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 $\bf q\ne 1$.