Is this simple equivalence on Euler's identity true?

Question

Is the following equivalence true?

$$\cos(x\cdot a)+i\cdot \sin(x\cdot a)=e^{ix\cdot a}=(e^{ix})^a=(\cos(x)+i\cdot \sin(x))^{a}$$

why?

Answer 1

Try $x = 2\pi$ and $a=\frac{1}{2}$. Then:

$$ \cos(x\cdot a)+i\cdot \sin(x\cdot a)= \cos\pi+i\cdot\sin\pi = -1 + i\cdot 0 = -1 $$ and $$ (\cos(x)+i\cdot \sin(x))^{a}=\big(\cos (2\pi) + i\sin(2\pi)\big)^{1/2} =1^{1/2} = 1 $$ Now it is true that $-1$ is a square root of $1$. But normally when we write $1^{1/2}$ we mean the principal square root of $1$, namely $1$.

Answer 2

Yes, what you have done is actually proven De Moivre's theorem, assuming Euler's identity is correct. De Moivre's theorem states $(\cos \theta+i\sin \theta)^n=\cos n\theta+i\sin n\theta$. So, if $e^{i\theta}=(\cos \theta+i\sin \theta)$ (call this Equation $1$) is known, then following your method, $e^{in\theta}=(\cos \theta+i\sin \theta)^n$, and applying Equation $1$ again with $\theta$, replaced by $n\theta$ (to the left-hand side) gives the theorem of De Moivre. At least when $n$ is a integer, we don't need to worry about what $(e^{i\theta})^n$ means.