SOLUTION: I wondered how to check and see if my answer is correct. The problem is (3+4i)/4i, so I divided the 3 and 4i by 4i and ended up with 1+3/4i. Then I multiplied top and bottom by

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: I wondered how to check and see if my answer is correct. The problem is (3+4i)/4i, so I divided the 3 and 4i by 4i and ended up with 1+3/4i. Then I multiplied top and bottom by      Log On


   



Question 313420: I wondered how to check and see if my answer is correct.
The problem is (3+4i)/4i, so I divided the 3 and 4i by 4i and ended up with 1+3/4i. Then I multiplied top and bottom by -4i, and I got 1-12i/16, which reduced down to 1-3i/4. Is there a way to check and see if my answer is correct?

Found 2 solutions by CharlesG2, Fombitz:
Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
"I wondered how to check and see if my answer is correct.
The problem is (3+4i)/4i, so I divided the 3 and 4i by 4i and ended up with 1+3/4i. Then I multiplied top and bottom by -4i, and I got 1-12i/16, which reduced down to 1-3i/4. Is there a way to check and see if my answer is correct?"
(3 + 4i)/4i
3/(4i) + 4i/4i (should be equivalent to above line)
3/(4i) + 1
(3i)/(4i^2) + 1 (multiplied top and bottom of first term by i)
(3i)/(-4) + 1 (i^2 = -1)
(-3/4)i + 1
1 - (3/4)i (switched terms around so it would be in a + bi form)
check:
1 - (3/4)i
(4 - 3i)/4 (should be equivalent to above line)
(4i - 3i^2)/4i (multiplied top and bottom by i)
(4i + 3)/4i (i^2 = -1)
(3 + 4i)/4i (switched terms around so numerator would be in a + bi form)

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
%283%2B4i%29%2F4i=3%2F%284i%29%2B1
%283%2B4i%29%2F4i=%283%2F%284i%29%29%2A%28%284i%29%2F%284i%29%29%2B1
%283%2B4i%29%2F4i=%2812i%2F%2816i%5E2%29%29%2B1
%283%2B4i%29%2F4i=1-%2812i%29%2F16
%283%2B4i%29%2F4i=1-%283i%29%2F4
Yes, you're correct.
Just reduce your answer to simplest terms.