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The following is an old qualifying exam problem,

Construct a conformal map from the unit disk to the parabolic region $ \{ z \in \mathbb{C}: \Im z > (\Re z)^2 \} $.

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I strongly believe that either you or the problem statement made a mistake and the question was intended to be the the outside of the parabola, viz. $Q := \{x+iy \in \mathbb{C} : y < x^2\}$. Indeed, in this case, the map is fairly easy to find: the obvious $z \mapsto z + iz^2$ is a conformal isomorphism from the lower-half plane to $Q$, and $w \mapsto i \frac{w+1}{w-1}$ is one from the unit disk to the lower-half plane, so by composing them we get a conformal isomorphism from the unit disk to $Q$.

But if you really mean $P := \{x+iy \in \mathbb{C} : y>x^2\}$, then here is a paper (Carroll & Hayman, "Conformal Mapping of Parabola-Shaped Domains", Computational Methods and Function Theory 4 (2004) 111–126) which answers the question (take $\alpha = 1/2$ and compose the maps $\phi^{-1}$ and $g^{-1}$ of the paper and multiplication by $i$). Considering how difficult it is, I think it is unreasonable that this was really meant as a qualifying exam problem.

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    $\begingroup$ How do you show that the map $z+ i z^2$ is injective on the lower half plane? Is there an easy way to see that is surjective onto $Q$ without the tedious algebraic calculations? $\endgroup$ – wellfedgremlin Jul 11 '14 at 21:41
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    $\begingroup$ Injectivity: if $z + iz^2 = z' + iz^{\prime2}$ for $z,z'\in\mathbb{C}$, then $(z'-z) + i(z'-z)(z'+z) = 0$ so either $z'=z$ or $z'+z = i$, and the latter is impossible for $z,z'$ in the lower-half plane. This shows that $z \mapsto z + iz^2$ is a conformal isomorphism to the image. But since it extends continuously to the boundary (the equator of the Riemann sphere), the boundary of the image will be the image of the boundary, hence, the parabola. So now it's just a question of finding the right component, and a single point suffices. $\endgroup$ – Gro-Tsen Jul 12 '14 at 23:54
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    $\begingroup$ Alternatively, I believe the following is true: if $U$ is the interior of a Jordan curve $\gamma$, and $f$ is holomorphic on $U$ and extends continuously to (the image of) $\gamma$, and $f'$ does not vanish on $U$ and (crucially) $f\circ\gamma$ is also a Jordan curve, then $f$ is a conformal isomorphism from $U$ to the interior of $f\circ\gamma$ (the crucial point being: $f$ is injective). I'll see if I can provide a reference for this. $\endgroup$ – Gro-Tsen Jul 12 '14 at 23:58
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    $\begingroup$ Sketch of proof of previous comment's statement: the winding number of $f\circ\gamma$ around a point $w$ of its interior can only be $1$ (with the right orientation), see, e.g., here, but this counts the number of times $f$ takes the value $w$, with multiplicity, so $f$ takes exactly once the value $w$. (And, it seems, the nonvanishing of $f'$ is automatic.) $\endgroup$ – Gro-Tsen Jul 13 '14 at 0:24
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    $\begingroup$ Maybe you'll find the following way of stating it more convincing: writing $z+iz^2$ as $i(z-i/2)^2+i/4$, it's obvious that $z\mapsto z+iz^2$ gives a conformal isomorphism, and in particular a homeomorphism, of the region $D=\{\Im(z)<1/2\}$ to $E$=the plane minus the half-line $\Im(w)\geq1/4$ on the purely imaginary axis. Now the real axis separates $D$ into two connected components (the lower-half plane and the strip $\{0<\Im(z)<1/2\}$), so its image, the parabola, separates $E$ into two connected components, each one being the image of one at the source, and it's obvious which is which. $\endgroup$ – Gro-Tsen Jul 13 '14 at 16:14

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