I came across this equation,
$$
e^{ix} = \cos(x) + i\sin(x)
$$
This is the simplified version, the real one is more complex but this part is the one I have a question about.
The right side clearly has a $i$ term in it, and is thus imaginary, but the left, $e^{x}$ is entirely real. How is this feasible? I couldn't find many good sources on the derivation of this equation, so maybe I am missing something?
NOTE: I came across the equation from a physics source, so I am not extremely adept with the mathematics behind this.
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user330848
- 11
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1Did you notice the $i$ in the exponent? Also: if $\sin(x) = 0$, then the imaginary part on the right vanishes. – StackTD Apr 12 '16 at 15:05
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2It is $e^{ix}$ not $e^{x}$ – GoodDeeds Apr 12 '16 at 15:05
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To the OP. Here is a nice thought: Can you figure out the value of $i^i$? It might surprise you, but that answer is also real... – imranfat Apr 12 '16 at 15:31
2 Answers
1
This identity can be derived using the Taylor Series expansions for $e^{ix}$, $\sin(x)$, and $\cos(x)$:
$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$ $$i\sin(x)=i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots)$$ $$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots=1+ix-\frac{x^2}{2!}+\frac{ix^3}{3!}+\frac{x^4}{4!}-\cdots$$
Adding the series for $\cos(x)$ and $i\sin(x)$ together, we obtain the the series for $e^{ix}$.
Ethan Hunt
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$e^y$ is a function whose range (output) are positive real numbers when the real numbers is its domain. However when we say $e^{ix}$ we are using the function $e^y$ on the complex number $ix$, thus we suddenly have a bit of a different function, acting on complex numbers, and thus we do not necessarily need to have a real number as output.
Ove Ahlman
- 4,329