# Question #5d8b2

Jun 16, 2016

#### Answer:

No. $n$ must be an integer for the given formula to hold.

#### Explanation:

De Moivre's theorem, which typically is stated as
${\left[r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)\right]}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right)$
does not apply to non integer values for $n$. This is because non integer powers of complex numbers are multiple-valued, whereas the trigonometric representation is single-valued.

For example, letting $z = i$ and $n = \frac{1}{2}$, we have

$i = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = \cos \left(\frac{5 \pi}{2}\right) + i \sin \left(\frac{5 \pi}{2}\right)$

Using Euler's formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, we can see that these representations, despite both being equal to $i$, produce different results when taken to the power of $\frac{1}{2}$:

${\left({e}^{i \frac{\pi}{2}}\right)}^{\frac{1}{2}} = {e}^{i \frac{\pi}{4}} = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$

and

${\left({e}^{i \frac{5 \pi}{2}}\right)}^{\frac{1}{2}} = {e}^{i \frac{5 \pi}{4}} = \cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right) = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i$

Thus ${\left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)}^{\frac{1}{2}}$ has two values, whereas $\left(\cos \left(\frac{1}{2} \cdot \frac{\pi}{2}\right) + i \sin \left(\frac{1}{2} \cdot \frac{\pi}{2}\right)\right)$ has only one.

As De Moivre's theorem requires the left hand side to be single valued to match with the right hand side, $n$ is restricted to integers.