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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. Calvin Lin Staff · 4 years, 7 months ago

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

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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:

\[ \begin{align} ( \cos \theta + i \sin \theta)^{k+1} & = ( \cos \theta + i \sin \theta)^{k} \times ( \cos \theta + i \sin \theta)^{1} \\ & = [\cos (k \theta) + i \sin (k\theta) ] \times ( \cos \theta + i \sin \theta)^{1} \\ &= [\cos (k\theta) \cos \theta - \sin (k\theta) \sin \theta] + i [ \sin (k \theta) \cos \theta + \cos (k \theta) \sin \theta] \\ &= \cos (k+1) \theta + i \sin (k+1) \theta. \end{align} \] Calvin Lin Staff · 4 years, 7 months ago

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