Question

AJUDAAAAAAAAAAAAAAA
Efetue Números complexos:

(1-2i)^5



(2+2i)^2

Answer 1

(1-2i)^5 = (1-2i)².(1-2i)²(1-2i)

(1 - 4i + 4i²).(1 - 4i + 4i²).(1-2i) ==>(1 - 4 - 4i)(1 - 4 - 4i)(1-2i)

( - 3 - 4i)( - 3 - 4i)(1-2i)==>(9 + 24i + 16i²)(1 - 2i)
 
( - 7 + 24i)(1- 2i) ==> - 7 + 14i + 24i - 48i²

41 + 38i 


(2+2i)^2 ==>  4  + 8i + 4i² ==> 4 + 8i - 4 ==> 8i

Answer 2

Olá Beatriz!
(1-2i)^5=
lembrando que :
[i²=-1,  i³=-i,  i^4=1, e 1^5=i]

Resolvendo:
(1-2i)^5= (1-2i)²(1-2i)²(1-2i)
(1-2i)5^=[(1²-4i+4i²)(1²-4i+4i²)(1-2i)
agora vamos substituiir i²=-1
(1-2i)^5={[1-4i+4(-1)][1-4i+4(-1)](1-2i)}
(1-2i)^5={(1-4i-3)(-4i-3)(-1-2i)}
(1-2i)^5=[16i²+24i+9](1-2i)
novamente i²=-1, vamos substituir
(1-2i)^5=[-16+24i+9](1-2i)
(1-2i)^5=(-7+24i)(1-2i)
(1-2i)^5=(-7+14i-7-48i²)
(1-2i)^5=[-14+14i-48(-1)]
(1-2i)^5=(-14+14i+48)
(1-2i)^5=[-7+38i-48(-1)]
(1-2i)^5=(-7+38i+48)
(1-2i)^5= 41+38i
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(2+2i)^2
(2+2i)^2=2²+8i+4i²
(2+2i)²= 4+8i+4i²
sendo i²=-1 substituindo temos
(2+2i)²=4+8i+4(-1)
(2+2i)²=4+8i-4
(2+2i)²=8i