# De Moivre's Theorem for Rational Exponents

## Main Question or Discussion Point

$cosθ + isinθ = cos(θ + 2kπ) + isin(θ + 2kπ)$ 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 $p ∈ ℤ$

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

mfb
Mentor
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 $p ∈ ℤ$

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

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 $e^{i\theta}$

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.

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.

$e^{i(\theta+2k\pi)}$ 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