Open set is complex numbers

Question

I want to show that each of the inequalities $(1): Re(z)>0$ and $|z+z^2|<1$ defines an open set in $C$. I showed that $|z|<1$ is open. I tried to apply any inequality like triangle inequality but did not work. I appreciate any help in this regard.

Answer 1

For every continuous function $\eta\colon\mathbb{C}\longrightarrow\mathbb R$ and every open subset $A$ of $\mathbb R$, $\eta^{-1}(A)$ is an open subset of $\mathbb C$. So, consider the functions $\operatorname{Re}$ and $f(z)=\lvert z+z^2\rvert$.