De Moivre’s Theorem

De Moivre’s Theorem #

  Statement:

1. \( \bf \left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta \) when \(\bf n\in \Z \)
2. \(\bf \cos n\theta+i\sin n\theta \) is one of the value of \( \bf \left(\cos\theta+i\sin\theta\right)^{n}\) when \(\bf n=\frac{p}{q} \) where \(\bf q\ne 1 \)

Proof:\( \\ \)

 Case-1: Let \(\bf n=0 \)  Then the \begin{align*} \bf L.H.S=\left(\cos\theta+i\sin\theta\right)^{0}=1 \end{align*}  And the \begin{align*} \bf R.H.S=\cos 0.\theta+i\sin 0.\theta =1 \end{align*}  Therefore L.H.S=R.H.S \( \\ \)  Case-2: Let \(\bf n \) is a positive integer.\( \\ \)  We prove it by the principal of mathematical induction .\( \\ \)   For \(\bf n=1\)  Then the \begin{align*} \bf L.H.S & \bf=\left(\cos\theta+i\sin\theta\right)^{1} \\ &\bf=\cos\theta+i\sin\theta \end{align*}  And the \begin{align*} \bf R.H.S=\cos \theta+i\sin \theta \end{align*}  Therefore L.H.S=R.H.S \( \\ \)  Let the statement is true for \(\bf n=k\), the we have $$ \bf \left(\cos\theta+i\sin\theta\right)^{k}=\cos k\theta+i\sin k\theta $$  Now \begin{align*} & \bf \left(\cos\theta+i\sin\theta\right)^{k+1} \\ &\bf=\left(\cos\theta+i\sin\theta\right)^{k}.\left(\cos\theta+i\sin\theta\right)\\ &\bf=\left(\cos k\theta+i\sin k\theta\right).\left(\cos\theta+i\sin\theta\right)\\ &\bf=\left(\cos k\theta\cos\theta- \sin k\theta\sin\theta \right)\\ &\bf +i\left(\cos k\theta\sin\theta+ \sin k\theta\cos\theta \right)\\ &\bf=\cos (k+1)\theta+i\sin (k+1)\theta\\ \end{align*}  Therefore the statement is true for \(\bf n=k+1\) when the statement is true for \(\bf n=k\). Then by the principal of mathematical induction the statement is true for \(\bf n\in \N \). \( \\ \)  Hence \( \bf \left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta \) when \(\bf n \) is a positive integer.\( \\ \)  Case-3: Let \(\bf n \) is a negative integer.\( \\ \)  Let \(\bf m=-n \), then \(\bf m \) is positive ineteger.\( \\ \)  Then by Case-2, we have $$ \bf \left(\cos\theta+i\sin\theta\right)^{m}=\cos m\theta+i\sin m\theta $$  Therefore \begin{align*} & \bf \left(\cos\theta+i\sin\theta\right)^{n} \\ & \bf =\left(\cos\theta+i\sin\theta\right)^{-m} \\ & \bf = \frac{1}{\left(\cos\theta+i\sin\theta\right)^{m}}\\ & \bf = \frac{1}{\left(\cos m\theta+i\sin m\theta\right)}\\ & \bf =\cos m\theta-i\sin m\theta \\ & \bf =\cos (-m)\theta+i\sin (-m)\theta\\ & \bf =\cos n\theta+i\sin n\theta \end{align*}  Hence \( \bf \left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta \) when \(\bf n \) is a negative integer.\( \\ \)  Case-4: Let \(\bf n=\frac{p}{q} \) where and \(\bf q\ne 1 \) . \begin{align*} & \bf \left(\cos \frac{p\theta}{q}+i\sin\frac{p\theta}{q}\right)^{q}\\ & \bf =\left(\cos q.\frac{p\theta}{q}+i\sin q.\frac{p\theta}{q}\right)~by Case-2\\ & \bf =\left(\cos p\theta+i\sin p\theta\right)\\ & \bf =\left(\cos\theta+i\sin\theta\right)^{p}~by~Case-2~or~3\\ \end{align*}  Therefore \begin{align*} & \bf \left(\cos \frac{p\theta}{q}+i\sin\frac{p\theta}{q}\right)^{q}=\left(\cos\theta+i\sin\theta\right)^{p}\\ \end{align*} Implies \(\bf \left(\cos \frac{p\theta}{q}+i\sin\frac{p\theta}{q}\right)\) is one of the value of \(\bf \left(\cos\theta+i\sin\theta\right)^{\frac{p}{q}}\)\( \\ \) That is \( \bf \left(\cos\theta+i\sin\theta\right)^{n}=\cos n\theta+i\sin n\theta \) when \(\bf n=\frac{p}{q} \) and \(\bf q\ne 1 \).