$|1+z|^4 \leq 1+4 \operatorname{Re} z+8|z|^2+3|z|^4$

hbghlyj

33225230.pdf (1 MB, Downloads: 34)

2024-9-15 14:54 Upload

Click on the file to download the attachment

$z\inC,$
\begin{aligned}
& |1+z|^4 \geq 1+4 \operatorname{Re} z+|z|^2+\frac{1}{5}|z|^4 \\
& |1+z|^4 \leq 1+4 \operatorname{Re} z+8|z|^2+3|z|^4
\end{aligned}

Aluminiumor

令 $z=a+bi,a,b\in\mathbb{R}$
式一即
$$[(a+1)^2+b^2]^2\geq1+4a+a^2+b^2+\frac15(a^2+b^2)^2$$
$$\Longleftrightarrow \frac45\left(a^2+b^2+\frac52a\right)^2+b^2\geq0$$
式二即
$$[(a+1)^2+b^2]^2\leq1+4a+8(a^2+b^2)+3(a^2+b^2)^2$$
$$\Longleftrightarrow 2\left(a^2+b^2-a\right)^2+6b^2\geq0$$