Why is the region formed by $|z|<1$ and $|z-2|<1$ not connected? Why is the a region $D$ formed by $|z| < 1$ and $|z-2|<1$ not connected?
After thinking about this I arrived on two answers
1) because the two circles don't intersect so the region is formed by two disconnected regions.
2) If we take a point $a$ in $|z|<1$ and a point $b$ in $|z-2|<1$ then there is no straight line in $D$ that connects these two points.
Which of these is the correct explanation?
 A: Both your answers are mathematically poor.


because the two circles don't intersect so the region is formed by two disconnected regions.

This sentence is mathematically bad because you are operating with poorly determined concepts. So, before you say a sentence like this, you need to explain:


*

*What is a region?

*What does "disconnected" mean?

*How do you know the two "regions" are "disconnected"

*How does the two regions being disconnected prove that the entire set is disconnected?




If we take a point $a$ in $|z|<1$ and a point $b$ in $|z-2|<1$ then there is no straight line in $D$ that connects these two points.

This statement is mathematically bad, because


*

*You didn't prove the statement.

*The same statement is true for the set $\{z\in \mathbb C: |z|=1\}$, but this set is connected.



In order to trully answer the question, you need to first define what "connected" means and then, following that definition, prove that your region is not connected.
A: In general topology, the definition of connected topological space is:
Let $(X, \tau)$ be a topological space. Then $(X, \tau)$ is connected if the only sets $B \in \tau$ satisfying $B^C \in \tau$ are $X$ and $\emptyset$.
Now, let $D$ be your $X$, and $\tau = \tau_e$, the euclidean topology. $D = D_1\cup D_2$, where $D_1 = \left\{z \in \mathbb C:|z| < 1\right\} \in \tau$ and $D_2 = \left\{z \in \mathbb C : |z - 2| < 1\right\} \in \tau$.
Thus, by definition, $D$ is not connected.
Now, you're two questions lack of details:


*

*I can see what you mean, but it's just an intuition what you wrote.

*This is even worse, mathematically it isn't correct. Instead of straight line, the rigorous definition of the idea written by you as second answer is arc connected space, which definition can be found here.

A: Your first answer is essentially correct, and it's possible that that's all your professor is really looking for, if this say an intro complex analysis class and you haven't studied topology yet.
Here's an example of what a proper mathematical proof of something like this might look like.
The definition of a connected set is something like this:

A set is connected iff given $x$, $y$ in the set, there is a curve from $x$ to $y$ which stays inside the set.

Except that "curve" has to be replaced with a proper definition:

A curve from $x$ to $y$ is a continuous function $f$ from $[0, 1]$ to $\mathbb C$ such that $f(0)=x$ and $f(1)=y$.

Remember that even "continuous function" isn't just an intuitive concept, it too has a precise technical definition involving $\epsilon$-s and $\delta$-s! All of these technicalities are going to have to show up in your proof in some way, either explicitly or in the form of theorems which themselves are proven using them.
You want an argument that basically says that a continuous curve can't go from the first disk to the second disk without going through the gap in between. In this case the simplest way is probably to use the intermediate value theorem:
Let $x$ be in the disk of radius $1$ around $0$ and $y$ be in the disk of radius $1$ around $2$. Let $f(t)$ be some curve from $x$ to $y$. Then the function $|f(t)|$ is a continuous real valued function (because it's a composition of continuous functions). $|f(0)|=|x|<1$, but $|f(1)|=|y|>1$, where you can prove the second inequality by the triangle inequality:
$$|y|=|2+(y-2)|\geq|2|-|y-2|>1$$
(I used a slightly less common, but equivalent form of the triangle inequality, which is $|a+b|\geq|a|-|b|$). Anyway, by the intermediate value theorem, there exists some $t\in[0, 1]$ such that $|f(t)|=1$. But then $f(t)$ can't be in either disk. It clearly isn't in the first disk, and by using the triangle inequality again you can prove it isn't in the second disk either.
Notice how we got lucky here because the formal definition of your set, using the definition of a disk, happened to be very mathematically simple. If we'd had some crazy blob shapes instead of disks, it might not have been so easy.
