# Find the maximum of $\left|z_1-z_2\right|+\left|z_2-z_3\right|+\left|z_3-z_1\right|$

Let $$z_1, z_2, z_3$$ be complex numbers satisfying the condition $$\left|z\right|^2\leq 4\left|z+\overline{z}\right|+33$$. Find the maximum value of $$P=\left|z_1-z_2\right|+\left|z_2-z_3\right|+\left|z_3-z_1\right|$$

If $$z=a+bi$$ then $$a^2+b^2\leq 8\left|a\right|+33 \Leftrightarrow \left(\left|a\right|-4\right)^2+b^2\leq 49$$. This follows that $$z_i$$ belongs to the union of two circles $$\left(x-4\right)^2+y^2\leq 49$$ (when $$Re z_i>0$$) and $$\left(x+4\right)^2+y^2\leq 49$$ (when $$Re z_i<0$$). Let $$A,B,C$$ be the points responding to $$z_1,z_2,z_3$$. Then $$P=AB+BC+CA$$. I have no idea to find the maximum value of $$P$$. But when I choose $$z_1=-8+\sqrt{33}i, z_2=-8-\sqrt{33}i, z_3=11$$, I can see $$P=2\sqrt{33}+2\sqrt{394}$$. I cannot find any other triple $$\left(z_1,z_2,z_3\right)$$ so that $$P$$ is greater. Help me some ideas. Thank you.

• Why $P=AB+BC+CA$?
– user
Commented Apr 14, 2021 at 17:59
• @user Presumably the notation $AB$ is the geometric notation for distance between $A$ and $B$, which is precisely $|z_1-z_2|$. Commented Apr 14, 2021 at 18:01
• @user because $|z_1-z_2|$=dist(A,B)="AB". Commented Apr 14, 2021 at 18:01
• @JeanMarie I see. I thought $AB$ stays for product of $A$ and $B$.
– user
Commented Apr 14, 2021 at 18:33
• @JeanMarie yes, I mean $z_2=-8-\sqrt{33}i$. Thank you Commented Apr 15, 2021 at 0:19

use $$z_1=q+\bigg[\sqrt{49-(q+4)^2}\bigg]i, \\z_2=q-\bigg[\sqrt{49-(q+4)^2}\bigg]i,\\ z_3=11$$ you should find $$q_{max} = -8$$