i 12 and i 2 are assumed to be 12i and 2i respectively.
(2-i)(5+12i)=10+19i+12=22+19i.
(1+2i)^3=1+6i+3(2i)^2+(2i)^3=1+6i-12-8i=-11-2i=-(11+2i); -1/(11+2i)=-(11-2i)/(121+4)=-(11-2i)/125.
Z=-(22+19i)(11-2i)/125=-(242+165i+38)/125=-(280+165i)/125=-(56+33i)/25.
|Z|=√((-56)^2+(-33)^2)/25=√4225/25=65/25=13/5.