Difference between revisions of "De Moivre's Theorem"

Latest revision as of 09:49, 31 August 2024

De Moivre's Theorem is a very useful theorem in the mathematical fields of complex numbers. It allows complex numbers in polar form to be easily raised to certain powers. It states that for $x\in\mathbb{R}$ and $n\in\mathbb{Z}$ , $\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)$ .

Proof

This is one proof of de Moivre's theorem by induction.

If $n\ge0$ :

If $n=0$ , the formula holds true because $\cos(0x)+i\sin(0x)=1+0i=1=z^0.$

Assume the formula is true for $n=k$ . Now, consider $n=k+1$ :

$\begin{align*} (\cos x+i \sin x)^{k+1} & =(\cos x+i \sin x)^{k}(\cos x+i \sin x) & \text { by Exponential laws } \\ & =[\cos (k x)+i \sin (k x)](\cos x+i \sin x) & \text { by our assumption } \\ & =\cos (k x) \cos x-\sin (k x) \sin x+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\ & =\operatorname{cis}((k+1)(x)) & \text { by various trigonometric identities } \end{align*}$

Therefore, the result is true for all nonnegative integers $n$ .

If $n<0$ , one must consider $n=-m$ when $m$ is a positive integer.

$\begin{align*} (\operatorname{cis} x)^{n} &=(\operatorname{cis} x)^{-m} \\ &=\frac{1}{(\operatorname{cis} x)^{m}} \\ &=\frac{1}{\operatorname{cis}(m x)} \\ &=\cos (m x)-i \sin (m x) & \text { rationalization of the denominator } \\ &=\operatorname{cis}(-m x) \\ &=\operatorname{cis}(n x) \end{align*}$

And thus, the formula proves true for all integral values of $n$ . $\blacksquare$

Generalization

Note that from the functional equation $f(x)^n = f(nx)$ where $f(x) = \cos x + i\sin x$ , we see that $f(x)$ behaves like an exponential function. Indeed, Euler's identity states that $e^{ix} = \cos x+i\sin x$ . This extends de Moivre's theorem to all $n\in \mathbb{R}$ .

@@ Line 1: / Line 1: @@
-'''DeMoivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{Z}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.
+'''De Moivre's Theorem''' is a very useful theorem in the mathematical fields of [[complex numbers]]. It allows complex numbers in [[polar form]] to be easily raised to certain powers. It states that for <math>x\in\mathbb{R}</math> and <math>n\in\mathbb{Z}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.
 == Proof ==
-This is one proof of De Moivre's theorem by [[induction]].
+This is one proof of de Moivre's theorem by [[induction]].
-*If <math>n>0</math>, for <math>n=1</math>, the case is obviously true.
+*If <math>n\ge0</math>:
-:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:
+:If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin(0x)=1+0i=1=z^0.</math>
-\begin{align*}
+:Assume the formula is true for <math>n=k</math>. Now, consider <math>n=k+1</math>:
+<cmath>\begin{align*}
 (\cos x+i \sin x)^{k+1} & =(\cos x+i \sin x)^{k}(\cos x+i \sin x) & \text { by Exponential laws } \\
-& =[\cos (k x)+i \sin (k x)](\cos x+i \sin x) & \text { by the Assumption in Step II } \\
+& =[\cos (k x)+i \sin (k x)](\cos x+i \sin x) & \text { by our assumption } \\
-& =\cos (k x) \cos x-\sin (k x)+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\
+& =\cos (k x) \cos x-\sin (k x) \sin x+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\
-& =\operatorname{cis}(k+1) & \text { Various Trigonometric Identities }
+& =\operatorname{cis}((k+1)(x)) & \text { by various trigonometric identities }
-\end{align}
+\end{align*}</cmath>
-:Therefore, the result is true for all positive integers <math>n</math>.
+:Therefore, the result is true for all nonnegative integers <math>n</math>.
-*If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin (0x)=1+i0=1</math>. Since <math>z^0=1</math>, the equation holds true.
 *If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.
-:[[Image:DeMoivreInductionP2.gif]]
+<cmath>\begin{align*}
+(\operatorname{cis} x)^{n} &=(\operatorname{cis} x)^{-m}  \\
-And thus, the formula proves true for all integral values of <math>n</math>. <math>\Box</math>
+&=\frac{1}{(\operatorname{cis} x)^{m}}  \\
+&=\frac{1}{\operatorname{cis}(m x)}  \\
+&=\cos (m x)-i \sin (m x) & \text { rationalization of the denominator } \\
+&=\operatorname{cis}(-m x)  \\
+&=\operatorname{cis}(n x)
+\end{align*}</cmath>
-Note that from the functional equation <math>f(x)^n = f(nx)</math> where <math>f(x) = \cos x + i\sin x</math>, we see that <math>f(x)</math> behaves like an exponential function. Indeed, [[Euler's identity]] states that <math>e^{ix} = \cos x+i\sin x</math>. This extends De Moivre's theorem to all <math>n\in \mathbb{R}</math>.
+And thus, the formula proves true for all integral values of <math>n</math>. <math>\blacksquare</math>
 ==Generalization==
+Note that from the functional equation <math>f(x)^n = f(nx)</math> where <math>f(x) = \cos x + i\sin x</math>, we see that <math>f(x)</math> behaves like an exponential function. Indeed, [[Euler's identity]] states that <math>e^{ix} = \cos x+i\sin x</math>. This extends de Moivre's theorem to all <math>n\in \mathbb{R}</math>.
+==See Also==
 [[Category:Theorems]]
 [[Category:Complex numbers]]

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Difference between revisions of "De Moivre's Theorem"

Latest revision as of 09:49, 31 August 2024

Proof

Generalization

See Also