Question 543659
z*(2-i) = 3 + i
Divide both sides by (2-i)
z = {{{((3+i))/((2-i))}}}
multiply by the conjugate of the denominator over itself, (same as mult by 1)
z = {{{((3+i))/((2-i))}}} * {{{((2+i))/((2+i))}}} 
FOIL
z = {{{((6+3i+2i+i^2))/((4-i^2))}}}
we know i^2 = -1
z = {{{((6+5i-1))/((4-(-1)))}}} = {{{((5+5i))/((4+1)))}}}
which is
z = {{{((5+5i))/(5))}}}
cancel the 5
z = 1 + i
:
:
Confirm this; replace (1+i) for z in the original equation
(1+i)*(2-i) = 3 + i
FOIL
2-i+2i-i^2 = 3 + i
2 + i -(-1) = 3 + i
3 + i + 2 = 3 + i
3 + i = 3 + i; success!