Cho số phức \(z\) thỏa mãn \(\left| {{z}^{2}}-2z+5 \right|=\left| \left( z-1+2i \right)\left( z-1+3i \right) \right|\) và \(w=z-2+2i\) giá trị nhỏ nhất của \(\left| w \right|\) bằng ?

Ta có \({{z}^{2}}-2z+5={{\left( z-1 \right)}^{2}}+4={{\left( z-1 \right)}^{2}}-{{\left( 2i \right)}^{2}}=\left( z-1+2i \right)\left( z-1-2i \right)\)

Do đó \(\left| {{z}^{2}}-2z+5 \right|=\left| \left( z-1+2i \right)\left( z+3i-1 \right) \right|\)

\(\Leftrightarrow \left| \left( z-1+2i \right)\left( z-1-2i \right) \right|=\left| \left( z-1+2i \right)\left( z+3i-1 \right) \right|\)

$\begin{array}{l}
\Leftrightarrow \left| {z - 1 + 2i} \right|\left| {z - 1 - 2i} \right| = \left| {z - 1 + 2i} \right|\left| {z + 3i - 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
\left| {z - 1 + 2i} \right| = 0\\
\left| {z - 1 - 2i} \right| = \left| {z + 3i - 1} \right|
\end{array} \right.
\end{array}$

TH1:

$\begin{array}{l}
\left| {z - 1 + 2i} \right| = 0 \Leftrightarrow z = 1 - 2i\\
\Rightarrow w = 1 - 2i - 2 + 2i = - 1\\
\Rightarrow \left| w \right| = 1
\end{array}$

TH2: 

\(\left| z-1-2i \right|=\left| z+3i-1 \right|\)

\(\Leftrightarrow \left| x-1+\left( y-2 \right)i \right|=\left| x-1+\left( y+3 \right)i \right|\)

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

\(\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( x-1 \right)}^{2}}+{{\left( y+3 \right)}^{2}}\)

\(\Leftrightarrow y=-\dfrac{1}{2}\) \(\Rightarrow z=x-\dfrac{1}{2}i.\)

Khi đó \(\left| w \right|=\left| z-2+2i \right|=\left| x-2+\dfrac{3}{2}i \right|=\sqrt{{{\left( x-2 \right)}^{2}}+\dfrac{9}{4}}\ge \dfrac{3}{2}.\)

Vậy \(\min \left| w \right|=1\).