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Difference between revisions of "De Moivre's Theorem"

m (→‎Proof)
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: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>:
  
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:[[Image:DeMoivreInductionP1.gif]]
+
\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 <math>n</math>.
 
:Therefore, the result is true for all positive integers <math>n</math>.

Revision as of 02:03, 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.
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 identity states that $e^{ix} = \cos x+i\sin x$. This extends De Moivre's theorem to all $n\in \mathbb{R}$.

Generalization

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