How find this maximum of the $|(z-a)^2(z+b)|$

Question

let $a,b$ is give postive numbers,let $z\in C$, and such $$|z|=1$$ Find the maximum $$u=|(z-a)^2(z+b)|,a,b>0$$

My try: since $$z=x+yi,|z|=1\Longrightarrow x^2+y^2=1$$ then we have $$(|((x-a)+yi)^2((x+b)+yi)|$$ and then it's very ugly,Have someone nice methods? Thank you

score 0 · Answer 1 · answered Dec 28 '13 at 15:10

0

Since $|z|=1,$ then $z$ can be represented in the exponential form as $z= e^{i\theta}.$ Then $$|(z-a)^2(z+b)|=|(e^{i\theta}-a)^2 (e^{i\theta}+b)|=|e^{i\theta}-a|^2 \cdot |e^{i\theta}+b|.$$ RHS is a function depending only on $\theta.$

answered Dec 28 '13 at 15:10

M. Strochyk

8,397

2

Square it and differentiate it. – Berci Dec 28 '13 at 15:11

How find this maximum of the $|(z-a)^2(z+b)|$

1 Answers1