I got the question of trying to determine wheter the subset of $C$: $G=\{z:|z|<1\}$ is open, closed or both, I know the answer is that the set is open since there is an open ball around every point of G that is a subset of G. I am however a bit confused on how constructing a radius works.
My thoughts: For $z \in G$ let $\epsilon_{z}=1-|z|$ because if we draw a straight line from the origin to $z$ we get a line of length $|z|$ and if we continue this line to the unit circle, then the length from the point where the line intersect the circle and the point $z$ should be $1-|z|$ and so whenever $|z-x|< 1-|z|=\epsilon_{z}$, $x$ should be in the circle.
Also known is that the union of open subsets is open, aswell as the fact that each open ball is open and hence that $G$ is open since it is the union of open balls of radii $\epsilon_{z}$ centered around each $z \in G$.
Is this correct for an "to show that G is open argument"? and is it bad to be precise in what radius one chooses? I namely checked the solutions manual which chooses $\frac{1-|z|}{2}$ as their $\epsilon_{z}$ which confuses me as $1-|z|$ feels like the obvious candidate (atleast geometrically)
The book in question is "Functions of One Complex Variable 1 -John B Conway" and the questions is question 2a chapter 2
What might be an easier way to show that your radius works is the following.
Let $x\in G$, and consider $B_{1-|x|}(x)$ the open ball of radius $1-|x|$ around $x$. Then for any $z\in B_{1-|x|}(x)$ we find that $$|z|\leq |z-x|+|x|<1-|x|+|x|=1,$$ so $z\in G$. This shows that $B_{1-|x|}(x)\subset G$.