Let $\quad z=(x+i y)(1-2 i)$
$=x+i y-2 x i-2 i^{2} y$
$=x+i y-2 x i+2 y$
$\Rightarrow \quad z \quad z=x+2 y+i(y-2 x)$
$\therefore \quad \overline{\mathrm{z}}=\mathrm{x}+2 \mathrm{y}-\mathrm{i}(\mathrm{y}-2 \mathrm{x})$
According to the question,
$$
\begin{gathered}
\mathrm{z}=1+\mathrm{i} \\
\Rightarrow \quad \mathrm{x}+2 \mathrm{y}-\mathrm{i}(\mathrm{y}-2 \mathrm{x})=1+\mathrm{i}
\end{gathered}
$$
On equating the real and imaginary parts from both sides, we get
and $\begin{aligned}&x+2 y=1 \\&y-2 x=-1\end{aligned} \Leftrightarrow 2 x+4 y=2$
On adding Eqs. (i) and (ii), we get
$$
5 y=1 \Rightarrow y=\frac{1}{5}
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
&2 \mathrm{x}+4\left(\frac{1}{5}\right)=2 \\
&\Rightarrow \quad 2 \mathrm{x}=2-\frac{4}{5} \\
&\Rightarrow \quad 2 \mathrm{x}=\frac{6}{5} \Rightarrow \mathrm{x}=\frac{3}{5} \\
&\text { Taking } \mathrm{z}=\mathrm{x}+\mathrm{iy}=\frac{1-\mathrm{i}}{1-2 \mathrm{i}} \times \frac{1+2 \mathrm{i}}{1+2 \mathrm{i}} \\
&=\frac{1+2 \mathrm{i}-\mathrm{i}-2 \mathrm{i}^{2}}{1-4 \mathrm{i}^{2}} \\
&\RightaLet $\quad z=(x+i y)(1-2 i)$
$=x+i y-2 x i-2 i^{2} y$
$=x+i y-2 x i+2 y$
$\Rightarrow \quad z \quad z=x+2 y+i(y-2 x)$
$\therefore \quad \overline{\mathrm{z}}=\mathrm{x}+2 \mathrm{y}-\mathrm{i}(\mathrm{y}-2 \mathrm{x})$
According to the question,
$$
\begin{gathered}
\mathrm{z}=1+\mathrm{i} \\
\Rightarrow \quad \mathrm{x}+2 \mathrm{y}-\mathrm{i}(\mathrm{y}-2 \mathrm{x})=1+\mathrm{i}
\end{gathered}
$$
On equating the real and imaginary parts from both sides, we get
and $\begin{aligned}&x+2 y=1 \\&y-2 x=-1\end{aligned} \Leftrightarrow 2 x+4 y=2$
On adding Eqs. (i) and (ii), we get
$$
5 y=1 \Rightarrow y=\frac{1}{5}
$$
$\therefore$ From Eq. (i), we get
$$
\begin{aligned}
&2 \mathrm{x}+4\left(\frac{1}{5}\right)=2 \\
&\Rightarrow \quad 2 \mathrm{x}=2-\frac{4}{5} \\
&\Rightarrow \quad 2 \mathrm{x}=\frac{6}{5} \Rightarrow \mathrm{x}=\frac{3}{5} \\
&\text { Taking } \mathrm{z}=\mathrm{x}+\mathrm{iy}=\frac{1-\mathrm{i}}{1-2 \mathrm{i}} \times \frac{1+2 \mathrm{i}}{1+2 \mathrm{i}} \\
&=\frac{1+2 \mathrm{i}-\mathrm{i}-2 \mathrm{i}^{2}}{1-4 \mathrm{i}^{2}} \\
&\Rightarrow \quad \mathrm{z}=\frac{3+\mathrm{i}}{5}=\frac{3}{5}+\mathrm{i} \frac{1}{5} \\
&\quad \mathrm{z}=\frac{3}{5}+\mathrm{i} \frac{1}{5}, \text { which is true. }
\end{aligned}
$$