$\text{If} 1+2{\left|z\right|}^2={\left|z^2+1\right|}^2+2{\left|z^2+1\right|}^2\text{,then the calue of }\left|z\left(z+1\right)\right|\text{is}$

$1+2{\left|z\right|}^2={\left|z^2+1\right|}^2+2{\left|z^2+1\right|}^2$

$\Rightarrow 1+2z\overline{z}=\left(z^2+1\right)\left({\overline{z}}^2+1\right)+2\left(z+1\right)\left(\overline{z}+1\right)$

$\begin{array}{l}\Rightarrow (z\overline{z}{)}^2+2(z+\overline{z})+z^2+{\overline{z}}^2+2=0\\ \Rightarrow (z\overline{z}-1{)}^2+(z+\overline{z}+1{)}^2=0\\ \Rightarrow z\overline{z}=1\\ \Rightarrow \overline{z}=\frac{1}{z}\end{array}$

$\begin{array}{l}\text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}z+\overline{z}+1=0\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}z+\frac{1}{z}+1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{z}^2+z=-1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|z^2+z\right|=1\end{array}$