Question

求复数Z=(3-i)/(3+2i)的实部,虚部,共轭复数,模

Answer 1

Z=(3-i)/(3+2i)

=(3-i)(3-2i)/(3+2i)(3-2i)

=（9+2i^2-9i)/(9-4i^2)

=7/13-9/13*i

所以

实部是7/13,虚部-9/13*i,共轭复数7/13+9/13*i,模根号[（7/13)^2+(-9/13)^2]=根号130/169