Question

Números Complexos

Determine o número complexo z, tal que
(3 - i) . z + 2i + 8 - i = 8 + 11i

Answer 1

 Desenvolvendo,

$(3-i)z+2i+8-i=8+11i\\(3-i)z=8+11i-2i-8+i\\(3-i)z=10i\\\\z=\frac{10i}{3-i}\times\frac{3+i}{3+i}\\\\z=\frac{10i(3+i)}{9-i^2}\\\\z=\frac{10i(3+i)}{9-(-1)}\\\\z=\frac{10i(3+i)}{10}\\\\z=i(3+i)\\z=3i+i^2\\\boxed{z=-1+3i}$

Answer 2

(3 - i) . z + 2i + 8 - i = 8 + 11i

(3 - i) . z + 2i + 8 - i = 8 + 11i - 2i - 8 + i

(3 - i) . z  =  10i

z =    10i  
        3 - i 


z =    10i(3+i)     ==> z = 30i + 10i²
        (3 - i )(3+i)                  9 - i²

z = 30i - 10   ==> z = 10(3i -1)
          10                       10

z = 3i - 1