((-8)^2)^(1/3) <> (-8)^(2/3) <> (-8)^(1/3))^2

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  • #1
nomadreid
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((-8)^2)^(1/3)=64^(1/3)=4
(-8)^(1/3))^2 = (-2)^2=4
but
(-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
I understand why the latter one works. Why don't the first two?
Thanks.
 

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  • #2
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((2/3)(cos (π) + i*sin(π))
Why does 2/3 act as a multiplicative constant instead of an exponential? Moreover ##\sin \pi =0##.
 
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  • #4
nomadreid
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Oops, too quickly typed the last equation. I meant to multiply the two-thirds times the angles, not times the whole expression (or equivalently use it as an exponent), using Euler's formula. Anyway, my question is not about Euler's formula, which as I say I do get when I am not typing while my bus is about to leave, but rather why the first two formulas are invalid.
 
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HallsofIvy
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((-8)^2)^(1/3)=64^(1/3)=4
(-8)^(1/3))^2 = (-2)^2=4
but
(-8)^(2/3) = 4(-1)^(2/3) = 4((2/3)(cos (π) + i*sin(π)) and so forth, giving a complex number.
No, it doesn't. sin(π)= 0. The other two cube roots are non-real, of course.
I understand why the latter one works. Why don't the first two?
Thanks.
The first two do also, you just have restricted yourself to real numbers. 64 has three complex cube roots, 4 is one, the other two are non-real. 8 has 3 cube roots, one is 2 and the other two are non-real.
 
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  • #6
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They are valid. They give you one of possible three answers, the real one.
 
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  • #7
nomadreid
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Thanks, all of those who answered, everything clear now, and I apologize for the typos.
 

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