Prove De Moivre’s theorem:[latex] (\cos \theta+i \sin \theta)^n=\cos n \theta+i \sin n \theta [/latex] where n is any positive integer.

We use the principle of mathematical induction. Assume that the result is true for the particular positive integer k, i.e., assume [latex] (\cos \theta+i \sin \theta)^k=\cos k \theta+i \sin k \theta [/latex]. Then, multiplying both sides by cosθ + isinθ, we find [latex] (\cos \theta+i \sin \theta)^{k+1}=(\cos k \theta+i \sin k \theta)(\cos \theta+i \sin \theta)=\cos (k+1) \theta+i \sin (k+1) \theta [/latex] by Problem 1.19 . Thus, if the result is true for n=k, then it is also true for n=k+1. But, since the result is clearly true for n=1, it must also be true for n=1+1=2 and n=2+1=3, etc., and so must be true for all positive integers. The result is equivalent to the statement [latex] \left(e^{i \theta}\right)^n=e^{n i \theta} [/latex]

Question 1.20: Prove De Moivre’s theorem: (cos θ + isinθ )^n = cos nθ + isi...

Schaum’s Outline of Complex Variables

Prove De Moivre’s theorem: $(\cos \theta+i \sin \theta)^n=\cos n \theta+i \sin n \theta$ where n is any positive integer.

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We use the principle of mathematical induction. Assume that the result is true for the particular positive integer k, i.e., assume $(\cos \theta+i \sin \theta)^k=\cos k \theta+i \sin k \theta$ . Then, multiplying both sides by cosθ + isinθ, we find

(\cos \theta+i \sin \theta)^{k+1}=(\cos k \theta+i \sin k \theta)(\cos \theta+i \sin \theta)=\cos (k+1) \theta+i \sin (k+1) \theta

by Problem 1.19. Thus, if the result is true for n=k, then it is also true for n=k+1. But, since the result is clearly true for n=1, it must also be true for n=1+1=2 and n=2+1=3, etc., and so must be true for all positive integers.

The result is equivalent to the statement $\left(e^{i \theta}\right)^n=e^{n i \theta}$