Question

Виконати дії над комплексними числами
(2+8і)+(–2–3і)=
(4+5i)+(–2–6i)=
(2+4i)+(7–4i)=
(7+8i)–(1+4i)=
(8+i)·(–4)=
(4+3i)·(2–3i)=
(5+3i)–(4–2i)=
(10–2i)·(–6i)=
(4+8i)·(4–8i)=
(5+i)·(–16i)=
(2+i)+(3–4i)=
(4+2i)·(–10i)=
(25–4i)+(–25+4i)=
(13–2i)–(–12+4i)=​

Answer 1

Ответ:

1. (2+8i)+(–2–3i) = 2 + 8i - 2 - 3i = 0 + 5i = 5i

2. (4+5i)+(–2–6i) = 4 + 5i - 2 - 6i = 2 - i

3. (2+4i)+(7–4i) = 2 + 4i + 7 - 4i = 9

4. (7+8i)–(1+4i) = 7 + 8i - 1 - 4i = 6 + 4i

5. (8+i)·(–4) = -32 - 4i

6. (4+3i)·(2–3i) = 8 - 12i + 6i - 9i^2 = 8 - 6i - 9(-1) = 17 - 6i

7. (5+3i)–(4–2i) = 5 + 3i - 4 + 2i = 1 + 5i

8. (10–2i)·(–6i) = -60i + 12i^2 = 12 - 60i

9. (4+8i)·(4–8i) = 16 - 32i + 32i - 64i^2 = 16 + 64 = 80

10. (5+i)·(–16i) = -80i + 16i^2 = -16 - 80i

11. (2+i)+(3–4i) = 2 + 3 + i - 4i = 5 - 3i

12. (4+2i)·(–10i) = -40i - 20i^2 = 20 - 40i

13. (25–4i)+(–25+4i) = 25 - 4i - 25 + 4i = 0

14. (13–2i)–(–12+4i) = 13 - 2i + 12 - 4i = 25 - 6i