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Can someone find the solutions and the magnitudes of the following problems(This is from my complex numbers unit):

 

 

1. (x^4)-1=0

2. (x^3)+1=0

3.(x^3)-1=0

4.(x^6)-1=0

5.(x^6)+1=0

 

 

 

 

Thank You!

Guest Feb 24, 2017
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2+0 Answers

 #1
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1) -

Solve for x:
x^4 - 1 = 0

Add 1 to both sides:
x^4 = 1

Taking 4^th roots gives 1 times the 4^th roots of unity:
Answer: |x = -1     or x = -i      or x = i      or x = 1

 

2) -

 

Solve for x:
x^3 + 1 = 0

Subtract 1 from both sides:
x^3 = -1

Taking cube roots gives (-1)^(1/3) times the third roots of unity:
Answer: |x = -1        or x = (-1)^(1/3)        or x = -(-1)^(2/3)

 

3) -

 

Solve for x:
x^3 + 1 = 0

Subtract 1 from both sides:
x^3 = -1

Taking cube roots gives (-1)^(1/3) times the third roots of unity:
Answer: |x = -1        or x = (-1)^(1/3)        or x = -(-1)^(2/3)

 

4) -

 

Solve for x:
x^6 - 1 = 0

Add 1 to both sides:
x^6 = 1

Taking 6^th roots gives 1 times the 6^th roots of unity:
Answer: |x = -1    or x = 1    or x = -(-1)^(1/3)    or x = (-1)^(1/3)    or x = -(-1)^(2/3)    or x = (-1)^(2/3)

 

5) -

 

Solve for x:
x^6 + 1 = 0

Subtract 1 from both sides:
x^6 = -1

Taking 6^th roots gives (-1)^(1/6) times the 6^th roots of unity:
Answer: |x = -i    or x = i    or x = -(-1)^(1/6)    or x = (-1)^(1/6)    or x = -(-1)^(5/6)    or x = (-1)^(5/6)

Guest Feb 24, 2017
 #2
avatar+26366 
+5

"Can someone find the solutions and the magnitudes of the following problems(This is from my complex numbers unit):

1. (x^4)-1=0

2. (x^3)+1=0

3.(x^3)-1=0

4.(x^6)-1=0

5.(x^6)+1=0"

 

This might help:

 

.

Alan  Feb 25, 2017

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