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3 1 $begingroup$ Suppose $f$ is entire and $lim_{ztoinfty}f(z)=infty$ . Show that $f(mathbb{C})=mathbb{C}$ . First of all I don't really understand this question. I know $ztoinfty$ means $|z|toinfty$ , but what does $f(z)toinfty$ means? Does it mean $|f(z)|to infty$ ? Also, I just learnt about the one point compactification $infty$ to the complex plane. So the reason we write $ztoinfty$ instead of $|z|to infty$ is because we are referring to $infty$ as a point in the extended complex plane $bar{mathbb{C}}$ ? So $f(z)toinfty$ is also referring to $infty$ in $bar{mathbb{C}}$ ? complex-analysis share | cite | improve this question asked 42 mins ago ...