Difference between revisions of "De Moivre's Theorem"

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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 formula]] 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>.
 
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 formula]] 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>.
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==Generalization==
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[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Complex numbers]]
 
[[Category:Complex numbers]]

Revision as of 22:28, 27 March 2017

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 $n=k+1$:
DeMoivreInductionP1.gif
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.
DeMoivreInductionP2.gif

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 formula states that $e^{ix} = \cos x+i\sin x$. This extends De Moivre's theorem to all $n\in \mathbb{R}$.

Generalization