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
m
Line 6: Line 6:
 
*If <math>n>0</math>:
 
*If <math>n>0</math>:
  
−
:For <math>n=1</math>, the proposition is obviously true.
+
:If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin(0x)=1+i0=1=z^0.</math>
  
−
:Assume true for the case <math>n=k</math>. Now, for <math>n=k+1</math>:
+
:Assume the formula is true for <math>n=k</math>. Now, consider <math>n=k+1</math>:
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 17: Line 17:
 
\end{align*}</cmath>
 
\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.
 
*If <math>n<0</math>, one must consider <math>n=-m</math> when <math>m</math> is a positive integer.

Revision as of 10:20, 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 $n>0$:
If $n=0$, the formula holds true because $\cos(0x)+i\sin(0x)=1+i0=1=z^0.$
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 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 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 $n$. $\Box$

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 $n\in \mathbb{R}$.

See Also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=De_Moivre%27s_Theorem&oldid=226081"