Question

números complexos:
Sejam Z1= -1+6i e z2=3-2i. Calcule:

a)z1+z2
b)z1-z2
c)z1.z2
d)z1²
e)z2²

Answer 1

 -1+6i + 3- 2i = 2 + 4i
 
-1+6i - (3-2i) = -1 + 6i - 3 + 2i ==> - 4 + 8i 

 (-1+6i)(3-2i)= -3 +2i +18i - 12i^2==> -3+12+2i+18i ===> 9 + 20i 

(-1+6i)^2= 1 -2.1.6i + 36i^2 ==> 1 - 36 - 12i ==> - 35 - 12i

(3-2i)^2 = 9 -2.3.2i +4i^2 ==> 9 - 4 - 12i ==> 5 - 12i

Answer 2

a) Z1 + Z2

$-1+6i +3-2i. = \boxed{2+4i}$

b) Z1-Z2

$-1+6i -3+2i. = \boxed{-4+8i}$

c) Z1 X Z2

$(-1+6i )\cdot(3-2i) = -3+2i+18i-12i^2\\\\ \boxed{i^2 = -1}\\\\ \boxed{9+20i}$

d) Z1^2

$(-1+6i)\cdot(-1+6i) = 1-6i-6i+36i^2\\\\ \boxed{-35-12i}$

e) z^2

$(3-2i.)\cdot(3-2i) = 9-6i-6i+4i^2\\\\ \boxed{5-12i}$