Ta có $z^{2017}\left(11z+10i\right)=11-10iz\iff z^{2017}=\frac{11-10iz}{11z+10i}\Rightarrow {\left|z\right|}^{2017}=\left|\frac{11-10iz}{11z+10i}\right|$

Đặt $z=a+bi$ có $\left|\frac{11-10iz}{11z+10i}\right|=\left|\frac{11-10i\left(a+bi\right)}{11\left(a+bi\right)+10i}\right|=\sqrt{\frac{{\left(10b+11\right)}^2+100a^2}{121a^2+{\left(11b+10\right)}^2}}=\sqrt{\frac{100\left(a^2+b^2\right)+220b+121}{121\left(a^2+b^2\right)+220b+100}}$

Đặt $t=\left|z\right| \left(t\ge 0\right)$ ta có phương trình $t^{2017}=\sqrt{\frac{100t^2+220b+121}{121t^2+220b+100}}$

Nếu $t\ge 1\Rightarrow VT\ge 1; VP\le 1$

Nếu $t\le 1\Rightarrow VT\le 1; VP\ge 1$