Lời giải:
Ta có:
\(|z^2+1|=4|z|\Leftrightarrow \frac{|z^2+1|^2}{|z|^2}=16\)
\(\Leftrightarrow 16=\frac{(z^2+1)(\overline{z^2}+1)}{|z|^2}=\frac{|z|^4+z^2+\overline{z^2}+1}{|z|^2}\)
\(\Leftrightarrow 16=\frac{|z|^4+(z+\overline{z})^2-2|z|^2+1}{|z|^2}\geq \frac{|z|^4-2|z|^2+1}{|z|^2}\)
Đặt \(|z|^2=t\Rightarrow 16\geq \frac{t^2-2t+1}{t}\)
\(\Leftrightarrow t^2-18t+1\leq 0\Leftrightarrow 9-4\sqrt{5}\leq t\leq 9+4\sqrt{5}\)
\(\Rightarrow \sqrt{5}-2\leq |z|\leq \sqrt{5}+2\) hay \(|z|_{\min}=\sqrt{5}-2;|z|_{\max}=\sqrt{5}+2\)
Tổng quát: Nếu \(|z+\frac{1}{z}|=k\Rightarrow |z|_{\max}=\frac{\sqrt{k^2+4}+k}{2};|z|_{\min}=\frac{\sqrt{k^2+4}-k}{2}\)
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