SOLUTION: I need help with finding the complex number z that satisfies this equation: z*(2-i)=3+i Thank you :)

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Question 543659: I need help with finding the complex number z that satisfies this equation: z*(2-i)=3+i Thank you :)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
z*(2-i) = 3 + i
Divide both sides by (2-i)
z = %28%283%2Bi%29%29%2F%28%282-i%29%29
multiply by the conjugate of the denominator over itself, (same as mult by 1)
z = %28%283%2Bi%29%29%2F%28%282-i%29%29 * %28%282%2Bi%29%29%2F%28%282%2Bi%29%29
FOIL
z = %28%286%2B3i%2B2i%2Bi%5E2%29%29%2F%28%284-i%5E2%29%29
we know i^2 = -1
z = %28%286%2B5i-1%29%29%2F%28%284-%28-1%29%29%29 = %28%285%2B5i%29%29%2F%28%284%2B1%29%29%29
which is
z = %28%285%2B5i%29%29%2F%285%29%29
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!