SOLUTION: I'm not sure that I'm under the right heading because I really don't even know what my question is asking. The instructions are: perform the indicated operation. Write the result

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Question 142542: I'm not sure that I'm under the right heading because I really don't even know what my question is asking. The instructions are: perform the indicated operation. Write the result in the form of a + bi.
6/3 + i where 6 is over the entire second portion of the problem. I don't know where to start or even what this question is asking. This is a learning sheet that I'm doing for myself to better myself before my next semester in college, so I already have the answer of 9/5 - 3/5i. I just need to know what steps to take to get there. Thanks in advance :)
Found 3 solutions by stanbon, Earlsdon, scott8148:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
6/(3 + i)
Multiply numerator and denominator by 3-i to get:
= [6(3-i)]/[(3+i)(3-i)]
= [6(3-i)]/[9+1]
= [3(3-i)]/[5]
= (9/5) - (3/5)i
=====================
Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Rationalise:
6%2F%283%2Bi%29 Multiply the numerator and denominator by the complex conjugate of (3+i) which is (3-i):
6%283-i%29%2F%283%2Bi%29%283-i%29 Simplify.
6%283-i%29%2F%289-i%5E2%29 But i%5E2+=+-1, so...
6%283-i%29%2F%289-%28-1%29%29
6%283-i%29%2F10
3%283-i%29%2F5
%289-3i%29%2F5 or %289%2F5%29-%283%2F5%29i

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the operation is 6 divided by 3+i

the fraction is an answer, but complex numbers are a no-no in the denominator

multiplying by the conjugate (3-i) on the top and bottom is the way to go

[6(3-i)]/[(3+i)(3-i)] __ (18-6i)/(9-i^2) __ 18/10-6i/10 __ 9/5-3i/5