Exponentiation rules

Question

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?

Answer 1

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 $z_1$.

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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 $1^{x\xi}=1$, $\forall (x,\xi)\in \mathbb{R^2}$

Answer 2

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.

Answer 3

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 $90=270$?

Answer 4

"$e^{2k\pi i}=1=e^0$ 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 $z = w \pm 2k\pi i$.