[z^n]^(1/m) = and = [z^(1/m)]^n

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In summary, the conversation discusses proving the equality between two sets of values of z^{n/m} and (z^n)^{1/m} when n/m is an irreducible fraction. The suggested approach is to use polar form and show that each member of z^{1/m} raised to the power n is a member of (z^n)^{1/m}. The side question also involves finding all possible results of an expression involving complex numbers.
  • #1
cscott
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Homework Statement



Show that if m and n are positive integers, [itex]m \ne 0[/itex], and if n/m is an irreducible fraction, then the set of values of [itex]z^{n/m}[/itex] defined by [tex](z^{1/m})^n[/itex] is identical to the set of value of [itex](z^n)^{1/m}[/itex]

I need to prove the case of a reducible fraction as well, where the two expressions aren't equal.

The Attempt at a Solution



I've been staring at this for a day now and I don't see where to start this beyond messing with the expressions in polar form... hints? Thanks.

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Side question:

[itex](8^{2/3})(8^{-2/3})[/itex]

Does finding all three roots of each factor and then multiplying them in all combinations give all possible results of the above expression? Thanks.
 
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  • #2
Don't forget that an complex equality like the one you have is, in fact, an equality between sets. Starting from the polar expression for z is a good idea, but remember that [tex]z^{1/m}[/tex] is the set of the m-th roots of z, and you must prove that each of its members, when raised to the power n, is a member of the set [tex]\left(z^n\right)^{1/m}[/tex], whose elements are the m-th roots of [tex]z^n[/tex].
 
  • #3
I get (I'm going to use cis() notation):

[tex]z^n = r^n cis \left(n\theta \right)[/tex]
[tex]z^{1/m} = r^{1/m} cis \left (\frac{\theta}{m} + \frac{2k\pi}{m} \right)[/tex]

[tex](z^n)^{1/m} = r^{n/m} cis \left (\frac{n}{m}\theta + \frac{2k\pi}{m} \right)[/tex] k = 0,1,2,...,|m|-1

[tex](z^{1/m})^{n} = r^{n/m} cis \left (\frac{n}{m}\theta + \frac{2nk\pi}{m} \right)[/tex]

I don't think my expression for [itex](z^{1/m})^{n}[/itex] is right with the [itex]2nk[/itex] in it...
 

FAQ: [z^n]^(1/m) = and = [z^(1/m)]^n

What does the equation [z^n]^(1/m) = [z^(1/m)]^n mean?

The equation represents a mathematical operation where a complex number z is raised to the nth power and then the resulting value is raised to the 1/m power. This is equivalent to raising the original complex number to the n/m power.

Can the equation be simplified?

Yes, the equation can be simplified to z^(n/m). This is known as the power rule for exponents.

What does the n and m represent in the equation?

The n represents the number of times the complex number is multiplied by itself, while the m represents the number of times the resulting value is taken to the power of 1/m.

Is the equation valid for all values of z, n, and m?

Yes, the equation is valid for all complex numbers z and all positive integers n and m.

How is this equation useful in science?

This equation is useful in various fields of science, such as physics and engineering, where complex numbers are used to represent quantities with both magnitude and direction. It allows for simplification of complex calculations involving powers of complex numbers.

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