Lời giải:Đặt \(z=a+bi\left( a,b\in \mathbb{R} \right)\). Do \(\left| z \right|=1\) nên \({{a}^{2}}+{{b}^{2}}=1\).
Sử dụng công thức: \(\left| u.v \right|=\left| u \right|\left| v \right|\) ta có: \(\left| {{z}^{2}}-z \right|=\left| z \right|\left| z-1 \right|=\left| z-1 \right|=\sqrt{{{\left( a-1 \right)}^{2}}+{{b}^{2}}}=\sqrt{2-2a}\).
\(\left| {{z}^{2}}+z+1 \right|=\left| {{\left( a+bi \right)}^{2}}+a+bi+1 \right|=\left| {{a}^{2}}-{{b}^{2}}+a+1+\left( 2ab+b \right)i \right|=\sqrt{{{\left( {{a}^{2}}-{{b}^{2}}+a+1 \right)}^{2}}+{{\left( 2ab+b \right)}^{2}}}\)
\(=\sqrt{{{a}^{2}}{{(2a+1)}^{2}}+{{b}^{2}}{{\left( 2a+1 \right)}^{2}}}=\left| 2a+1 \right|\) (vì \({{a}^{2}}+{{b}^{2}}=1\)).
Vậy \(P=\left| 2a+1 \right|+\sqrt{2-2a}\).
TH1: \(a<-\frac{1}{2}\).
Suy ra \(P=-2a-1+\sqrt{2-2a}=\left( 2-2a \right)+\sqrt{2-2a}-3\le 4+2-3=3\) (vì \(0\le \sqrt{2-2a}\le 2\)).
TH2: \(a\ge -\frac{1}{2}\)
Suy ra \(P=2a+1+\sqrt{2-2a}=-\left( 2-2a \right)+\sqrt{2-2a}+3=-{{\left( \sqrt{2-2a}-\frac{1}{2} \right)}^{2}}+3+\frac{1}{4}\le \frac{13}{4}\)
Xảy ra khi \(a=\frac{7}{16}\)
