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Is there a neat way of determining the domain that maps to the upper half plane by the map $f(z)={z^2+4\over z}$? i can see that the map is holomorphic everywhere except at $0$. And therefore the mapping is conformal everywhere but at 0.

Also, what if the map were something more complicated, say the $\sinh (az)$?

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@JonasMeyer: Thanks for pointing out my mistake. How do I proceed knowing this information? –  Rocky May 17 '12 at 23:28
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If $z=x+iy$, the inequality $\mathrm{Im}(f(z))>0$ reduces to $y(x^2+y^2-4)>0$. Since $x^2+y^2-4=0$ is the equation of a circle, the inequalities give you either the inside or outside of the circle. Keeping in mind that $ab>0$ if and only if $a$ and $b$ have the same (nonzero) sign, this allows you to describe the preimage of the upper half plane in terms of whether you're inside or outside the circle, and whether you're above or below the line $y=0$.

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