What are the values of $r$ for which the set $\lbrace z\in\mathbb{C} : |z^2+az+b|
Connectedness of \lbrace z\in\mathbb{C} : |z^2+az+b|
3 Answers
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 $4r\gt|a^2-4b|$.
$r>|(z+a/2)^2-(a^2/4-b)|\ge |(z+a/2)^2|-|(a^2/4-b)|$, so $r+|(a^2/4-b)|\ge |(z+a/2)^2|$ Now, $|(z+a/2)^2|$ is a circle of centre at $-a/2$ and radius $r+|(a^2/4-b)|$ so $r+|(a^2/4-b)|\ge 0$ gives the connectedness.
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0@joriki Since r>0 the condition $r+|(a^2/4-b)|\geq 0$ is always satisfied. I ask Patience to correct the signs, I stop the correcting. – 2012-08-04
Following @Patience it is possible to simplify the original question. $ r>|z^2+az+b|=|(z+a/2)^2-(a^2/4-b)|. $ Denote $w:=z+a/2$ and $c:=a^2/4-b$. The translation does not change the connectedness, so it is enough to consider the set $ \{w\in\mathbf{C} : |w^2-d|
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0@joriki I have to admit that I was not precise enough, the word "translation" probably not the best. Try to delete "The translation ...ness" and start with "So ...". Hopefully it will be clearer. – 2012-08-04