SOLUTION: How would. You simplify this problem? {{{(7-6i)+(9+11i)}}} I added 6i to both sides and got 17i then I brought the 9 and 7 down to get 7+9+17. Then added all three numbers

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: How would. You simplify this problem? {{{(7-6i)+(9+11i)}}} I added 6i to both sides and got 17i then I brought the 9 and 7 down to get 7+9+17. Then added all three numbers       Log On


   

Question 534174: How would. You simplify this problem?
%287-6i%29%2B%289%2B11i%29
I added 6i to both sides and got 17i then I brought the 9 and 7 down to get 7+9+17. Then added all three numbers together and got 33. I'm not sure if this is correct
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
%287-6i%29%2B%289%2B11i%29
Multiply like any 2 binomials
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= 7*(9+11i) - 6i*(9+11i)
= 63+%2B+77i+-+54i+-+66i%5E2
= 63 + 23i + 66
= 129 + 23i

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I'm confused. Maybe something got mixed up in the text of the problem or the previous answer.
Were you supposed to add the two complex numbers or multiply them?
If adding
%287-6i%29%2B%289%2B11i%29=%287%2B9%29%2Bi%28-6%2B11%29=16%2B5i
When working with complex numbers, you work as if the i was a variable x.
You cannot mix the real part of the number, with the imaginary part that has the i, just like you cannot add 16+5x.
The only difference is that the even powers of i are real numbers, so multiplications of complex numbers can always simplify to something as short as 16%2B5i
i%5E2=-1, i%5E4=1 and all i%5Eevennumber powers simplify to 1 (if the exponent is a multiple of 4) or to -1 (if not a multiple of 4).