Difference between revisions of "De Moivre's Theorem"

Latest revision as of 10:49, 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 4: / Line 4: @@
 This is one proof of de Moivre's theorem by [[induction]].
-*If <math>n>=0</math>:
+*If <math>n\ge0</math>:
-:If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin(0x)=1+i0=1=z^0.</math>
+:If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin(0x)=1+0i=1=z^0.</math>
 :Assume the formula is true for <math>n=k</math>. Now, consider <math>n=k+1</math>:
@@ Line 12: / Line 12: @@
 <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) \sin x+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\
-& =\operatorname{cis}((k+1)(x)) & \text { Various Trigonometric Identities }
+& =\operatorname{cis}((k+1)(x)) & \text { by various trigonometric identities }
 \end{align*}</cmath>
@@ Line 30: / Line 30: @@
 \end{align*}</cmath>
-And thus, the formula proves true for all integral values of <math>n</math>. <math>\Box</math>
+And thus, the formula proves true for all integral values of <math>n</math>. <math>\blacksquare</math>
 ==Generalization==

Page

Toolbox

Search

Difference between revisions of "De Moivre's Theorem"

Latest revision as of 10:49, 31 August 2024

Proof

Generalization

See Also