State and prove demoivers theorem for integral index 

Statement: -For any real number
and any integer n 

${\left(\cos \theta +i\sin \theta \right)}^n=\cos n\theta +i\sin n\theta$

 

Proof: Let n be a positive integer we prove that the theorem using the principle of mathmatical induction on N 

  $$\begin{gathered} Let S\left(n\right)={\left(\cos \theta +i\sin \theta \right)}^n=\cos n\theta +i\sin n\theta \\ case-1: If n=1 \\ L.H.S={\left(\cos \theta +i\sin \theta \right)}^1=\cos \theta +i\sin \theta \\ R.H.S=\cos \left(1\right)\theta +i\sin \left(1\theta \right) \\ \cos n\theta +i\sin n\theta \\ L.H.S=R.H.S \therefore S\left(1\right)is true \\ Assume that S\left(k\right) is true. \end{gathered}$$

$$\begin{gathered} {\left(\cos \theta +i \sin \theta \right)}^k=\cos k\theta -i\sin k\theta \to \left(1\right) \\ Case-ii: To prove that S(K+1) is true \\ S(K+1)=L.H.S\equiv {\left(\cos \theta +i\sin \theta \right)}^{K+1} \\ from\left(1\right) \\ \Rightarrow \left(\cos K\theta +i\sin K\theta \right)\left(\cos \theta +i\sin \theta \right) \\ \Rightarrow \cos \left(K\theta \right)\cos \theta +i\cos \left(K\theta \right)\sin \theta +i\sin K\theta \cos \theta +i^2\sin K\theta \sin \theta \\ \Rightarrow \left(\cos K\theta \cos \theta -\sin K\theta \sin \theta \right)+i\left(\cos k\theta \sin \theta +\sin k\theta .\cos \theta \right) \\ \left(\therefore i^2=-1\right) \\ =\cos \left(k\theta +\theta \right)+\sin \left(k\theta +\theta \right) \\ =\cos \left(k+1\right)\theta +i\sin \left(k+1\right)\theta =R.H.S \\ \therefore {\left(\cos \theta +i\sin \theta \right)}^{k+1} \\ \cos \left(k+1\right)\theta +i\sin \left(k+1\right)\theta \\ \therefore S\left(k+1\right)is true By the priniciple of Mathematical Induction S\left(n\right) is true \\ \forall n\in N \end{gathered}$$