Maximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$. 
If $z\in C$ satisfies $|z^4+z^3-z-1| = |z^4-z^3+z-1|.$ Then minimum value of $|z-1-2i|$ and maximum value of $|z^2-1-2i|$ is . . .

$\bf{Attempt}$ from $|z^4+z^3-z-1| = |z^4-z^3+z-1|$
$$|(z+1)(z^3-1)| = |(z-1)(z^3+1)|$$
$$\bigg|\frac{z-1}{z+1}\bigg|=\bigg|\frac{z^3-1}{z^3+1}\bigg|=k$$
Could some help me how to solve it, thanks
 A: Hint:  Your factorizations are incomplete.
$$  z^3 - 1 = (z-1)(z^2+z+1)  $$
$$  z^3 + 1 = (z+1)(z^2-z+1)  $$
This leaves $|z^2 -z +1| = |z^2 + z + 1|$ (having already extracted the factors $z-1$ and $z+1$).
A: You've gone just a bit astray.
First of all, since you don't know what $z$ is, it's dangerous to divide by $|z+1|,$ as it could well be equal to $0.$ In fact, $z=-1$ is a solution to the given equation, so that is entirely reasonable!
Next, you haven't quite finished factoring. Observing that $z=1$ is clearly a root of $z^3-1=0,$ then $z-1$ is a factor of $z^3-1.$ In particular, $$z^4+z^3-z-1=(z+1)(z-1)\left(z^2+z+1\right),$$ and similarly, $$z^4-z^3+z-1=(z-1)(z+1)\left(z^2-z+1\right).$$ Now, the given equation becomes $$|z+1||z-1|\left|z^2+z+1\right|=|z-1||z+1|\left|z^2-z+1\right|,$$ or equivalently, $$|z+1||z-1|\left(\left|z^2+z+1\right|-\left|z^2-z+1\right|\right)=0.$$ Thus, we can conclude that $z\in\{1,-1\}$ or that $\left|z^2+z+1\right|=\left|z^2-z+1\right|.$ It is easily verified that $z^2+z+1$ and $z^2-z+1$ are never both equal to $0,$ and neither is equal to $0$ when $z\in\{1,-1\}.$ Thus, assuming that the given equation holds and that $z\notin\{1,-1\},$ we have that $$\left|z^2+z+1\right|=\left|z^2-z+1\right|,\tag{1}$$ and you could divide through by one side of the equation safely.
However, we still don't have to do so. Since the modulus function is non-negative, then $(1)$ holds if and only if $$\left|z^2+z+1\right|^2=\left|z^2-z+1\right|^2\tag{2}$$ holds. Recall now the following facts (which hold for any $u,v,w\in\Bbb C$): 


*

*$|w|^2=w\cdot\overline w,$ where $\overline w$ is the complex conjugate of $w.$ In particular, if $w=x+iy$ for $x,y\in\Bbb R,$ then $|w|^2=x^2+y^2.$

*$\overline{u+v}=\overline u+\overline v.$

*$\overline{u\cdot v}=\overline u\cdot\overline v.$

*For any $\alpha\in\Bbb R,$ we have $\overline{\alpha}=\alpha.$

*Applying facts 2-4 shows that $\overline{u-v}=\overline u-\overline v.$

*$\overline{\left(\overline w\right)}=w.$

*The real part of $w$ is $\Re(w)=\frac12\left(w+\overline w\right).$


With these facts at our disposal, we see that $$\begin{eqnarray}\left|z^2+z+1\right|^2 &=& \left(z^2+z+1\right)\cdot\overline{(z^2+z+1)}\\ &=& \left[\left(z^2+1\right)+z\right]\cdot\left[\overline{(z^2+1)}+\overline z\right]\\ &=& \left|z^2+1\right|^2+\overline z\cdot\left(z^2+1\right)+z\cdot\overline{\left(z^2+1\right)}+|z|^2\\ &=& \left|z^2+1\right|^2+|z|^2+2\Re\left(\overline z\cdot\left(z^2+1\right)\right),\end{eqnarray}$$ and similarly that $$\left|z^2-z+1\right|^2=\left|z^2+1\right|^2+|z|^2-2\Re\left(\overline z\cdot\left(z^2+1\right)\right),$$ whence we see that $(2)$ is equivalent to $$\Re\left(\overline z\cdot\left(z^2+1\right)\right)=0.\tag{3}$$ From there, $$\begin{eqnarray}0 &=& 2\Re\left(\overline z\cdot\left(z^2+1\right)\right)\\ &=& \overline z\cdot\left(z^2+1\right)+z\cdot\overline{\left(z^2+1\right)}\\ &=& \overline z\cdot\left(z^2+1\right)+z\cdot\left(\left(\overline z\right)^2+1\right)\\ &=& |z|^2\cdot z+z+|z|^2\cdot\overline z+\overline z\\ &=& \left(|z|^2+1\right)\cdot z+\left(|z|^2+1\right)\cdot\overline z\\ &=& 2\left(|z|^2+1\right)\Re(z),\end{eqnarray}$$ so since $|z|^2+1\ne 0,$ then we have that $(3)$ is equivalent to $$\Re(z)=0.\tag{4}$$
Thus, letting $A:=\{1,-1\}\cup\{\alpha i:\alpha\in\Bbb R\},$ the given equation holds if and only if $z\in A.$

At this point, we need only minimize $|z-1-2i|$ and maximize $\left|z^2-1-2i\right|$ among these points.
Suppose that $z=\alpha i$ for some real $\alpha.$ Observe that $z$ minimizes $|z-1-2i|$ if and only if $z$ minimizes $$|z-1-2i|^2=|-1+i(\alpha-2)|^2=(-1)^2+(\alpha-2)^2=(\alpha-2)^2+1.$$ Since $(\alpha-2)^2\ge0$ for all real $\alpha,$ then $\alpha=2$ clearly minimizes $(\alpha-2)^2+1.$ Thus for $z\in A,$ the minimum value of $|z-1-2i|$ occurs at some point $z\in\{1,-1,2i\}.$ I leave it to you to determine said minimum value.
Similarly, if $z=\alpha i$ for some real $\alpha,$ then $z$ maximizes $\left|z^2-1-2i\right|$ if and only if $z$ maximizes $$\left|z^2-1-2i\right|=(\alpha^2+1)^2+4.$$ Is this possible?
