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Given the transformation $C(z)=\frac {z-i}{z+i}$ , $z\in \mathbb{C}$, I'm told that

$f(\infty)$=1. I understand that $f(-i)=\infty$ but what does $f(\infty)$ even mean?.

Generally I have trouble determerning the image of a subset under a complex function?

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    Probably means that as $|z|\to\infty$ we have $f(z)\to1$. – 2017-01-10
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    By definition, $$f(\infty)=\lim_{|z|\to\infty}f(z)$$ when the limit on the RHS exists. Thus, $f(\infty)=1$ means $$\forall\epsilon>0\quad\exists K\quad\forall z\in\mathbb C\quad|z|>K\implies|f(z)-1|<\epsilon$$ – 2017-01-10
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    In the context of Mobius transformations, it is customary to extend the complex plane to the Riemann sphere, $\Bbb C\cup \{\infty\}$ with the properties that $\frac z0 = \infty$ for $z \neq 0$ and $\frac{z}{\infty} = 0$ for $z \neq \infty$ (and possibly a few more I might've forgotten). The indeterminate forms $\frac\infty\infty$ and $\frac00$ are decided through limits the usual way. – 2017-01-10

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