Julia P.

# Find all complex numbers whose sixth power equals 64

Basically, written out, looks like

(a + bi)^6 = 64

## 2 Answers By Expert Tutors

By:

Tutor
4.9 (890)

Math Tutor--High School/College levels

Mark M.

tutor
Here is another way to do this problem without DeMoivre's Theorem:

If z is a 6th root of 64, then z6 = 64.  So, z6 - 64 = 0

Factoring this as a difference of squares, we get (z3 - 8)(z+ 8) = 0

Apply the formulas for factoring a sum or difference of cubes to get:

(z - 2)(z2 + 2z + 4)(z + 2)(z2 - 2z + 4) = 0

Set each factor to 0 and solve to get the 6th roots of 64.
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03/18/15

Julia P.

Thank you so much! This helped me, and I was able to solve the problem correctly.
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03/18/15 Tutor
5.0 (243)

Mathematics Teacher - NCLB Highly Qualified

Julia P.

Well I understand this, but how can I find all complex numbers (i.e. a + bi) which are equal to +-2?
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03/18/15 Mark M.

Do you know De Moivre's Theorem?
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03/18/15

Julia P.

No, and googling it didn't help me much either.
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03/18/15

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