We know that z₁=1+3i and z₂=1-3i are solution, so:
[tex](z-z_1)(z-z_2)=0\\\\\big(z-(1+3i)\big)\big(z-(1-3i)\big)=0\\\\(z-1-3i)(z-1+3i)=0\\\\
\big((z-1)-3i\big)\big((z-1)+3i\big)=0[/tex]
Now we use [tex](a-b)(a+b)=a^2-b^2[/tex] where [tex]a=z-1[/tex] and [tex]b=3i[/tex]:
[tex]\big((z-1)-3i\big)\big((z-1)+3i\big)=0\\\\
(z-1)^2-(3i)^2=0\\\\z^2-2z+1-9i^2=0\qquad [i^2=-1]\\\\
z^2-2z+1+9=0\\\\\boxed{z^2-2z+10=0}[/tex]
