Let $\omega \in \mathbb{C}$, and $\left | \omega \right | = 1$. Find the maximum length of $z = \left( \omega + 2 \right) ^3 \left( \omega - 3 \right)^2$.
Problem
Source: 2018 CGMO Day 2 Problem 5
Tags: algebra, maximum value, complex
14.08.2018 09:56
How do we find the maximum of a complex number? $z = \left( \omega + 2 \right) ^3 \left( \omega - 3 \right)^2$ can be complex number, like $\left( i + 2 \right) ^3 \left( i - 3 \right)^2=82+76i$.
14.08.2018 09:59
JiaHo wrote: How do we find the maximum of a complex number? $z = \left( \omega + 2 \right) ^3 \left( \omega - 3 \right)^2$ can be complex number, like $\left( i + 2 \right) ^3 \left( i - 3 \right)^2=82+76i$. Find the maximum value of $|z|.$
14.08.2018 14:25
Consider a $\triangle ABC$ with $BC=5$ and a point $D$ on $BC$ such that $BD=2 \, , \, DC=3$. Let $AB=x \, , \, AC=y$. We are essentially asked given $AD=1$ find the maximum value of $x^3 y^2$. By Stewart's we get: $$35=3x^2+2y^2$$And then by GM-QM we get: $$\sqrt[5]{x^3 y^2} \leq \sqrt{\frac{3x^2+2y^2}{5}}=\sqrt{7} \Rightarrow x^3 y^2 \leq 7^{\frac{5}{2}}$$With equality when $x=y=\sqrt{7}$
14.08.2018 14:56
To simplify things we consider the square of the form, hence $|z+2|^6 \cdot |z-3|^4$. With $\omega=:e^{it}$ we have $|z+2|^2=5+4\cos t, |z-3|^2=10-6\cos t$ and the form, modulo a factor, becomes $f(t)=(5+4\cos t)^3 (5-3\cos t)^2$. This has a maximum over ${\bf R}$ which also is a local maximum, hence a $t$ for which this form is maximal must satisfy $f'(t)=0$. We have $${{f'(t)}\over {f(t)}}={{-12\sin t}\over {5+4\cos t}}+{{6\sin t}\over {5-3\cos t}}$$and solving $f'(t)=0$ gives $\sin t=0\vee 10-6\cos t=5+4\cos t$. Hence $\sin t=0\vee \cos t={1\over 2}$. Plugging in the values shows that we have a maximum when $\cos t=1/2$. Then $|z+2|^2=5+4\cos t=7, |z-3|^2=10-6\cos t=7$ and $|z+2|^3 |z-3|^2=(\sqrt{7})^5=7^{5/2}$. I'm not sure if it is a coincidence that points on the circle which give this maximum lie on the perpendicular bisector of $-2,3$, probably because the ratio $2:3$ appears in the exponents also.
14.08.2018 18:47
Congratulation sbealing and alexheinis, both are nice solutions!
16.08.2018 20:29
Let $ \omega = cos{x} + i sin{x}$ and $a = cos{x}$. Plugging back, we get $|z| = \sqrt{(5+4a)^3(10-6a)^2}$ $ \leq (\frac{3(5+4a)+2(10-6a)}{5})^{\frac{5}{2}}$ (AM-GM inequality) $ = 49 \sqrt{7}$ The inequality holds when $a = \frac{1}{2} = cos{x}$
16.08.2018 22:45
Note that $\overline{\omega} = \omega^{-1}$. Note that $\Re(\omega) \in [-1, 1]$. \begin{align*} |z|^2 &= (\omega + 2)^3(\omega - 3)^2\overline{(\omega + 2)^3(\omega - 3)^2} = ((\omega + 2)(\omega^{-1} + 2))^3((\omega - 3)(\omega^{-1} - 3))^2\\ &= (5 + 2(\omega + \omega^{-1}))^3(10 - 3(\omega + \omega^{-1}))^2 = (5 + 4\Re(\omega))^3(10 - 6\Re(\omega))^2\\ &\leq \left(\frac{3(5 + 4\Re(\omega)) + 2(10 - 6\Re(\omega))}{5}\right)^5 = 7^{5} \end{align*}Therefore $|z| \leq 49\sqrt{7}$, equality when $\Re(\omega) = \frac{1}{2}$.
29.03.2019 01:04
17.07.2019 14:26
sbealing wrote: Consider a $\triangle ABC$ with $BC=5$ and a point $D$ on $BC$ such that $BD=2 \, , \, DC=3$. Let $AB=x \, , \, AC=y$. We are essentially asked given $AD=1$ find the maximum value of $x^3 y^2$. By Stewart's we get: $$35=3x^2+2y^2$$And then by GM-QM we get: $$\sqrt[5]{x^3 y^2} \leq \sqrt{\frac{3x^2+2y^2}{5}}=\sqrt{7} \Rightarrow x^3 y^2 \leq 7^{\frac{5}{2}}$$With equality when $x=y=\sqrt{7}$ Can you please explain why you did that/how that works in complex numbers? Please, I'm really curious
18.07.2019 18:17
trumpeter wrote:
where did this come from?