Finding $z \in \mathbb{C}$ such that $|z^{6}+z^{3}+2|= 4+|4z^{2}+4|$

Question

everyone. I don't know how to start to solve this exercise that asks for the $z \in \mathbb{C}$ such that $$|z^6+z^3+2|= 4 +|4z^2+4|$$ And $$|z|=1$$

I tried finding the roots of $z^6+z^3+2$ but that didn't help me to find the region that describes $|z^6+z^3+2|= 4 +|4z^2+4|$. Thanks.

score 5 · Answer 1 · answered Sep 05 '17 at 02:00

5

Hint:$\;\,$Since $|z|=1$, the LHS can't be greater than $4$.

answered Sep 05 '17 at 02:00

quasi

Why ${}{}{}$...? – Nosrati Sep 05 '17 at 02:04
1

The absolute value of a sum is less than or equal to . . . – quasi Sep 05 '17 at 02:04
1

@MyGlasses triangle inequality – operatorerror Sep 05 '17 at 02:20
Can someone tell me what you are trying to solve with |z| = 1? – Satish Ramanathan Sep 05 '17 at 02:22
@qbert Really in the best case $4z^2+4$ at the RHS will be $0$ or $z=\pm i$ which give $\sqrt{2}=4$. – Nosrati Sep 05 '17 at 02:24
+1: Very slick! – Cameron Buie Sep 07 '17 at 00:42

1 Answers1