Polynomial/Euclidean division/(4+3i)x^3+x^2+5i over (1+i)x^2+x -3+2i/Example

We perform the Euclidean division

${\displaystyle P=(4+3{\mathrm {i} })X^{3}+X^{2}+5{\mathrm {i} }{\text{ divided by }}T=(1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }.}$

The inverse of ${\displaystyle {}1+{\mathrm {i} }}$ is ${\displaystyle {}{\frac {1}{2}}-{\frac {1}{2}}{\mathrm {i} }}$, and therefore we have

${\displaystyle {}{\begin{aligned}(4+3{\mathrm {i} })(1+{\mathrm {i} })^{-1}&=(4+3{\mathrm {i} }){\left({\frac {1}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}\\&=2+{\frac {3}{2}}-2{\mathrm {i} }+{\frac {3}{2}}{\mathrm {i} }\\&={\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }.\end{aligned}}}$

Hence, ${\displaystyle {}Q}$ starts with ${\displaystyle {}{\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X}$, and we have

${\displaystyle {}{\begin{aligned}((1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }){\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X&=(4+3{\mathrm {i} })X^{3}+{\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-{\frac {19}{2}}+{\frac {17}{2}}{\mathrm {i} }\right)}X\\\end{aligned}}}$

We have to subtract this term from ${\displaystyle {}P}$ and we obtain

${\displaystyle {}{\begin{aligned}P-{\left((4+3{\mathrm {i} })X^{3}+{\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-{\frac {19}{2}}+{\frac {17}{2}}{\mathrm {i} }\right)}X\right)}&={\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left({\frac {19}{2}}-{\frac {17}{2}}{\mathrm {i} }\right)}X+5{\mathrm {i} }\\\,\end{aligned}}}$

We apply the same procedure to this polynomial (which we call ${\displaystyle {}P'}$). We compute

${\displaystyle {}{\begin{aligned}{\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}{\left({\frac {1}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}&=-1+{\frac {3}{2}}{\mathrm {i} }\\\end{aligned}}}$

Therefore, the constant term of ${\displaystyle {}Q}$ equals ${\displaystyle {}-1+{\frac {3}{2}}{\mathrm {i} }}$, and we obtain

${\displaystyle {\left((1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }\right)}{\left(-1+{\frac {3}{2}}{\mathrm {i} }\right)}={\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-1+{\frac {3}{2}}{\mathrm {i} }\right)}X-{\frac {13}{2}}{\mathrm {i} }\,.}$

We subtract this from ${\displaystyle {}P'}$ and get

${\displaystyle {}P'-{\left({\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-1+{\frac {3}{2}}{\mathrm {i} }\right)}X-{\frac {13}{2}}{\mathrm {i} }\right)}={\left({\frac {21}{2}}-10{\mathrm {i} }\right)}X+{\frac {23}{2}}{\mathrm {i} }\,.}$

This term is the remainder ${\displaystyle {}R}$, the Euclidean division altogether is

${\displaystyle {}{\begin{aligned}(4+3{\mathrm {i} })X^{3}+X^{2}+5{\mathrm {i} }&=((1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }){\left({\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X-1+{\frac {3}{2}}{\mathrm {i} }\right)}+{\left({\frac {21}{2}}-10{\mathrm {i} }\right)}X+{\frac {23}{2}}{\mathrm {i} }\\\,\end{aligned}}}$