Consider the open set $\{z \in \mathbb{C}: |z-i|<1\}$. Write down an explicit formula for a neighbourhood of $1/2 + i/4$ contained entirely in the open set.
I am not sure on how to complete this problem. I tried using the equation to a circle.
Consider the open set $\{z \in \mathbb{C}: |z-i|<1\}$. Write down an explicit formula for a neighbourhood of $1/2 + i/4$ contained entirely in the open set.
I am not sure on how to complete this problem. I tried using the equation to a circle.
Try thinking of $\mathbb{R^2}$ instead of $\mathbb{C}$. Making a sketch should help.
Hint assume the radius of the circle is $\delta$ and notice that
$ \left|i-(\frac{1}{2}+\frac{i}{4})\right|+\delta < 1 \,.$
Make use of the triangle inequality. A neigborhood is a set of points that are within $r$ of $1/2 + i/4$. You want all of those points to also be within $1$ of $i$.
So the distance from $i$ to $1/2+i/4$ is a known constant, the distance from $1/2+i/4$ to $x$ is less than $r$, and the distance from $i$ to $x$ is less than 1. Relate those quantities by the triangle inequality, then find a value of $r$ which makes them all true.