Tìm giá trị lớn nhất của $P=\left| {{z}^{2}}-z \right|+\left|...

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: $a<-\dfrac{1}{2}$.
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: $a\ge -\dfrac{1}{2}$.
Suy ra $P=2a+1+\sqrt{2-2a}=-\left( 2-2a \right)+\sqrt{2-2a}+3=-{{\left( \sqrt{2-2a}-\dfrac{1}{2} \right)}^{2}}+3+\dfrac{1}{4}\le \dfrac{13}{4}$.
Xảy ra khi $a=\dfrac{7}{16}$.