Connected Preimage of Complex Polynomial Let $p(z)=z^2+z+i$. For what $R>0$ is the set $S_R=\{z\in\mathbb{C}:|p(z)|<R\}$ connected in $\mathbb{C}$?

If $R=\infty$, then $S_{\infty}=\mathbb{C}$ which is connected. Also, the roots of $p(z)$ are separated by the real line, so if $S_R\cap\mathbb{R}=\emptyset$, then $S_R$ is not connected in $\mathbb{C}$. For any $x\in\mathbb{R}$, Im$(p(x))=i$, so if $R\leq1$, then $S_R$ is not connected. I'm not sure about $1<R<\infty$.
 A: For users of different levels to understand easily, I will present my solution in a visual intuition-based style, but certainly it can be converted to a rigorous proof.
Firstly, if $z=x+iy$, then

$$|p(z)|^2:=f(x,y)=y^4+(2x^2+2x+1)y^2+(4x+2)y+(x^4+2x^3+x^2+1)$$

Then, 

$$|p(z)|<R\implies f(x,y)-R^2<0$$

Let's plot the inequality in Desmos for different values of $R$.
When $R=1$:
When $R=1.03$:
When $R=1.04$:
When $R=1.45$:
The observations are:


*

*The solution set to the inequality is composed of two 'leaves'.

*The 'leaves' are respectively centered at the two roots of $p(z)$ on the complex plane.

*Therefore, any one leaf can be transformed to the another by flipping across the $x$-axis and $x=-\frac12$.

*The leaves glue together at some values of $R$ between $1.03$ and $1.04$.

*If we shift the leaves to the right by $\frac12$, then when the leaves merely touch each other at the origin, $R$ is at its critical value.


Consider $\displaystyle{f^*(x,y)=f\left(x-\frac12,y\right)}$. After doing a little algebra, we have $$f^*(x,y)=y^4+\left(2x^2+\frac12\right)y^2+4xy+\left(x^4-\frac{x^2}2+\frac{17}{16}\right)$$
By the intuition illustrated above, we should have

$$f^*(0,0)-R_{\text{crit}}^2=0\implies R_{\text{crit}}=\frac{\sqrt {17}}{4}\approx1.030776\qquad{(*)}$$

which agrees with observation #4.

(*): In case you do not understand how the equation is obtained, here is the explanation: 
The boundary of the Desmos plot above represents $f(x,y)-R^2=0$. 
When the $\frac12$-right-shifted leaves are in mere contact, $(0,0)$ lies on the boundary of the graph $f^*(x,y)-R^2_{\text{crit}}<0$. 
Thus instead of inequality, equality holds at $(0,0)$: $f^*(0,0)-R^2_{\text{crit}}=0$. 
For better intuition, please also plot $f^*(x,y)-R^2=0$ on Desmos for various values of $R$.
A: Here is a rigorous proof:
If $z=x+iy$, then

$$|p(z)|^2:=f(x,y)=y^4+(2x^2+2x+1)y^2+(4x+2)y+(x^4+2x^3+x^2+1)$$

whose roots are precisely the same as the roots of $p(z)$ on the complex plane, which are $(x,y)=r_1=(-1.300,0.625)$ and $(x,y)=r_2=(0.300,-0.625)$ (approx.).
Define

$$f^*(x,y)=f(x-1/2,y)=y^4+\left(2x^2+\frac12\right)y^2+4xy+\left(x^4-\frac{x^2}2+\frac{17}{16}\right)$$

which is a shifted version of $f$ that has better symmetry properties.
Furthermore, define $S_R=\{z:|p(z)|<R\}=\{z:f(x,y)-R^2<0\}$, $S^*_R=\{z:f^*(x,y)-R^2<0\}$.

Theorem 1:

$S^*_R$ is disconnected for $R\le \frac{\sqrt {17}}4$.

Proof:
We will prove that $S^*_R$ is disconnected when $R\le \frac{\sqrt {17}}4$, by showing that elements in $S^*_R$ is separated by the line $y=x$:


*

*$S^*_R$ necessarily contains the roots of $p(z)$.

*There exists $s\in S^*_R$ such that $s$ lies below $y=x$, e.g. $s=r_2$.

*There exists $s\in S^*_R$ such that $s$ lies above $y=x$, e.g. $s=r_1$.

*There exists no $s\in S^*_R$ such that $s$ lies on $y=x$, because $\{z:f^*(x,x)-R^2=4x^2(x^2+1)+\frac{17}{16}-R^2<0\}$ is an empty set.


Hence $S^*_R$ is disconnected whenever $R\le\frac{\sqrt {17}}4$.

Theorem 2:

When $R>\frac{\sqrt{17}}{4}$, $\partial S^*_R$, which is equivalent to $f^*(x,y)-R^2=0$, is one single continuous closed curve, and thus $S^*_R$ (which is the interior of such curve) is connected.

Proof:
Converting $f^*(x,y)$ to $f^*(r,\theta)$, we get
$$f^*(r,\theta)=r^4+\left(2\sin 2\theta-\frac{\cos2\theta}{2}\right)r^2+\frac{17}{16}-R^2$$
Hence, $$f^*(r,\theta)=0 \implies r^4+\left(2\sin 2\theta-\frac{\cos2\theta}{2}\right)r^2+\frac{17}{16}-R^2=0$$ which can be inverted as 
$$r^2=\frac{\cos2\theta}4-\sin2\theta\pm \sqrt{\left(\frac{\cos2\theta}4-\sin2\theta\right)^2-\frac{17}{16}+R^2}$$
First, the square root is defined for all $\theta$ only when $-\frac{17}{16}+R^2>0\implies R>\frac{\sqrt{17}}{4}$.
Second, the negative root must be rejected as otherwise the RHS will be negative for all values of $\theta$, contradicting the basic fact that $r$ is real. 
Therefore, 
$$f^*(x,y)-R^2=0 \quad \text{and} \quad r(\theta)=\sqrt{\frac{\cos\theta}4-\sin\theta+ \sqrt{\left(\frac{\cos\theta}4-\sin\theta\right)^2-\frac{17}{16}+R^2}}$$
describe the same curve for $R>\frac{\sqrt{17}}4$. $r(\theta)$ clearly represents one single continuous closed curve, thus $S^*_R$, the interior of $r(\theta)$, is connected.

To sum up, $S^*_R$ is connected only for $R>\frac{\sqrt{17}}4$.
Since connectedness is invariant upon shifting, $S_R$ is connected only for $R>\frac{\sqrt{17}}4$. $\blacksquare$
