SOLUTION: Hello, at first we have the complex number z = -1 + i sqrt(3). Secondly we have the fraction z/(1 + 2i). If we substitute z in the numerator we get: {(-1 + i sqrt(3)}/(1+2i).

Question 529655: Hello, at first we have the complex number z = -1 + i sqrt(3).
Secondly we have the fraction z/(1 + 2i).
If we substitute z in the numerator we get: {(-1 + i sqrt(3)}/(1+2i).
How do we calculate this fraction into the form a + bi, (a,b belong to R)?
Thanks a lot in advance
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
clear the complex number from the denominator by FOILing the numerator AND denominator with the conjugate, (1 - 2i)

remember, i^2 is -1
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