I have the following set:
$$Q = \{z\in \mathbb{C}: |z-z_0|+|z-z_i|<r\}$$
and I must prove that it's connected, where the definition of connected is the following: for each $2$ points on the set, we can connect them with segments (I don't know how to translate it to english, but here they're called 'poligonals' which are segments connected to the tip of the other). I must first visualize this set of points in order to have an intuition about which poligonal will work.
I know that in the reals, this is the equation of a square:
$$|x|+|y|<r$$
I can solve for $y$ an break down to $4$ cases, in which each solution will give a line that is a side of the quase. But how can I make it for complex numbers? Like, if I isolate $|z-z_1|$ I get:
$$|z-z_1|<r-|z-z_0|$$
but when it comes to opening the absolute values in cases, I can't, because complex numbers don't have the notion of order, so I can't use $<$ and so. I think it will be a square too, but I don't know how to prove it.
If I suppose $z_0 = a_0+ib_0, z_1 = a_1+b_1i$ e $z = x+iy$
I end up with:
$$\sqrt{(x-x_0)^2+(y-y_0)^2}+\sqrt{(x-x_1)^2+(y-y_1)^2}<r$$
this seems hard to deal with too.
So, how to find the graph of this, and how to prove that it's path connected?
Hint: State your goal like this: you want to prove that if $a,b\in Q$, then $(1-t)a+tb\in Q$ for any $t\in [0,1]$. (Those are the points on the line segment connecting $a$ and $b$.) So, given that $$|a-z_0|+|a-z_1|<r\quad\quad |b-z_0|+|b-z_1|<r$$ you want to conclude that $$|(1-t)a+tb-z_0|+|(1-t)a+tb-z_1|<r$$ for any $t\in[0,1]$. Stick to the triangle inequality - don't pull things apart into coordinates and square roots.
The user "symplectomorphic" also gave a good hint in the comment above. The shape of $Q$ is very familiar.