De Moivre's Theorem for Rational Exponents

[PFuser1232]

$cosθ + isinθ = cos(θ + 2kπ) + isin(θ + 2kπ)$ for $k ∈ ℤ$
$[cosθ + isinθ]^n = [cos(θ + 2kπ) + isin(θ + 2kπ)]^n$
$cos(nθ) + isin(nθ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)$
$cos(nθ + 2mπ) + isin(nθ + 2mπ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)$ for $m ∈ ℤ$

Now consider the special case $n = 1/p$ for $p ∈ ℤ$

$cos(\frac{θ}{p} + 2mπ) + isin(\frac{θ}{p} + 2mπ) = cos(\frac{θ}{p} + \frac{2kπ}{p}) + isin(\frac{θ}{p} + \frac{2kπ}{p})$

Is this a contradiction?

 

[mfb]

Where do you see a contradiction?

 

[mathman]

$cosθ + isinθ = cos(θ + 2kπ) + isin(θ + 2kπ)$ for $k ∈ ℤ$
$[cosθ + isinθ]^n = [cos(θ + 2kπ) + isin(θ + 2kπ)]^n$
$cos(nθ) + isin(nθ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)$
$cos(nθ + 2mπ) + isin(nθ + 2mπ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)$ for $m ∈ ℤ$

Now consider the special case $n = 1/p$ for $p ∈ ℤ$

$cos(\frac{θ}{p} + 2mπ) + isin(\frac{θ}{p} + 2mπ) = cos(\frac{θ}{p} + \frac{2kπ}{p}) + isin(\frac{θ}{p} + \frac{2kπ}{p})$

Is this a contradiction?

No. The left side and the right side are two different things. The left side simply expresses the fact $e^{2in\pi}=1$ while the right side simply says there are p pth roots of $e^{i\theta}$

 

[PFuser1232]

No. The left side and the right side are two different things. The left side simply expresses the fact $e^{2in\pi}=1$ while the right side simply says there are p pth roots of $e^{i\theta}$

The very fact that the RHS and LHS seem to represent two different things is the reason I'm confused as to whether the equality holds.

 

[PFuser1232]

No. The left side and the right side are two different things. The left side simply expresses the fact $e^{2in\pi}=1$ while the right side simply says there are p pth roots of $e^{i\theta}$

Also, could you please elaborate on what the RHS represents?

 

[mathman]

$e^{i(\theta+2k\pi)}$ is the same number for all integer values of k. However $e^{\frac{i(\theta+2k\pi)}{p}}$ will have p possible values for k =0,1,...,p-1. These are the pth roots.

 

[PFuser1232]

$e^{i(\theta+2k\pi)}$ is the same number for all integer values of k. However $e^{\frac{i(\theta+2k\pi)}{p}}$ will have p possible values for k =0,1,...,p-1. These are the pth roots.

But this means that, technically, they're not equal. Right? They're only equal in special cases where k/p is an integer.

 

[mathman]

But this means that, technically, they're not equal. Right? They're only equal in special cases where k/p is an integer.

I hate to keep repeating myself. The left side of the expression represents one particular pth root. The right side is any pth root, depending on k, which is different, unless k/p is an integer.