Xét các số phức $z$,w thỏa mãn $|z|=|w|=1$. Khi $|z-2 w-3-4 i|$...

Giả sử $z=a+bi,\ w=c+di,a,b,c,d\in \mathbb{R}$.
$|z|=|w|=1\Rightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=1 \\
& {{c}^{2}}+{{d}^{2}}=1 \\
\end{aligned} \right.$
$|z-2w-3-4i|=|a-2c-3+\left( b-2d-4 \right)i|=\sqrt{{{\left( a-2c-3 \right)}^{2}}+{{\left( b-2d-4 \right)}^{2}}}$
$|z-2w-3-4i|=\sqrt{{{\left( a-2c-3 \right)}^{2}}+{{\left( b-2d-4 \right)}^{2}}}\le \sqrt{{{a}^{2}}+{{b}^{2}}}+\sqrt{4\left( {{c}^{2}}+{{d}^{2}} \right)}+\sqrt{{{3}^{2}}+{{4}^{2}}}=8$.
Dấu " =" xảy ra khi $\left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=1,{{c}^{2}}+{{d}^{2}}=1 \\
& \dfrac{a}{b}=\dfrac{-2c}{-2d}=\dfrac{-3}{-4} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=1,{{c}^{2}}+{{d}^{2}}=1 \\
& \dfrac{a}{b}=\dfrac{-2c}{-2d}=\dfrac{-3}{-4} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{3}{5},b=\dfrac{4}{5} \\
& c=\dfrac{-3}{5},d=\dfrac{-5}{5} \\
\end{aligned} \right.$.
Do đó $|z-w|=\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-d \right)}^{2}}}=\sqrt{{{\left( \dfrac{6}{5} \right)}^{2}}+{{\left( \dfrac{8}{5} \right)}^{2}}}=2$