we know these properties of exponentiating $$x^bx^a=x^{a+b}$$ $$(x^a)^b=x^{ab}$$ Now, I've been introduced to complex numbers and this amazing formula: $$e^{ix}=\cos{x}+i\sin{x}$$ Take $x=2k\pi,k\in Z$ to get $$e^{2k\pi i}=1=e^0$$ which would then imply that $\forall k\in Z:2k\pi i=0$ which kinda doesn't make sense to me. Another thing is... Take a fourier transform of the function given by (taken from https://en.wikipedia.org/wiki/Fourier_transform) $$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi ix\xi}dx$$ All this stuff involving $e^{2k\pi i}$ are kinda confusing to me. One could say the following: $$e^{-2\pi ix\xi}=(e^{-2\pi i})^{x\xi}=1^{x\xi}=1$$ which kinda doesn't make sense, does it? Why would one include the term $e^{-2\pi ix\xi}$ in a formula then. Could someone explain me, what is actually going on and when these basic properties of exponentiation are valid?
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The exponentiation "rules" that you referenced are not true in general. You've stumbled upon a counterexample or two. Note that $z_1^{z_2} \equiv e^{z_2 \log(z_1)}$ where $\log(z_1)=\log(|z_1|)+i\arg(z_1)$ and $\arg(z_1)$ is the multivalued argument of $z_1$. – Mark Viola Jan 11 '18 at 19:09
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"which would then imply that $\forall k\in Z:2k\pi i=0$ No, it wouldn't. I know I risk yet another moderator warning for being too blunt-spoken, but you shouldn't just learn enough mathematics to ask a question, it should be enough to understand the answer. – Jan 11 '18 at 19:14
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But if the op understood the answer, s/he wouldn't have ask the question. $a^x = a^y$ really does seem to imply $x = y$ and it's reasonable for the op to wonder what is going on. Is there some exception, or is the implication wrong? The answer in complex numbers exponents can be cyclic, logorithms are not unique and $a^x = a^y; x \ne y$ is possible. So the "rules" are ... false. – fleablood Jan 11 '18 at 19:22
4 Answers
"$e^{2k\pi i}=1=e^0$ which would imply $\forall k\in Z:2k\pi i=0$"
No more than $\sin (k + 2k\pi) = \sin k$ would imply $2k\pi = 0$ or that $(-x)^2 = x^2$ would imply $x = -x$.
Thing is: if $b \in \mathbb R$ and $b > 1$ (or $0 < c < 1$) and $x,y \in \mathbb R$ with $x < y$ then $b^x < b^y$ (or $c^x > c^n$). This implies that $f(x) = b^x$ (or $c^x$) is one to one and that $b^x = b^y \implies x = y$ and that there is a single-valued well defined function called $\log_b$ where $\log_b k = x \iff b^x = k$.
We already have exceptions. $1^x = 0$ and $\log_1 k$ is meaningless. And for $b < 0$ then $b^x$ is not defined for all values of $x$ and $\log_{b;b< 0} k$ is also meaningless. But we tacitly assume that the base is positive real and not equal to $1$ and that $b^x$ is one-to-one and $\log_b$ is a well defined single valued function.
$e^{ix} = \cos x = i\sin x$ allows us to define $b^z; z \in \mathbb C$ and everything is fine in that regards but as trig functions a periodic an not one to one the tacit assumptions are simply not at all true. $e^z$ is NOT one-to-one and $\ln $ is NOT single valued although as a multi-valued function it is well defined.
And that's the heart of it. If $e^z = e^w$ that does not imply $z =w$; it implies $z = w \pm 2k\pi i$.
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The exponentiation "rules" that you referenced are not true in general. You've stumbled upon a counterexample or two. Note that as a definition
$$z_1^{z_2}\equiv e^{z_2 \log(z_1)}$$
where the complex logarithm function, $\log(z_1)$, is multivalued. We can express the multivalued logarithm in terms of its real and imaginary parts as
$$\underbrace{\log(z_1)}_{\text{Complex Logarithm}}=\underbrace{\log(|z_1|)}_{\text{Natural Logarithm for Real Numbers}}+i\arg(z_1)$$
where $\arg(z_1)$ is the multivalued argument of $z_1$.
Note that $e^{-2\pi ix\xi}\ne 1$ unless $x \xi\in \mathbb{Z}$.
Also, note that $1^{x\xi}=e^{i2n\pi x\xi }$ is actually multivalued unless $x\xi\in \mathbb{Z}$. When working on $\mathbb{R}$, $n=0$ and hence $1^{x\xi}=1$, $\forall (x,\xi)\in \mathbb{R^2}$
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So $1^{x\xi}$ is always 1 for reals $x,\xi$ but whenever i take anything to a complex power, so if $x,\xi \in C$ then i would run into this trouble and i would have to say that it is periodic with period $2k\pi i$ Right? – Michal Dvořák Jan 11 '18 at 20:18
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If one restricts the range of $1^{x\xi}$ to the real numbers, then $1^{x\xi}=1$. If not, then $1^{x\xi}=e^{i2n\pi x\xi}\ne 1$ unless either $n=0$, $x\xi\in \mathbb{Z}$, or both. – Mark Viola Jan 11 '18 at 20:21
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So just the assumption that $e^{ab}=(e^{a})^b$ where $a,b\in C$ this generaly works only for real $a,b$, am i getting it correct? Because $1^{x\xi}=e^{x\xi\ln{1}}=e^{x\xi\ln{1}+2k\pi i}$... – Michal Dvořák Jan 11 '18 at 20:24
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It is true that for real analysis $(e^a)^b=e^{ab}$, but not generally true for complex $a$ and $b$. Also, it is incorrect to write $\log(1)=\log(1)=i2n\pi$, where the logarithm on the left-hand side is the complex logarithm while the logarithm on the right-hand side is the natural logarithm for the reals. But it seems that you have it. – Mark Viola Jan 11 '18 at 20:36
For two complex numbers $z$, for any real number $\alpha$, the exponent of a complex number $z^{\alpha}$ is multi-valued without fixing any branch.
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Yes, what I wrote is misleading, what I wanted to say is that, without fixing a branch, then $z^{\alpha}$ is multi-valued. – user284331 Jan 11 '18 at 19:07
You are correct that $$e^{2k\pi i}=1.$$ But why? It's because for $z\in\mathbb{C}$, $e^z$ is repetitive every $2\pi i$, meaning that for any integer $k$, $$e^{z+2k\pi i}=e^z$$ What happens when $z=0$?
Also let me ask you this question. We know that $$\sin90^\circ=\sin270^\circ$$ Does it mean that $90=270$?
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