Question

ajudem.....................

Answer 1

5) k+mi -2 +2i = 3+2i
K-2= 3(parte real) e m+2=2(parte imaginaria)

K= 5 e m= 0

6) 4-m= m-3 (isolando parte imaginaria)
M= -1/2

Answer 2

5) 

z₁-z₂ = 3+2i
z₁ = k+mi
z₂ = 2-2i

z₁-z₂ = (k+mi)-(2-2i) = k+mi-2+2i = k-2+mi+2i = 3+2i
(k-2)+(m+2)i = 3+2i

k-2 = 3
k = 3+2
k = 5

m+2 = 2
m = 2-2
m = 0

Logo, temos k=5 e m=0

z₁ = 5+0i = 5
z₂ = 2-2i

z₁-z₂ = (5+0i)-(2-2i) = 5-2+2i = 3+2i

6) 

z₁ = 2+mi
z₂ = 3+4i

z₂-z₁ = z₁-1-3i
z₂-z₁-z₁ = -1-3i
z₂-2z₁ = -1-3i

(3+4i)-2(2+mi) = -1-3i
3+4i-4-2mi = -1-3i
3-4+4i-2mi = -1-3i
-1+(4-2m)i = -1-3i

4-2m = -3
-2m = -3-4
-2m = -7
m = -7/-2
m = 7/2

z₁ = 2+7i/2
z₂ = 3+4i


z₂-2z₁= (3+4i)-2(2+7i/2) = 3+4i-4-7i = 3-4+4i-7i = -1-3i