De Moivre's Theorem: Difference between revisions

Revision as of 02:06, 6 February 2022

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 $x\in\mathbb{R}$ 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 $n>0$ , for $n=1$ , the case is obviously true.

Assume true for the case $n=k$ . Now, the case of $n=k+1$ :

$\begin{align*} (\cos x+i \sin x)^{k+1} & =(\cos x+i \sin x)^{k}(\cos x+i \sin x) & \text { by Exponential laws } \\ & =[\cos (k x)+i \sin (k x)](\cos x+i \sin x) & \text { by the Assumption in Step II } \\ & =\cos (k x) \cos x-\sin (k x)+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\ & =\operatorname{cis}(k+1) & \text { Various Trigonometric Identities } \end{align*}$

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

If $n=0$ , the formula holds true because $\cos(0x)+i\sin (0x)=1+i0=1$ . Since $z^0=1$ , the equation holds true.

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 $n$ . $\Box$

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 $n\in \mathbb{R}$ .

Generalization

@@ Line 22: / Line 22: @@
 <cmath>\begin{align*}
-(\operatorname{cis} x)^{n} &=(\operatorname{cis} x)^{-m} \\
+(\operatorname{cis} x)^{n} &=(\operatorname{cis} x)^{-m} & \\
-&=\frac{1}{(\operatorname{cis} x)^{m}} \\
+&=\frac{1}{(\operatorname{cis} x)^{m}} & \\
-&=\frac{1}{\operatorname{cis}(m x)} \\
+&=\frac{1}{\operatorname{cis}(m x)} & \\
-&=\cos (m x)-i \sin (m x) \quad \text { rationalization of the denominator } \\
+&=\cos (m x)-i \sin (m x) & \text { rationalization of the denominator } \\
-&=\operatorname{cis}(-m x) \\
+&=\operatorname{cis}(-m x) & \\
-&=\operatorname{cis}(n x)
+&=\operatorname{cis}(n x) &
 \end{align*}</cmath>

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De Moivre's Theorem: Difference between revisions

Revision as of 02:06, 6 February 2022

Proof

Generalization