Question

1. Sendo Z₁ = 4 + 5i e Z₂ = 2 + 6i: Z₁ + Z₂ = (4+51)+(2+6i) = 4+5i +2 +6i = 6+1 Z₁-Z₂ = (4+5i) - (2 + 6i)= 4 + 5i-2-6i = 2-i​

Answer 1

Resposta: Multiplicação de Números Complexos

Os números complexos são multiplicados com base na propriedade distributiva, sempre lembrando que um numeral complexo é formado por uma parte real e uma imaginária. Veja:

4 + 3i → Re(z) = 4 e Im(z) = 3

2 – 5i → Re(z) = 2 e Im(z) = –5

4 + 3i → Re(z) = 4 e Im(z) = 3

12 – 9i →Re(z) = 12 e Im(z) = –9

Multiplicando os complexos

Exemplos

a) (2 + 5i) * (1 – 2i)

2 – 4i + 5i – 10i² (lembrando que i² = – 1)

2 – 4i + 5i – 10 * (–1)

2 – 4i + 5i + 10

12 + i

b) (4 + 3i) * (2 + 6i)

8 + 24i + 6i + 18i² (lembrando que i² = – 1)

8 + 24i + 6i + 18 * (–1)

8 + 24i + 6i – 18

–10 + 30i

c) (6 – 3i) * (–3 + 7i)

–18 + 42i + 9i – 21i² (lembrando que i² = – 1)

–18 + 42i + 9i – 21 * (–1)

–18 + 42i + 9i + 21

3 + 51i

d) (10 + 10i) * (10 – 10i)

100 – 100i + 100i – 100i² (lembrando que i² = – 1)

100 – 100i + 100i – 100 * (–1)

100 + 100 + 0i

200 + 0i

200

e) 4 + 3i + (1 – 2i) * (3 + i)

4 + 3i + (3 + i – 6i – 2i²)

4 + 3i + 3 + i – 6i – 2i² (lembrando que i² = – 1)

4 + 3i + 3 + i – 6i – 2 * (–1)

4 + 3i + 3 + i – 6i + 2

9 – 2i

f) (2 – 3i) * (1 – 5i) – 4i – 8

2 – 10i – 3i + 15i² – 4i – 8 (lembrando que i² = – 1)

2 – 10i – 3i + 15 * (–1) – 4i – 8

2 – 10i – 3i – 15 – 4i – 8

2 – 15 – 8 – 10i – 3i – 4i

–21 – 17i

g) (–12 – 5i) * (5 + 5i) – 4i + 7

–60 – 60i – 25i – 25i² – 4i + 7 (lembrando que i² = – 1)

–60 – 60i – 25i – 25 * (–1) – 4i + 7

–60 – 60i – 25i + 25 – 4i + 7

–60 + 25 + 7 – 60i – 25i – 4i

–60 + 32 – 89i

–28 – 89i

h) (4 + 3i) * (2 – 5i) + (4 – 3i) * (2 + 5i)

8 – 20i + 6i – 15i² + (8 + 20i – 6i – 15i²)

8 – 20i + 6i – 15i² + 8 + 20i – 6i – 15i²

8 + 8 – 20i + 20i + 6i – 6i – 15i² – 15i²

16 – 30i² (lembrando que i² = – 1)

16 – 30 * (– 1)

46

i) (3 + 30i) * (2 – 3i) + 4 – 5i

6 – 9i + 60i – 90i² + 4 – 5i (lembrando que i² = – 1)

6 – 9i + 60i – 90 * (–1) + 4 – 5i

6 – 9i + 60i + 90 + 4 – 5i

6 + 90 + 4 – 9i + 60i – 5i

100 + 46i

 

j) (20 – 4i) * (2 + 5i) + (8 + 9i) * (7 – 10i) + 4 + 6i

40 + 100i – 8i – 20i² + (56 – 80i + 63i – 90i²) + 4 + 6i

40 + 100i – 8i – 20i² + 56 – 80i + 63i – 90i² + 4 + 6i (lembrando que i² = – 1)

40 + 100i – 8i – 20 * (–1) + 56 – 80i + 63i – 90 * (–1) + 4 + 6i

40 + 92i + 20 + 56 – 17i + 90 + 4 + 6i

(40 + 20 + 56 + 90 + 4) + (92i – 17i + 6i)

210 + 81i

Explicação passo a passo:

Espero ter ajudado