Difference between revisions of "De Moivre's Theorem"
| Line 21: | 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. | ||
| − | \begin{align*} | + | %%\begin{align*} |
(\operatorname{cis} x)^{n} &=(\operatorname{cis} x)^{-m} \\ | (\operatorname{cis} x)^{n} &=(\operatorname{cis} x)^{-m} \\ | ||
&=\frac{1}{(\operatorname{cis} x)^{m}} \\ | &=\frac{1}{(\operatorname{cis} x)^{m}} \\ | ||
| Line 28: | Line 28: | ||
&=\operatorname{cis}(-m x) \\ | &=\operatorname{cis}(-m x) \\ | ||
&=\operatorname{cis}(n x) | &=\operatorname{cis}(n x) | ||
| − | \end{align*} | + | \end{align*}%% |
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> | ||
Revision as of 02:05, 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 
and 
, 
.
Proof
This is one proof of De Moivre's theorem by induction.
- If
, for 
, the case is obviously true.
, for 
, the case is obviously true.- Assume true for the case
. Now, the case of 
:
. Now, the case of 
: & \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*}](http://latex.artofproblemsolving.com/0/d/e/0de8d2aa3dd1f8e99bc3a590d031647560802b11.png)
- Therefore, the result is true for all positive integers
.
- If
, the formula holds true because 
. Since 
, the equation holds true.
, the formula holds true because 
. Since 
, the equation holds true.- If
, one must consider 
when 
is a positive integer.
, one must consider 
when 
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) \quad \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 
. 
Note that from the functional equation 
where 
, we see that 
behaves like an exponential function. Indeed, Euler's identity states that 
. This extends De Moivre's theorem to all 
.
