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# demoivres theorem

could anyone tell me whether demoivres theorem holds for even complex powers and if so could you sketch out a brief proof for it. thanks in advance

Note by Pranav Chakravarthy
4 years, 7 months ago

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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. Staff · 4 years, 7 months ago

I think Pranav means $$(cos\theta + isin\theta)^n$$ where n is complex · 4 years, 7 months ago

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{align} (\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{align}

For the complete proof, proceed by induction:

\begin{align} ( \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{align} Staff · 4 years, 7 months ago