Difference between revisions of "De Moivre's Theorem"
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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>: | ||
− | <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 the Assumption in Step II } \ | ||
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*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. | ||
− | <cmath>\begin{align* | + | <cmath>\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}} & \ |
Revision as of 02:06, 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.
- Assume true for the case . Now, the case of :
- Therefore, the result is true for all positive integers .
- If , the formula holds true because . Since , the equation holds true.
- If , one must consider when is a positive integer.
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 .