SOLUTION: 2 + 3i/4 + i I have tried the following: (2+3i/4+i)*(4-i/4-i) (8+10i-3i^2)/16-i^2 11-10i/5 I think I'm soo confused.

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: 2 + 3i/4 + i I have tried the following: (2+3i/4+i)*(4-i/4-i) (8+10i-3i^2)/16-i^2 11-10i/5 I think I'm soo confused.       Log On


   

Question 67733: 2 + 3i/4 + i
I have tried the following:
(2+3i/4+i)*(4-i/4-i)
(8+10i-3i^2)/16-i^2
11-10i/5
I think I'm soo confused.
Found 2 solutions by Earlsdon, ankor@dixie-net.com:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify:
%282%2B3i%29%2F%284%2Bi%29 Multiply by %284-i%29%2F%284-i%29
%288-2i%2B12i-3i%5E2%29%2F%2816-i%5E2%29 Simplify.
%288%2B10i%2B3%29%2F%2816%2B1%29
%2811%2B10i%29%2F17

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
2 + 3i/4 + i
I have tried the following:
(2+3i/4+i)*(4-i/4-i)
(8+10i-3i^2)/16-i^2
11-10i/5
:
You are on the right track here:
This part is right:
%28%288+%2B+10i+-+3i%5E2%29%29%2F%28%2816-+i%5E2%29%29
:
You know that i^2 = -1, replace the i^2 with -1, like you did in the numerator:
%28%288+%2B+10i+-+3%28-1%29%29%29%2F%28%2816-+%28-1%29%29%29
:
%28%288+%2B+10i+%2B3+%29%29%2F%28%2816+%2B+1%29%29 = %28%2811+%2B+10i%29%29%2F%28%2817%29%29
:
I mean you had it!! Just a small math error in the denominator.
you're not confused