Có bao nhiêu số phức z thỏa mãn $\left( 1+i \right)\left| z+4-2i...

Giả sử $\text{z}=a+bi\left( a,b\in \mathbb{R} \right)$, ta có $\left( 1+i \right)\left| z+4-2i \right|+\left( 1-i \right)\left| z \right|=6+4i$
$\Leftrightarrow \left( 1+i \right)\left| a+4+\left( b-2 \right)i \right|+\left( 1-i \right)\sqrt{{{a}^{2}}+{{b}^{2}}}=6+4i$
$\Leftrightarrow \left( 1+i \right)\sqrt{{{\left( a+4 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}+\left( 1-i \right)\sqrt{{{a}^{2}}+{{b}^{2}}}=6+4i$
$\Leftrightarrow \sqrt{{{\left( a+4 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}+\sqrt{{{a}^{2}}+{{b}^{2}}}+\left[ \sqrt{{{\left( a+4 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}-\sqrt{{{a}^{2}}+{{b}^{2}}} \right]i=6+4i$
$\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{\left( a+4 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}+\sqrt{{{a}^{2}}+{{b}^{2}}}=6 \\
& \sqrt{{{\left( a+4 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}-\sqrt{{{a}^{2}}+{{b}^{2}}}=4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{{{a}^{2}}+{{b}^{2}}}=1 \\
& \sqrt{{{\left( a+4 \right)}^{2}}+{{\left( b-2 \right)}^{2}}}=5 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}=1 \\
& {{a}^{2}}+{{b}^{2}}+8\text{a}-4b+20=25 \\
\end{aligned} \right.\Rightarrow 1+8\text{a}-4b+20=25\Leftrightarrow 8\text{a}-4b=4\Leftrightarrow b=2\text{a}-1$
$\Rightarrow {{a}^{2}}+{{\left( 2\text{a}-1 \right)}^{2}}=1\Leftrightarrow 5{{\text{a}}^{2}}-4\text{a}=0\Leftrightarrow \left[ \begin{aligned}
& a=0 \\
& a=\dfrac{4}{5} \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& b=-1\Rightarrow z=-i \\
& b=\dfrac{3}{5}\Rightarrow z=\dfrac{4}{5}+\dfrac{3}{5}i \\
\end{aligned} \right.$
Do đó có 2 số phức z thỏa mãn bài toán.