SOLUTION: could someone please help me with this problem? (3-i) / (1+i) I know you use the conjugate and multiply by (3-i) / (1+I) I think,but I don't think I did something right.Thanks

Algebra ->  Real-numbers -> SOLUTION: could someone please help me with this problem? (3-i) / (1+i) I know you use the conjugate and multiply by (3-i) / (1+I) I think,but I don't think I did something right.Thanks       Log On


   

Question 131945: could someone please help me with this problem?
(3-i) / (1+i) I know you use the conjugate and multiply by (3-i) / (1+I) I think,but I don't think I did something right.Thanks for your help in advance.
Sincerely,LH
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
A conjugate of a complex number a%2Bbi is formed by changing the sign on the imaginary part. So the conjugate of a%2Bbi is a-bi.

To simplify an expression with a complex denominator, you need to multiply the entire expression by 1 in the form of the conjugate of the denominator divided by itself. That means if you have %28a%2Bbi%29%2F%28c%2Bdi%29, you want to multiply by 1 in the form of %28c-di%29%2F%28c-di%29.

Your problem:

%283-i%29+%2F+%281%2Bi%29

The conjugate of the denominator is %281-i%29, so you need to multiply the entire expression by %281-i%29%2F%281-i%29, thus:

%28%283-i%29%2F%281%2Bi%29%29%28%281-i%29%2F%281-i%29%29

Now multiply numerator times numerator and denominator times denominator just like multiplying any other pair of fractions.

Using FOIL, the numerator product becomes: 3%2B3i-i%2B1=4%2B2i (remember i%5E2=-1 so -i%2Ai=-%28i%5E2%29=-%28-1%29=1)

The denominator product is easier because, by using the conjugate, you have the factors of the difference of two squares, so %281%2Bi%29%281-i%29=1-i%5E2=1-%28-1%29=2

Putting the pieces back together we have %284%2B2i%29%2F2=2%2Bi

Done.