How to prove De Moivre's theorem inductively

Question

It is given that $(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$ where $n\in Z^+$.

I can show it works for $n=1$ but I am stuck in showing it inductively. I have got as far as below but are stuck in the rearranging:

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

How can I rearrange the left to look like the right?

Answer 1

$$(\cos \theta + i\sin \theta)^k = \cos k\theta + i\sin k\theta$$ $$\text{So for } (\cos \theta + i\sin \theta)^{k+1} = (\cos \theta + i\sin \theta)^k \cdot (\cos \theta + i\sin \theta) = (\cos k\theta + i\sin k\theta)\cdot (\cos \theta + i\sin \theta)$$ $$=\cos k\theta\cos\theta +i\sin\theta\cos k\theta +i\sin k\theta\cos\theta +i^2\sin{k\theta}\sin\theta$$ $$=(\cos{k\theta}\cos\theta - \sin k\theta\sin\theta)+i(\sin\theta\cos k\theta +\sin k\theta\cos\theta)$$ $$=\cos(k+1)\theta + i\sin(k+1)\theta$$

Hence, this proves that $(\cos\theta +i\sin\theta)^n = \cos n\theta + i\sin n\theta$.

Answer 2

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

Answer 3

$$(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)=\\ \cos k\theta\cos\theta-\sin k\theta\sin\theta+i(\cos k\theta\sin\theta+\sin k\theta\cos\theta)\\ =\cos(k+1)\theta+i\sin(k+1)\theta $$

Answer 4

By induction,

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

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

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

For $k=n \in \mathbb{N}$

$$(\cos \theta + i\sin \theta )^n = \cos n\theta + i\sin n\theta$$

We can also prove this by simple exponentiation rule and Euler's form of complex numbers.

$z = \cos \theta + i\sin \theta = e^{i\theta}$

As, $e^{ix} = \cos x+ i\sin x\cdots(i)$

$z^n = (e^{i\theta})^n$

As, $(e^a)^b= e^{ab}$

$z^n = (e^{i\theta})^n =e^{in\theta}$

From (i), we can write,

$$z^n = (\cos \theta + i\sin \theta )^n = \cos n\theta + i\sin n\theta$$

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