SOLUTION: with the help of the quadratic formula, find all the complex roots of each of the polynomials given in exercise 1. iii) z^3 + i

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Question 194674This question is from textbook An Introduction to Abstract Algebra with Notes to the Future Teacher
: with the help of the quadratic formula, find all the complex roots of each of the polynomials given in exercise 1.
iii) z^3 + i This question is from textbook An Introduction to Abstract Algebra with Notes to the Future Teacher

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First, perform synthetic division on the polynomial z%5E3%2Bi using the test zero z=i. Note: you can easily solve for "z" and find that one solution is z=i 

i |  1  0  0   i
  |     i  -1  -i
  -----------
     1  i  -1   0


Since the first three numbers in the third row are 1, i, and -1, this means that

z%5E3%2Bi=%28z-i%29%28z%5E2%2Biz-1%29


Now simply use the quadratic formula to solve z%5E2%2Biz-1=0 for "z":


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%28i%29+%2B-+sqrt%28+%28i%29%5E2-4%281%29%28-1%29+%29%29%2F%282%281%29%29 Plug in a=1, b=i, and c=-1


z+=+%28-i+%2B-+sqrt%28+-1-4%281%29%28-1%29+%29%29%2F%282%281%29%29 Square i to get i%5E2=%28sqrt%28-1%29%29%5E2=-1.


z+=+%28-i+%2B-+sqrt%28+-1%2B4+%29%29%2F%282%29 Multiply


z+=+%28-i+%2B-+sqrt%28+3+%29%29%2F%282%29 Combine like terms.


So the other two solutions are z+=+%28-i+%2B+sqrt%28+3+%29%29%2F%282%29 or z+=+%28-i+-+sqrt%28+3+%29%29%2F%282%29


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Answer:

So the 3 solutions are z=i, z+=+%28-i+%2B+sqrt%28+3+%29%29%2F%282%29, or z+=+%28-i+-+sqrt%28+3+%29%29%2F%282%29