ATRACTOR

1)\(z=3+2i\;\;\; \;\;\;w=2-i\)

\(\begin{array}{l} \; \;\;2+3i=\left(2-i\right)\left(q_{1}+q_{2}i\right)+r_{1}+r_{2}i\Leftrightarrow\\ \Leftrightarrow\frac{2+3i}{2-i}-\left(q_{1}+q_{2}i\right)=r_{1}+r_{2}i\Leftrightarrow\\ \Leftrightarrow\frac{\left(2+3i\right)\left(2+i\right)}{5}-\left(q_{1}+q_{2}i\right)=r_{1}+r_{2}i\Leftrightarrow\\ \Leftrightarrow\frac{1}{5}-q_{1}+\left(\frac{8}{5}-q_{2}\right)i=r_{1}+r_{2}i \end{array}\)

It suffices to take \(q_{1}=0\) and \(q_{2}=0\).

We then have \(\left(2+3i\right)=\left(2-i\right)\left(2i\right)+r_{1}+r_{2}i\), hence \(r_{1}+r_{2}i=2+3i-\left(2+4i\right)=-i\)

We have \(2+3i=\left(2-i\right)2i-i\) (and \(v(-i)=1<5=v(2-i)\)).

Moreover, we can also take \(q_{1}=1\) and \(q_{2}=2\) (because \(\left(\frac{1}{5}-1\right)^{2}+\left(\frac{8}{5}-2\right)^{2}=\frac{16}{25}+\frac{4}{25}<1\) which gives

\(r_{1}+r_{2}i=2+3i-\left(2-i\right)\left(1+2i\right)=2+3i-\left(4+3i\right)=-2\) and we have \(2+3i=\left(2-i\right)\left(1+2i\right)-2\) (and \(v(-2)=4<5=v(2-i)\)).

2)\(z=3+6i\;\;\;\;\;\; w=3+i\)

\(\begin{array}{l} \; \;\;3+6i=\left(3+i\right)\left(q_{1}+q_{2}i\right)+r_{1}+r_{2}i\Leftrightarrow\\ \Leftrightarrow\frac{3+6i}{3+i}=q_{1}+q_{2}i+\frac{r_{1}+r_{2}i}{3+i}\Leftrightarrow\\ \Leftrightarrow\frac{\left(3+6i\right)\left(3-i\right)}{10}=q_{1}+q_{2}i+\frac{r_{1}+r_{2}i}{3+i}\Leftrightarrow\\ \Leftrightarrow\frac{15}{10}+\frac{15}{10}i=q_{1}+q_{2}i+\frac{r_{1}+r_{2}i}{3+i} \end{array}\)

It suffices to take \(q_{1}=1\) or \(2\) and \(q_{2}=1\) or \(2\).

In the case \(q_{1}=1\) and \(q_{2}=1\) we get:

\(r_{1}+r_{2}i=\left(3+6i\right)-\left(3+i\right)\left(1+i\right)=1+2i\), which means, \(3+6i=\left(3+i\right)\left(1+i\right)+1+2i\), with \(v(1+2i)=5<10=v(3+i)\).

The other cases are similar:

\(3+6i=\left(3+i\right)\left(1+2i\right)+2-i\)

\(3+6i=\left(3+i\right)\left(2+i\right)-2+i\)

\(3+6i=\left(3+i\right)\left(2+2i\right)-1-2i\)