Biết số phức z thỏa mãn \(\left| z-3-4i \right|=\sqrt{5}\) và biểu thức \(T={{\left| z+2 \right|}^{2}}-{{\left| z-i \right|}^{2}}\) đạt giá trị lớn nhất. Tính \(\left| z \right|\)?

Gọi \(z=x+yi\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}=5\,\,\left( 1 \right)\)\(T={{\left| x+yi+2 \right|}^{2}}-{{\left| x+yi-i \right|}^{2}}={{\left( x+2 \right)}^{2}}+{{y}^{2}}-{{x}^{2}}-{{\left( y-1 \right)}^{2}}=4x+2y+3=4\left( x-3 \right)+2\left( y-4 \right)+23\)Ta có: \({{\left[ 4\left( x-3 \right)+2\left( y-4 \right) \right]}^{2}}\le \left( {{4}^{2}}+{{2}^{2}} \right)\left[ {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}} \right]=180\Leftrightarrow 4\left( x-3 \right)+2\left( y-4 \right)\le 6\sqrt{5}\)Dấu bằng xảy ra \(\Leftrightarrow \dfrac{x-3}{4}=\dfrac{y-4}{2}\Leftrightarrow x-3=2y-8\Leftrightarrow x=2y-5\)Thay vào (1) ta có:
\(\begin{array}{l}{\left( {2y - 8} \right)^2} + {\left( {y - 4} \right)^2} = 5 \Leftrightarrow 5{\left( {y - 4} \right)^2} = 5 \Leftrightarrow \left[ \begin{array}{l}y - 4 = 1\\y - 4 = - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}y = 5\\y = 3\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x = 5\\y = 5\end{array} \right.\\\left\{ \begin{array}{l}x = 1\\y = 3\end{array} \right.\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}z = 5 + 5i\\z = 1 + 3i\end{array} \right.\\z = 5 + 5i \Rightarrow T = 33(TM)\\z = 1 + 3i \Rightarrow T = 13 < 33(KTM)\end{array}\)

\(=>z=5+5i\)\(=> \left| z \right| =5 \sqrt {2}\)