Question

1)Даны 2 комплексных числа. z1= -5+3i , z2=1+2i. Вычислить: 4z1-2z2, z1*z2, z2/z1, z1 в 2й степени, (z1*z2):(z1+z2).

Answer 1

z1=-5+3i     z2=1+2i   
используем i^2=-1

4z1-2z2=(-20+12i)-(2+4i)=-22+8i
z1*z2= (-5+3i)(1+2i)=-5+3i-10i-6=-11-7i

z2/z1=[(1+2i)(-5-3i)]/[(-5+3i)(-5-3i)]=(-5-10i-3i+6)/((-5)^2-(-3i)^2)=
= 1-13i/(25+9)=(1-13i)/36=1/36-13/36*i

z1^2=(-5+3i)^2=25-10*3i+(3i)^2=25-30i-9=16-30i
z1*z2/(z1+z2)= (-11-7i)/(-5+3i+1+2i)=
=(-11-7i)/(-4+5i) =(-11-7i)(-4-5i)/[(-4)^2-(5i)^2]=
=(44+28i+55i-35)/(16+25)=9+83i/41 = 9/41+8/41*i