第4章 4.4

(25i)+(3+4i)=5i(2-5i)+(3+4i)=5-i

(25i)(3+4i)=19i(2-5i)-(3+4i)=-1-9i

(25i)(3+4i)=67i+20=267i(2-5i)(3+4i)=6-7i+20=26-7i

25i3+4i=(25i)(34i)(3+4i)(34i)=623i209+16=1423i25\dfrac{2-5i}{3+4i}=\dfrac{(2-5i)(3-4i)}{(3+4i)(3-4i)}=\dfrac{6-23i-20}{9+16}=\dfrac{-14-23i}{25}

解説: r31bn1z