Giving an equation: $$z^2-(m+2)z+4(m-1)=0$$I need to find the number of all the integer $m$ such that this equation has two complex solutions $z_1$, $z_2$ that satisfy: $$\vert z_1^2-m(z_1-4)\vert=\vert z_2^2-m(z_2-4)\vert$$My solution:
First, the equation has two different complex solutions if only if: $$\triangle=b^2-4ac<0$$In fact: $$(m+2)^2-16(m-1)<0$$$$\Leftrightarrow m^2-12m+20<0$$$$\Leftrightarrow 2<m<10$$The second condition:
Because $z_1$, $z_2$ are two solutions of the equation, then:$$\begin{cases}z_1^2-m(z_1-4)=2z_1+4\\z_2^2-m(z_2-4)=2z_2+4\end{cases}$$So I have: $$\vert z_1+2\vert=\vert z_2+2\vert$$What does this condition mean ? I need an explanation please.