TITLE: Can one determine where $\vert z^2-x \vert \le \vert z\vert$?
QUESTION [9 upvotes]: Please look at the nice attempt in the answer below, too before you answer:
Let $z \in \mathbb C$ have positive real part and $x>0$ a positive number.
I am trying to find the smallest real part of $z$ such that 
$$\vert z^2-x \vert \le \vert z\vert.$$
Does anyone know how to solve this?
The problem is that over $\mathbb C$ such inequalities become multi-valued since $z=a+ib$.
I should add that wolframalpha also gets some expressions but it is not completely transparent what the minimum possible real part is after all.
click me
Any comments are highly appreciated.

REPLY [5 votes]: This is not a full solution. It is instead a collection of graphs that should be useful for understanding the problem, along with a partial solution covering the easier cases.
In all following graphs, the blue curve is where $\left|\frac{z^2-x}{z}\right| = 1$, and the red curves are where $\left|\frac{z^2-x}{z}\right| = c$ for $c=0.1,0.2,\dots,0.9$. These graphs were generated by applying the quadratic formula to solve $z^2-x=re^{it}z$ for $z$ and graphing in Asymptote. Small ticks on the axes are at every $0.25$, and large ticks at every integer.
First, $x=0.35$ - an interesting case, where the minimum real part doesn't come on the axis:

Next, $x=0.5$ - large enough that the minimum actually is on the real axis. At least, I'm pretty sure of that.

Finally, $x=0.25$ - a key phase shift. For this and all smaller $x$, there are solutions on the imaginary axis.

So then, the key questions:

Where do we switch from having minima above and below the real axis to having one minimum on the real axis?  
In the region where the point on the real axis isn't the minimum, what is the minimum exactly?

After some more work, I might as well deal with the easier parts. That second phase shift, between a pair of off-axis minima to a single on-axis minimum, comes at $x=\frac{\sqrt{5}+1}{8}\approx 0.4045$. This was found by taking my parametric form $z(\theta)=\frac12\left(e^{i\theta}+\sqrt{4x+e^{2i\theta}}\right)$ and looking at $z''(\pi)$. That particular value of the second derivative (where the graph crosses the real axis) is always real. It's positive for $x>\frac{\sqrt{5}+1}{8}$ so the graph curves inward toward the center at $x$ as in my second picture, and negative for $x<\frac{\sqrt{5}+1}{8}$ so the graph curves outward as in my first picture. For larger $x$, we can then find the exact minimum real part by evaluating $z(\pi)$:
$$\min\left\{\text{Re}(z): \left|z^2-x\right|\le |z|\text{ and }\text{Re}(z)\ge 0\right\} = \frac12(\sqrt{4x+1}-1)\text{ if }x\ge \frac{\sqrt{5}+1}{8}$$
Also, for $x\le \frac14$, $z=i\sqrt{x}$ satisfies the inequality; $\left|z^2-x\right|=|-x-x|=2x \le \sqrt{x}=|z|$ and there is a solution on the imaginary axis:
$$\min\left\{\text{Re}(z): \left|z^2-x\right|\le |z|\text{ and }\text{Re}(z)\ge 0\right\} = 0\text{ if }x\le \frac14$$
That's it for the easy cases. Between $\frac14$ and $\frac{\sqrt{5}+1}{8}$, it gets a lot more complicated.