Respostas

2013-10-02T15:34:05-03:00
u=(1+7i, 2-6i)
v=(5-2i, -4i)

a) u+v=((1+7i)+(5-2i), (2-6i)+(-4i))

u+v=(1+7i+5-2i, 2-6i-4i)

u+v=(6+5i, 2-10i)

b) (3+i).v=(3+i).(5-2i, -4i)=((3+i).(5-2i),(3+i).(-4i))

(3+i).v=(3.5+3.(-2i)+i.5+i(-2i), 3.(-4i)+i.(-4i))

(3+i).v=(15-6i+5i-2i^2, -12i-4i^2)

(3+i).v=(15-i-2i^2, -12i-4i^2)

Como i^2=(\sqrt{-1})^2=-1, fazemos:

(3+i).v=(15-i-2(-1), -12i-4(-1))

(3+i).v=(15-i+2, -12i+4)

(3+i).v=(17-i, 4-12i)

c) (Produto escalar) u.v=(1+7i, 2-6i).(5-2i, -4i)

u.v=((5-2i).(1+7i), (-4i)(2-6i))

u.v=(5.1+5.7i-2i.1-2i.7i, -4i.2-4i(-6i))

u.v=(5+35i-2i-14i^2, -8i+24i^2)

u.v=(5+33i-14i^2, -8i+24i^2)

Como i^2=(\sqrt{-1})^2=-1, fazemos:

u.v=(5+33i-14(-1), -8i+24(-1))

u.v=(5+33i+14, -8i-24)

u.v=(19+33i, -24-8i)

d) (Tamanho do vetor ou módulo do vetor) ||v||=||(5-2i, -4i)||

||v||=\sqrt{(5-2i, -4i).(5-2i, -4i)}

||v||=\sqrt{(5-2i)^2+(-4i)^2}

||v||=\sqrt{5^2-2.5.2i+(2i)^2+(-4)^2.i^2}

||v||=\sqrt{25-20i+4i^2+16.i^2}

Como i^2=(\sqrt{-1})^2=-1, fazemos:

||v||=\sqrt{25-20i+4(-1)+16.(-1)}

||v||=\sqrt{25-20i-4-16}

||v||=\sqrt{25-20i-20}

||v||=\sqrt{5-20i}