# De Moivre's Theorem for Rational Exponents

1. Dec 18, 2014

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

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

2. Dec 18, 2014

### Staff: Mentor

Where do you see a contradiction?

3. Dec 18, 2014

### mathman

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 $e^{i\theta}$

4. Dec 18, 2014

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.

5. Dec 18, 2014

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

6. Dec 19, 2014

### 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.

7. Dec 19, 2014