Đặt
\(z=x+yi\Rightarrow2z+i.\overline{z}+1-i=2x+2yi+i\left(x-yi\right)+1-i=2x+y+1+\left(x+2y-1\right)i\)
\(\Rightarrow x+2y-1=0\)
a.
\(z'=2z+2-i=2\left(x+yi\right)+2-i=2x+2+\left(2y-1\right)i=2x+2-x.i\)
\(\Rightarrow\) Quỹ tích z' là đường thẳng có pt:
\(x+2y=2\Leftrightarrow x+2y-2=0\)
b.
\(T=\left|z+1-6i\right|=\left|x+1+\left(y-6\right)i\right|=\left|2-2y+\left(y-6\right)i\right|\)
\(=\sqrt{\left(2-2y\right)^2+\left(y-6\right)^2}=\sqrt{5\left(y-2\right)^2+20}\ge\sqrt{20}\)
Dấu "=" xảy ra khi \(y=2\Rightarrow x=1-2y=-3\)
\(\Rightarrow\left|z\right|=\sqrt{x^2+y^2}=\sqrt{13}\)
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