а) (1+i)*z = 1-i
б) (9-i)*z = 15-5i
2) Вычислить
а) i(4-3i)*i(4+3i) =
б) (1+i)^3+(3-4i)^2 =
[m]z=\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1-i)(1+i)}=\frac{1-2i+i^2}{1-i^2}=[/m]
так как [m] i^2=-1[/m],
[m]=\frac{1-2i-1}{1-(-1)}=\frac{-2i}{2}=-i[/m]
1б)
[m]z=\frac{15-5i}{9-i}=\frac{(15-5i)(9+i)}{(9-i)(9+i)}=\frac{135-45i+15i-5i^2}{81-i^2}=[/m]
так как [m] i^2=-1[/m],
[m]=\frac{140-30i}{82}=\frac{140}{82}-\frac{30}{82}i=\frac{70}{41}-\frac{15}{41}i[/m]
2a)
[m]i\cdot (4-3i)\cdot i\cdot (4+3i)=i^2\cdot (4^2-9i^2)=-1\cdot (16+9)=-25[/m]
2б)
[m](1+i)^3+(3-4i)^2=1+3i+3i^2-i^3+9-2\cdot 3\cdot 4i+16i^2=[/m]
[m]=1+3i-3+i+9-24i-16=-9-20i[/m]