Nonconstant entire functions have dense image

Suppose that \(f(\mathbb C)\) were not dense in \(\mathbb C\). Then there would be a point \(a\in\mathbb C\) and an \(r>0\) such that \[ B(a,r)\cap f(\mathbb C)=\varnothing. \] Thus \[ |f(z)-a|\ge r\qquad\text{for every }z\in\mathbb C. \] The function \[ g(z)=\frac1{f(z)-a} \] is therefore entire and bounded by \(1/r\). By Liouville’s theorem, \(g\) is constant. Hence \(f\) is constant, contradicting the hypothesis.

Therefore every nonempty open disk meets \(f(\mathbb C)\), so \(f(\mathbb C)\) is dense in the plane.