Let $\varepsilon\in(0,1)$ be given. Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is a continuous, open mapping and suppose $t:\mathbb{C}\rightarrow\mathbb{C}$ is a continuous function such that $|t(z)|<\varepsilon$ for all $z\in\mathbb{C}$. Define $g(z)=f(z)+f(z)t(z)$. Is it true that $g$ is an open map also?
Intuitively I believe this to be true, for since $f$ is open and $g$ is a 'small' variant of $f$ and so no ball can collapse under the mapping $g$. However I am struggling to show this computationally. I have tried showing this from definition but keep getting lost in the algebra. Here I have been considering $\mathbb{C}\cong\mathbb{R}^2$ equipped with the Euclidean norm. Any help would be appreciated.