SOLUTION: Please, I need help with complex number division: (4-i)/(5+i)

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Question 62092: Please, I need help with complex number division:
(4-i)/(5+i)
Found 2 solutions by funmath, Earlsdon:
Answer by funmath(2873) About Me  (Show Source):
You can put this solution on YOUR website!
%284-i%29%2F%285%2Bi%29 Multiply the top and bottom by the conjugate of the bottom, 5-i.
%284-i%29%285-i%29%2F%28%285%2Bi%29%285-i%29%29
%284%285%29%2B4%28-i%29-i%285%29-i%28-i%29%29%2F%285%285%29%2B5%28-i%29%2Bi%285%29%2Bi%28-i%29%29
%2820-4i-5i%2Bi%5E2%29%2F%2825-5i%2B5i-i%5E2%29 Remember, i^2=-1
%2820-4i-5i-1%29%2F%2825-5i%2B5i-%28-1%29%29
%2819-9i%29%2F26
19%2F26-%289%2F26%29i (Standard form a+bi)
Happy Calculating!!!

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Rather than do the actual division, you can get the same result by simplifying as follows:
%284-i%29%2F%285%2Bi%29 Multiply the numerator and the denominator by complex conjugate of the denominator.
The complex conjugate of %285%2Bi%29+=+%285-i%29
%284-i%29%285-i%29%2F%285%2Bi%29%285-i%29+=+%2820-9i%2Bi%5E2%29%2F%2825-i%5E2%29 Substitute i%5E2+=+-1
%2820-9i%2B%28-1%29%29%2F%2825-%28-1%29%29 Simplify:
%2819-9i%29%2F%2826%29 or %281%2F26%29%2819-9i%29