модули

[Гост]

Ако [tex]|z-\frac{4}{z}|=2[/tex], максимумът на [tex]|z|[/tex] е?

[Добромир Глухаров]

$\left|z-\frac{4}{z}\right|=2\Rightarrow|z^2-4|=2|z|(|z|=r;\ arg(z)=\varphi)\Rightarrow|r^2(cos2\varphi+i.sin2\varphi)-4|=2r\Rightarrow\sqrt{(r^2cos2\varphi-4)^2+r^4sin^22\varphi}=2r\Rightarrow r^4cos^22\varphi-8r^2cos2\varphi+16+r^4sin^22\varphi=4r^2$

$r^4-4(1+2cos2\varphi)r^2+16=0;\ a=-4(1+2cos2\varphi)\Rightarrow a\in[-12;4]$

$r^4+ar^2+16=0\Rightarrow D=a^2-64\geq0\Rightarrow a\in[-12;-8]$

$r_{1,2}^2=\frac{-a\pm\sqrt{a^2-64}}{2}\geq0$

$r_1^2\geq r_2^2$

$r\leq r_1=\sqrt{\frac{\sqrt{a^2-64}-a}{2}}$

$\underline{a<0}$

$\varphi(a)=\sqrt{a^2-64}-a$

$\varphi'(a)=\frac{2a}{2\sqrt{a^2-64}}-1<0$

$\Rightarrow\max_{a\in[-12;-8]}\varphi(a)=\varphi(-12)=\sqrt{80}+12=12+4\sqrt{5}=4(3+\sqrt{5})$

$max|z|=max\ r=\sqrt{2(3+\sqrt{5})}=\sqrt{6+2\sqrt{5}}=\sqrt{(\sqrt{5}+1)^2}=1+\sqrt{5}$

[Гост]

[tex]|z| = |z – (4/z) + (4/z)| ≤ |z – (4/z)| + |(4/z)|[/tex]
[tex]|z| ≤ 2 + |(4/z)|[/tex]
[tex]|z| ≤ 2 + (4 / |z|)[/tex]
[tex]|z|^2 ≤ 2|z| + 4[/tex]
[tex]|z|^2 – 2|z| + 1 ≤ |1 + 4|[/tex]
[tex](|z| – 1)^2 ≤ 5[/tex]
[tex]||z| – 1| ≤ √5[/tex]
[tex]– √5 ≤ |z| – 1 ≤ √5[/tex]
[tex]– √5 + 1 ≤ |z| ≤ √5 + 1[/tex]
[tex]\max |z| = √5 + 1[/tex]