TITLE: Connectedness of $\lbrace z\in\mathbb{C} : |z^2+az+b|<r\rbrace$
QUESTION [7 upvotes]: What are the values of $r$ for which the set $$\lbrace z\in\mathbb{C} : |z^2+az+b|<r\rbrace$$ is connected ? Here $a,b\in\mathbb{C}$ and $r\in\mathbb{R}$.

REPLY [4 votes]: Consider $D_s=\{z\in\mathbb C\,\mid\,|z^2-1|\lt s\}$ for some $s\gt0$. 

First assume that $s\gt1$. If $z$ is in $D_s$, $z^2$ and $0$ are both at distance $\lt s$ from $1$ hence $tz^2+(1-t)0$ is also at distance $\lt s$ from $1$, for every $t$ in $[0,1]$. Thus, if $z$ is in $D_s$, then $\sqrt{t}z$ is in $D_s$, for every $t$ in $[0,1]$. This proves that $D_s$ is star-shaped with center $0$, and in particular that $D_s$ is connected.
Assume now that $s\leqslant1$. Then, $1$ and $-1$ are in $D_s$, but the distance between $1$ and every point on the imaginary axis is at least $1$, hence $D_s$ contains no $z$ such that $\arg(z)=\pm\pi/4$. Thus $D_s$ is not connected.

Coming back to the domain $T=\{z\in\mathbb C\,\mid\,|z^2+az+b|\lt r\}$, note that $z^2+az+b=w^2-c^2$ with $w=z+a/2$ and $c^2=(a^2/4)-b$. If $c=0$, $T$ is the disk of equation $|z+a/2|^2\lt r$, with center $-a/2$ and radius $\sqrt{r}$, which is connected. If $c\ne0$, $T$ is defined by $|v^2-1|\lt s$ with $s=r/|c|^2$ and $v=w/c$. The transformation $z\mapsto v$ is affine and invertible hence $T$ is connected if and only if $D_s$ is connected, which happens if and only if $s\gt1$, that is $r\gt|c|^2$.
To sum up, the domain $T$ is connected if and only if $T$ is star-shaped with center $-a/2$ if and only if $4r\gt|a^2-4b|$.