Hướng dẫn giải (chi tiết)

Đặt $z = x + yi\,\left( {x,y \in \mathbb{R}} \right)$

$11{z^{2018}} + 10i{z^{2017}} + 10iz - 11 = 0$

$\Leftrightarrow {z^{2017}} = \frac{{11 - 10iz}}{{11z + 10i}} \Rightarrow {\left| z \right|^{2017}} = \left| {\frac{{11 - 10iz}}{{11z + 10i}}} \right|$

$\Leftrightarrow {\left| z \right|^{2017}} = \frac{{\sqrt {100\left( {{x^2} + {y^2}} \right) + 121 + 220y} }}{{\sqrt {121\left( {{x^2} + {y^2}} \right) + 100 + 220y} }}$

TH1: $\left| z \right| < 1 \Leftrightarrow {x^2} + {y^2} < 1$

$\Rightarrow 100\left( {{x^2} + {y^2}} \right) + 121 + 220y > 121\left( {{x^2} + {y^2}} \right) + 100 + 220y$$\Rightarrow \left| z \right| > 1\left( {{\rm{sai}}} \right)$

TH2: $\left| z \right| > 1 \Leftrightarrow {x^2} + {y^2} > 1$

$\Rightarrow 100\left( {{x^2} + {y^2}} \right) + 121 + 220y < 121\left( {{x^2} + {y^2}} \right) + 100 + 220y \Rightarrow \left| z \right| < 1\left( {{\rm{sai}}} \right)$

TH3: $\left| z \right| = 1 \Leftrightarrow {x^2} + {y^2} = 1$. $\Rightarrow 1 = \frac{{\sqrt {100 + 121 + 220y} }}{{\sqrt {121 + 100 + 220} }} \Leftrightarrow 1 = 1$

Vậy $\left| z \right| = 1$

Hướng dẫn giải (chi tiết)

Đặt $f\left( z \right) = {z^4} - 2{z^3} + 6{z^2} - 8z + 9 \Rightarrow f\left( z \right) = 0$

Ta có: ${z^2} + 4 = {z^2} - 4{i^2} = \left( {z + 2i} \right)\left( {z - 2i} \right)$

$\Rightarrow T = \left[ {\left( {{z_1} + 2i} \right)\left( {{z_2} + 2i} \right)\left( {{z_3} + 2i} \right)\left( {{z_4} + 2i} \right)} \right].\left[ {\left( {{z_1} - 2i} \right)\left( {{z_2} - 2i} \right)\left( {{z_3} - 2i} \right)\left( {{z_4} - 2i} \right)} \right]$

$= {\left[ {f\left( { - 2i} \right).f\left( {2i} \right)} \right]^4} = 1$