De Moivre's Theorem: Difference between revisions
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*If <math>n\ge0</math>: | *If <math>n\ge0</math>: | ||
:If <math>n=0</math>, the formula holds true because <math>\cos(0x)+i\sin(0x)=1+ | :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>: | :Assume the formula is true for <math>n=k</math>. Now, consider <math>n=k+1</math>: | ||
Revision as of 09:23, 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 
and 
, 
.
Proof
This is one proof of de Moivre's theorem by induction.
- If
:
:- If
, the formula holds true because 
, the formula holds true because 
- Assume the formula is true for
. Now, consider 
:
. Now, consider 
: & \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*}](http://latex.artofproblemsolving.com/d/e/6/de60109242847b17882883d070585f1a3d59ab96.png)
- Therefore, the result is true for all nonnegative integers
.
- If
, one must consider 
when 
is a positive integer.
, one must consider 
when 
is a positive integer.
And thus, the formula proves true for all integral values of 
. 
Generalization
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 
.
