demoivres theorem

I'm not certain what you mean by "even complex powers".

Demoivre's theorem states that $( \cos \theta + i \sin \theta)^n = \cos (n \theta) + i \sin (n \theta)$. This holds true for all real values of $\theta$ and integers $n$. If by "even complex powers", you mean $n = 2k$ for some integer $k$, then yes it does. It holds for all integer powers, regardless of parity.

The quick proof of it, assumes that you know Euler's formula which converts a complex number to it's trigonometric form. We have that $R e^{i \theta} = R(\cos \theta + i \sin \theta)$. This is sometimes denote as $R \mbox{Cis} \theta$. As such, we have

$\cos \theta + i \sin \theta)^n = ( e^{i \theta} )^n = e^{i n \theta} = \cos (n \theta) + i \sin (n \theta)$

If you have not learnt Euler's formula for complex numbers (or do not know that $e^{ \pi i } + 1 = 0$), you can still prove this by using the sum and product formulas for trigonometric functions. For example, for $n=2$, we have

$\begin{aligned} (\cos \theta + i \sin \theta)^2 & = \cos^2 \theta + 2 i \cos \theta \sin \theta + i^2 \sin^2 \theta\\ & = (\cos^2 \theta - \sin^2 \theta) + i (2 \sin \theta \cos \theta)\\ & = \cos (2 \theta) + i \sin (2 \theta). \end{aligned}$

For the complete proof, proceed by induction:

$\begin{aligned} ( \cos \theta + i \sin \theta)^{k+1} & = ( \cos \theta + i \sin \theta)^{k} \times ( \cos \theta + i \sin \theta)^{1} \\ & = [\cos (k \theta) + i \sin (k\theta) ] \times ( \cos \theta + i \sin \theta)^{1} \\ &= [\cos (k\theta) \cos \theta - \sin (k\theta) \sin \theta] + i [ \sin (k \theta) \cos \theta + \cos (k \theta) \sin \theta] \\ &= \cos (k+1) \theta + i \sin (k+1) \theta. \end{aligned}$

I think Pranav means $(cos\theta + isin\theta)^n$ where n is complex

Zi Song, that was my other interpretation. In order to make sense of that, there are other questions that should have raised first. I deal with them below.

First issue: Does the statement $a^b= X$ make sense if $b$ is a rational number?

We find that it doesn't necessarily make sense, as square roots can take multiple values. For example, we say that $1^{\frac {1}{2} } = 1$ or $-1$. Hence, we can't say that $a^b = X$ directly, but that it takes on several values.

With this in mind, let's consider DeMoivres again. What we can say, is that $(\cos \theta + i \sin \theta)^n$ has $\cos n\theta + i \sin n\theta)$ as one of its values. In the case of $\theta = 0, n = \frac {1}{2}$, we get the statement that one of the values of $1^{ \frac {1}{2} }$ is $1$ (and the other is $-1$).

Second issue: Does the statement "$a^b$ takes on several values" make sense if $b$ is an irrational real number?

What does $1^{\frac {1}{\pi}}$ mean? How many values does it take on? Just 1? Or infinitely many? Certainly it can't take on $\pi$ many values. I'm leaving this aside, and you can look it up if you're interested.

Third issue: If $n$ is complex, it means that $n \theta$ is complex. Hence, we need to define $\cos, \sin : \mathbb{C} \rightarrow \mathbb{C}$

One way to define $\cos, \sin$ is through the MacLaurin Series expansion, which applies to a complex argument. For example, we have $\sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \ldots$, and can substitute $x$ for any complex number. Technically, we say that this makes $\cos, \sin : \mathbb{C} \rightarrow \mathbb{C}$ an analytic continuation of $\cos, \sin : \mathbb{R} \rightarrow \mathbb{R}$

Having dealt with these issues, what we can say is that $(\cos w + i \sin w)^z$ has $\cos zw + i \sin zw$ as one of its values, for $z, w \in \mathbb{C}$. The easiest proof, is once again using Euler's formula, and checking that the trigonometric form holds even when $\theta$ is complex, and that exponentiation laws hold for complex numbers.

For example, $e^ {-1} = \cos i + i \sin i$. Note that since $\cos i, \sin i$ are no longer real values, we cannot compare real and imaginary parts and claim that $\cos i = \frac{1}{e}$ and that $\sin i = 0$ (because that's not true). However, you can use the MacLaurin expansion to calculate these values.