Extreme values of $|z|$ when $|z^2+1|=|z-1|$ Problem statement: Find the extreme values of $|z|$ when $$|z^2+1|=|z-1|,\ z\in \mathbb{C}-\{0\}$$
My try:
$$|z^2+1|=|z-1|\implies|z^2+1|^2=|z-1|^2\implies(z^2+1)(\overline{z}^2+1)=(z-1)(\overline{z}-1)$$
$$\implies|z|^4+(z+\overline{z})^2=3|z|^2-(z+\overline{z})\cdots\text{(I have skipped some algebra here..)}$$
Since $(z+\overline{z})$ is a real number, let $a=(z+\overline{z})$.Now,
$$|z|^4-3|z|^2+(a^2+a)=0$$
Using the quadratic formula, we have
$$|z|^2=\frac{3\pm\sqrt{9-4(a^2+a)}}{2}$$
Since the discriminant is strictly non-negative and the minimum value of $(a^2+a)$ is $\frac{-1}{4}$, we have,
$$\frac{-1}{4} \le(a^2+a) \le \frac{9}{4}\implies -1 \le4(a^2+a) \le9$$
Now, considering the two values of $|z|^2$ and accordingly maximising or minimising $(a^2+a)$ using the above inequality, we get
$$\max{|z|}=\sqrt{\frac{3+\sqrt{10}}{2}}\ \ \ \text{and}\ \ \ \min{|z|=0}$$
First of all, I would like to know whether my solution is correct. Secondly, I have trouble minimising $|z|$ because the problem statement says $|z|\ne0$, hence my answer contradicts the problem statement.
I would also like to know if there is any other way of solving this problem of finding extreme values of $|z|$ given an equation in $z$.
Thanks for any answers!!
 A: Some additional elements versus what you did.
Important to note is that $a = 2 \Re(z)$ where $\Re(z)$ stands for the real part of $z$. Indeed if $z \in S = \{z \in \mathbb C \setminus \{0\} \mid \vert z^2+1\vert=\vert z-1 \vert\} $ then (I haven't check your computations)
$$|z|^2=\frac{3\pm\sqrt{9-4(a^2+a)}}{2}.$$
And $\vert z \vert$ belongs to the interval $I= [0, \sqrt{\frac{3+\sqrt{10}}{2}}]$.
Now the question is are those values really reached?
We have $\vert z \vert = \sqrt{\frac{3+\sqrt{10}}{2}}$ for $a = 2\Re(z)= -1/2$ which represents a vertical line in the complex plane. As $\sqrt{\frac{3+\sqrt{10}}{2}} > 1/4$, we indeed have two complex satisfying the equation: the intersections of the vertical line with the circle $\vert z \vert = \sqrt{\frac{3+\sqrt{10}}{2}}$.
Let's now have a look at $\vert z \vert =0$.This would only be possible for $\sqrt{9-4(a^2+a)} = 3$, i.e. $a^2+a=0$ or $ a = 0,-1$. $a=0$ is to be ignored knowing the hypothesis $z \neq 0$. $a = -1$ cannot be if $\vert z \vert =0$. Therefore if a minimum value is attained for $\vert z \vert$, it cannot be $0$. However $0$ is a lower bound.
Another interesting question that could be studied is : is $\vert z \vert$ attaining a local minimum on $S$?
A: (this solution mostly is similar to what you have, which is fine, but you can avoid the quadratic equation and bounding)
Observe that 
$$|z^2+1|=|z-1| \iff (z^2 + 1)(\overline{z}^2 + 1) = (z - 1)(\overline{z}-1) \iff |z|^4 - |z|^2 + (z^2 + \overline{z}^2 + z + \overline{z}) = 0$$
Note that for all pairs of reals $(a, b)$, there exists a complex number $c$ such that $c+\overline{c} = a, c\overline{c} = b$ if $a^2 \leq 4b$. This is because $a = 2\text{Re}(c)$ and $b = \text{Re}(c)^2 + \text{Im}(c)^2 \geq \text{Re}(c)^2 = \frac{a^2}{4}$. If $a$, $b$ satisfy this condition, then we can easily contruct $c$ using this, so we're done.
If we let $s = z + \overline{z} = 2\text{Re}(z), p=z\overline{z} = |z|^2$, then $p^2 - 3p + s^2 + s = 0$ with $p$ and $s$ real, so $s^2 \leq 4p$.
This quadratic equation can be rewritten as $(p-\frac{3}{2})^2 + (s+\frac{1}{2})^2 = \frac{5}{2}$. Hence, $p$ is maximised when $s = -\frac{1}{2}$, so $p = \frac{3+\sqrt{10}}{2}$ (which we can easily observe to satisfy $s^2 \leq 4p$). This yields the maxima for $|z|$ as $\sqrt{\frac{3+\sqrt{10}}{2}}$ at $z = -\frac{1}{4} \pm \frac{\sqrt{23 + 8\sqrt{10}}}{4}i$.
Furthermore, $p$ is minimised when $p \to 0^+$, so $z \to 0^+$. Thus, there's no explicit minimum, but there is an infimum of $0$.
Hope that helps.
