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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 $r\in\mathbb{R}$.

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up vote 4 down vote accepted

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 $4r\gt|a^2-4b|$.

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@vesszabo: Thanks. – Did Aug 4 '12 at 12:06

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 $$ \{w\in\mathbf{C} : |w^2-d|<r\}. $$ The case $d=0$ is easy, so we may assume that $d\neq 0$. Denote $w:=u\sqrt{d}$. The dilation does not change the connectedness, so it is enough to consider the set $$ \{u\in\mathbf{C} : |u^2-1|<r/d=:R\}. $$ At this moment I have no further idea, but I hope someone else will be able to follow this. Or, maybe, this is well-known, and the answer can be found in a complex-function-theory book.

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Why does the translation not change the connectedness? If you have some arbitrarily shaped region and you translate it, you might move the connection between two parts outside the circle of radius $r$, no? – joriki Aug 4 '12 at 10:36
@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. – vesszabo Aug 4 '12 at 11:26

$r>|(z+a/2)^2-(a^2/4-b)|\ge |(z+a/2)^2|-|(a^2/4-b)|$, 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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For the first the radius should be $\sqrt{r+|a^2/4-b|}$. For the second you just proved that the set $\{z\in\mathbb{C}:|z^2+az+b|<r\}$ is contained in that circle. – Norbert Aug 4 '12 at 9:27
Also you switched from $z+a/2$ to $z-a/2$. – joriki Aug 4 '12 at 10:18
@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. – vesszabo Aug 4 '12 at 11:56

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