Xét các số phức $z, w$ thỏa mãn $\left| z \right|=1$ và $\left| w...

Ta có $\left| z+i\overline{w}+6+8i \right|\ge \left| \left| 6+8i \right|-\left| z+i\overline{w} \right| \right|\ge \left| \left| 6+8i \right|-\left| z \right|-\left| i\overline{w} \right| \right|=10-1-2=7$.
Dấu “ $=$ ” xảy ra khi
$\left\{ \begin{aligned}
& z+i\overline{w}={{t}_{1}}\left( 6+8i \right), {{t}_{1}}\le 0 \\
& i\overline{w}={{t}_{2}}z, {{t}_{2}}\ge 0 \\
& \left| z \right|=1 \\
& \left| w \right|=2 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& z=t\left( 6+8i \right),t\le 0 \\
& i\overline{w}={t}'\left( 6+8i \right),{t}'\le 0 \\
& \left| z \right|=1 \\
& \left| w \right|=2 \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& z=-\dfrac{1}{10}\left( 6+8i \right) \\
& i\overline{w}=-\dfrac{2}{10}\left( 6+8i \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& z=-\dfrac{1}{10}\left( 6+8i \right) \\
& \overline{w}=\dfrac{1}{5}\left( -8+6i \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& z=-\dfrac{1}{10}\left( 6+8i \right) \\
& w=\dfrac{1}{5}\left( -8-6i \right) \\
\end{aligned} \right.$.
Khi đó $\left| z-w \right|=\dfrac{\sqrt{29}}{5}$.