SOLUTION: please help to simplify 3i - 5 DIVIDED BY -i^3 + 3i^2

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Question 66240This question is from textbook An Incremental Development
: please help to simplify
3i - 5
DIVIDED BY
-i^3 + 3i^2 This question is from textbook An Incremental Development

Found 2 solutions by Edwin McCravy, ptaylor:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
please help to simplify

  3i - 5
———————————
 -i³ + 3i²

i² = -1 and i³ = i²·i = (-1)·i = -i

So replace i³ by -i and i² by -1

    3i - 5
———————————————
 -(-i) + 3(-1)

  3i - 5
 ————————
   i - 3

Write numerator and denominator
in standard A + Bi order:


  -5 + 3i
 —————————
  -3 + i

Multiply top and bottom by the conjugate
of the denominator, which is -3 - i

  (-5 + 3i)(-3 - i)
 ——————————————————— 
  (-3 + i)(-3 - i)

FOIL out top and bottom:

  15 + 5i - 9i - 3i²
 ————————————————————
   9 + 3i - 3i - i²

Combine terms:


  15 - 4i - 3i²
 ———————————————               
      9 - i²

Replace i² by -1

  15 - 4i - 3(-1)
 —————————————————
      9 - (-1)

  15 - 4i + 3
 —————————————
     9 + 1

 18 - 4i
—————————
   10

Make two fractions:

 18     4i 
———— - ————
 10     10

Reduce the fractions

 9     2i 
——— - ————
 5     5

Write in A + Bi form:

 9     2 
——— - ——— i
 5     5

Edwin
Answer by ptaylor(2198)   (Show Source): You can put this solution on YOUR website!
please help to simplify
3i - 5
DIVIDED BY
-i^3 + 3i^2
i=sqrt(-1)
i^2=-1
i^3=(i^2)(i)=-sqrt(-1)
-i^3=sqrt(-1)=i
3i^2=3(-1)=-3
Now we have
(3i-5)/(i-3) Multiply numerator and denominator by (i+3) to get rid of the radical in the denominator and we have:
((3i-5)(i+3))/((i-3)(i+3)) Multiplying these out we get:
(3i^2-5i+9i-15)/(i^2-3i+3i-9)
(-3+4i-15)/-1-9
(4i-18)/-10 reduce by dividing both numerator and denominator by -2
(9-2i)/5


Hope this helps-----ptaylor

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