Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "De Moivre's Theorem"
m (Undo revision 209937 by Yiyj1 (talk))
(Tag: Undo)
(26 intermediate revisions by 8 users not shown)
Line 1: Line 1:
−
{{WotWAnnounce|week=September 5- September 11}}
+
'''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>.
−
 
 
−
'''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{N}</math>, <math>\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)</math>.
 
  
 
== Proof ==
 
== Proof ==
Line 10: Line 8:
 
:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:
 
:Assume true for the case <math>n=k</math>. Now, the case of <math>n=k+1</math>:
  
−
:[[Image:DeMoivreInductionP1.gif]]
+
<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) \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 }
 +
\end{align*}</cmath>
  
 
:Therefore, the result is true for all positive integers <math>n</math>.
 
:Therefore, the result is true for all positive integers <math>n</math>.
Line 18: Line 21:
 
*If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.
 
*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}  \\
 +
&=\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>
  
 
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>\Box</math>
  
−
Note that from the functional equation <math>f(x) = 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, by [[Euler's formula]] (<math>e^{ix} = \cos x+i\sin x\right</math>). This extends De Moivre's theorem to all <math>n\in \mathbb{R}</math>.
+
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>.
 +
 
 +
==Generalization==
  
 +
==See Also==
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Complex numbers]]
 
[[Category:Complex numbers]]

Revision as of 00:42, 11 January 2024

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) \sin x+i[\cos (k x) \sin x+\sin (k x) \cos x] & \\ & =\operatorname{cis}((k+1)(x)) & \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

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&oldid=210459"