface is its universal cover. The only surface uniformized by $\hat{\mathbb{C}}$ is the sphere itself, because every nonidentity Möbius transformation has a fixed point and therefore cannot act as a deck transformation. The Riemann surfaces uniformized by the complex plane are

Let $U\subseteq \mathbb{C}$ have a complement containing at least two points. Its universal cover cannot be the sphere because $U$ is noncompact. If it were the plane, the covering map $\mathbb{C}\to U\hookrightarrow \mathbb{C}$ would be a nonconstant entire function omitting two values, contrary to the Little Picard theorem. The plane domain with two omitted points is hyperbolic, so

Now put $S=R\setminus \{P_1,\dots ,P_n\}$. The spherical possibility occurs only for $(g,n)=(0,0)$. By the preceding list, the plane-uniformized possibilities arewhich give respectively $2g-2+n=-1,0,0$. Every remaining pair has $2g-2+n>0$ and must have disc universal cover. Thus the uniformization of a punctured compact Riemann surface gives

For the complex torus $X=\mathbb{C}/\Lambda$, any analytic map $f:\mathbb{C}\to X$ lifts through the universal covering map to an entire function $F:\mathbb{C}\to \mathbb{C}$. If $F$ is constant, so is $f$. Otherwise Little Picard says that $F$ omits at most one complex number. For every $x\in X$, its lifts form an infinite coset $z+\Lambda$, so at least one lift is attained by $F$. Hence the holomorphic map from the complex plane to a one-dimensional complex torus satisfies

Finally, $\mathbb{C}$ and $\mathbb{D}$ are homeomorphic byThey are not conformally equivalent: a biholomorphism $\mathbb{C}\to \mathbb{D}$ would be a nonconstant bounded entire function, contradicting the Liouville theorem. This proves the claimed example of complex plane and unit disc are homeomorphic but not conformally equivalent.