Can one determine where $\vert z^2-x \vert \le \vert z\vert$? 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.
 A: 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.
