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Yes. Let $f$ be univalent, with image in the open right half-plane. Then $f^2$ is also univalent. Applying the standard growth estimate for univalent functions to $f^2$, we conclude that $f$ satisfies your growth condition.

To arrange that $f$ satisfies the second condition, notice that we only need $\limsup|f(z)|=\infty$ at every point of the unit circle (Then your condition easily follows)follows. For the point $1$, you first take a sequence of boundary points tending to $1$, then a sequence tending to each point of this sequence where $f\to\infty$, then a diagonal sequence.).

To ensure that $\limsup|f(z)|=\infty$ at every point, one can use geometric construction of the image. Begin with some disc in the right half-plane. Then dig narrow channels from the circumference to infinity, and continue digging channels from dense sets of boundary points... Or look at the pictures in Collingwood Lohwater, The theory of cluster sets, . Or look at the pictures of unbounded normality regions in transcendental dynamics, for example, http://en.wikipedia.org/wiki/Pierre_fatou; according to a theorem of Baker, all these regions have the property that infinity belongs to the impression of every prime end, which exactly means that your mapping function has the required property.

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Yes. Let $f$ be univalent, with image in the open right half-plane. Then $f^2$ is also univalent. Applying the standard growth estimate for univalent functions to $f^2$, we conclude that $f$ satisfies your growth condition.

To arrange that $f$ satisfies the second condition, notice that we only need $\limsup|f(z)|=\infty$ at every point of the unit circle (Then your condition easily follows).

To ensure that $\limsup|f(z)|=\infty$ at every point, one can use geometric construction of the image. Begin with some disc in the right half-plane. Then dig narrow channels from the circumference to infinity, and continue digging channels from dense sets of boundary points... Or look at the pictures in Collingwood Lohwater, The theory of cluster sets, Or look at the pictures of unbounded normality regions in transcendental dynamics, for example,

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Yes. Let $f$ be univalent, with image in the open right half-plane. Then $f^2$ is also univalent. Applying the standard growth estimate for univalent functions to $f^2$, we conclude that $f$ satisfies your growth condition.

To arrange that $f$ satisfies the second condition, notice that we only need $\limsup|f(z)|=\infty$ at every point of the unit circle (Then your condition easily follows).

To ensure that $\limsup|f(z)|=\infty$ at every point, one can use geometric construction of the image.