I came across this problem but could not see any simple solution. Give a counterexemple for the following claim: "If P is a non-constant complex polynomial and V is a simply connected subset of C( set of complex numbers), so P(V), the image of V under P, is also simply connected".
Apparently, it does not seem so difficult, but I still can not figure out the trick. Any suggestion?
As Jacky Chong already suggested, consider the following: Let $$A := \{ z \in \mathbb{C} \; | \; 1 < |z| < 2, \Re(z) > 0 \}$$ and $$p(z) := z^3$$ Clearly, $p$ is a non-constant polynomial (hence holomorphic on all of $\mathbb{C}$ and, in particular, on $A$) and $A$ is a simply connected subset of $\mathbb{C}$. Every element $z \in A$ has the form $z = r e^{i \phi}$ with $r \in (1,2)$ and $\phi \in (-\pi/2, \pi/2)$. Hence, $p(z)$ for $z \in A$ yields a complex number of the form $w = p(z)$ with $w = R e^{i \theta}$, $R \in (1, 8)$ and $\theta \in (-\pi, \pi]$. Hence, the image domain $p(A)$ is the annulus $$p(A) = \{ z \in \mathbb{C} \; | \; 1 < |z| < 8 \}$$ which is clearly not simply connected.