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$$ \begin{aligned} e^{i 2\pi f(x)} &= (e^{i\ 2\pi})^{f(x)} \\ e^{i 2\pi f(x)} &= (\cos {2\pi} + i\ \sin {2\pi})^{f(x)} \\ e^{i 2\pi f(x)} &= (1 + i 0)^{f(x)} \\ e^{i 2\pi f(x)} &= 1^{f(x)} \\ e^{i 2\pi f(x)} &= 1 \end{aligned} $$

True or False?

pico
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    The right of the first equation is not a function. It is multi-valued, unless you are able to choose a branch. The left is a function. – user647486 Apr 23 '19 at 15:22
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    Exponentials with complex numbers don't work like exponentials with real numbers. – saulspatz Apr 23 '19 at 15:22
  • False. $(e^a)^b\neq e^{ab},$ in general, when $a,b$ are complex. – Thomas Andrews Apr 23 '19 at 15:33
  • It's similar to the problem with $$-1=(-1)^1=\left((-1)^2\right)^{1/2}=1^{1/2}=1.$$ – Thomas Andrews Apr 23 '19 at 15:35
  • DSP books always play this trick: (See Understanding DSP, Lyons, 3rd Edition, appendix B)

    Geometric Series Identity: $$ \sum^{N-1}_{n=0} r^{n} = \frac{1-r^N}{1-r} $$ Thus apply complex exponential to above rule:

    $$ \sum^{N-1}{n=0} e^{-j2\pi n m / N} = \sum^{N-1}{n=0} (e^{-j2\pi m / N})^n = \frac{1-e^{-j2\pi m}}{1-e^{j2 \pi m / N}} $$

    Why is it valid in this case, when "a" is a complex number?

     – pico Apr 23 '19 at 16:04
  • wait a second: $e^{i 2 \pi}$ is a real number because it equals "1"? and I didn't specify that f(x) needed to be complex... so I would assume that f(x) is a real number function that doesn't equal $1 / 2 \pi$ … hmm... – pico Apr 23 '19 at 16:21
  • @pico When $n$ is an integer, $(e^{a})^n=e^{an}.$ for all complex $a.$ This is because $a^b$ is, for complex $a,b,$ best seen as a multi-valued function which one value when $b$ is an integer, finitely many values when $b$ is a rational, and infinitely many values when $b$ isan irrational complex number. Basically, $(a^b)^c$ and $a^{bc}$ share a value, but it is not always the principle value. The principle value if $1^x$ is $1,$ but there are many other values $e^{2\pi i m x}.$ for any integer $m.$ But when $x$ is an integer, all these values are $=1.$ – Thomas Andrews Apr 23 '19 at 19:06

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Long story short :

The property $(e^b)^c = e^{bc}$ holds for real numbers and in case of complex numbers you get such errors. (This is also the statement for any such exponent property). Wikipedia has an article-brunch analysis such false cases.

Also a side note, this question is broad. Even if it worked, without knowing what $f(x)$ is you can't say much. For example, if you allowed $f(x)$ to be equal to $\infty$ for some $x$ of its domain (Lebesgue Measure Theory cases), then $1^\infty$ is undefined.

Rebellos
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