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# There are six different sixth roots of 64. That is, there are six complex numbers that solve

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There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=0, find them.

Dec 22, 2017

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I think you meant x^6=64, nbot x^6=0... Here you go!

The obvious two solutions to this equation is 2 and -2 since, (±2)6 = 64 but the question asked for the complex numbers too, so there are more solutions.

The first step in order to approach this problem is to factor x= 64.

We can re-write this problem into x6 - 26 = 0
And also, (x3)2-(23)2=0

After using difference of cubes and sum of cubes(search it up if you don't know what it is)

I got: (x-2)(x^2+2x+2^2)*{(x+2)(x^2-2x+2^2)}=0

We need to solve ±(x-2)=0,

(x^2+2x+2^2)=0

(x^2-2x+2^2)=0

After you do, you have your six conjugates. I'll leave the solving for you!

Dec 22, 2017
edited by supermanaccz  Dec 22, 2017
edited by supermanaccz  Dec 22, 2017
#2
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64^(1/6) =

+ OR - 2

2 e^((i π)/3)≈1.0000 + 1.7321 i

2 e2 e^(-(2 i π)/3)≈-1.0000 + 1.7321 i

2 e^(-(2 i π)/3)≈-1.0000 - 1.7321 i

2 e^(-(i π)/3)≈1.0000 - 1.7321 i

Dec 22, 2017
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Thank you supermanaccz and guest,

Superman, I have never seen it done your way, there are usually several ways to solve the same problem and it is interesting to see alternate routes.

There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=0, find them.

The 6 points will be equally distributed on the complex plane at an absolute distance of 2 units.

A revolution is 2pi radians so the points will be   2pi/6 = pi/3 radians apart.

Also you know that one of them is   $$2 = 2e^0 = 2e^{(0\pi/3)i}$$

$$\sqrt[6]{2}=2e^{(0\pi/3)i}, \quad2 e^{1(\pi/3)i}, \quad2 e^{2(\pi/3)i}, \quad2 e^{3(\pi/3)i},\ \quad2 e^{4(\pi/3)i}, \quad2 e^{5(\pi/3)i}\\ \sqrt[6]{2}=2e^{(\pi/3)i}, \quad2 e^{(\pi/3)i}, \quad2 e^{(2\pi/3)i}, \quad2 e^{\pi i},\ \quad2 e^{(4\pi/3)i}, \quad2 e^{(5\pi/3)i}\\$$