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De Moivre's Theorem: Difference between revisions

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'''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 ==
== 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>
 
:Assume the formula is true for <math>n=k</math>. Now, consider <math>n=k+1</math>:


<cmath>\begin{align*}
<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 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] & \\
& =\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*}</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.
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\end{align*}</cmath>
\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>
 
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==
==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:Theorems]]
[[Category:Complex numbers]]
[[Category:Complex numbers]]

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 $n\in \mathbb{R}$.

See Also

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