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It is well known that $e^i$ is transcdental number and $| e^i|=1 $ , but if

i will try to write $e^i$ by other way :

Short Proof:

$e^i =[(e^i)^{2\pi)}]^\frac{1}{2\pi}=[{{e}^{2i\pi}}]^{\frac{1}{2\pi}}={1}^{\frac{1}{2\pi}}=1$ .

My question here : Why $e^i =1$ is not true by using the above proof ?

Note: $i$ is the unit imaginary part

Thank you for any help

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closed as off-topic by Jack D'Aurizio, Did, Rob Arthan, Cameron Williams, Daniel W. Farlow Oct 21 '16 at 2:43

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  • $\begingroup$ No $e^{i\pi}=-1$ not $1$ as you did. $\endgroup$ – hamam_Abdallah Oct 20 '16 at 22:24
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    $\begingroup$ Complex exponents don't really comply as expected to real exponents' laws. Sometimes they do, with certain conditions, in general they don't. $\endgroup$ – DonAntonio Oct 20 '16 at 22:26
  • $\begingroup$ sorry it's a wrong typo i meant $ 2\pi$ , now it fixed $\endgroup$ – zeraoulia rafik Oct 20 '16 at 22:26
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    $\begingroup$ $e^i=\cos(1)+i\sin(1)$ $\endgroup$ – Aweygan Oct 20 '16 at 22:27
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    $\begingroup$ $x = (x^a)^b$ only holds for positive real $x$ and real $a,b$. Otherwise, you could also prove that $-1 = \big((-1)^2\big)^{1/2} = 1^{1/2} = 1$. $\endgroup$ – Rahul Oct 20 '16 at 22:30
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Same proof method shows that $e = 1$:

$e = (e^{2i\pi})^\frac{1}{2i\pi} = 1^\frac{1}{2i\pi} = 1$

This should suggest it to you that there is something wrong with the proof method.

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    $\begingroup$ Very nice explanation. +1 With some minor changes we can prove using this that any number equals one... $\endgroup$ – DonAntonio Oct 20 '16 at 22:29
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    $\begingroup$ True, but it doesn't answer why either proof fails. $\endgroup$ – fleablood Oct 20 '16 at 22:37
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    $\begingroup$ @fleablood Perhaps, yet the idea conveyed here will give to any curious, intelligent student an urgent need to search for more info. That, imo, is what makes this answer a very good one...not to mention that a mathematical proof would probably need to go into complex numbers and even into the complex logarithm and complex exponents, branches and stuff like that. Too advanced, probably. $\endgroup$ – DonAntonio Oct 20 '16 at 22:40
  • $\begingroup$ @xyzzyz , thanks for the answer , but i asked why is not true and i never said it's true !!!! $\endgroup$ – zeraoulia rafik Oct 20 '16 at 22:46
  • $\begingroup$ You should use my answer to try to pinpoint the exact spot where the proof goes wrong. $\endgroup$ – xyzzyz Oct 20 '16 at 22:51
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What your argument shows, is that it is not true for complex numbers in general that $$\tag{*}(z^a)^b=z^{ab}.$$

For another example, $$ 1=(-1)^2=(i^{1/2})^2=i. $$ The law in $(*)$ holds for real numbers because of the way powers are defined. But when you try to define arbitrary powers of complex numbers (you have to define them, somehow, to use them), that property $(*)$ does not hold in general (it will still holds when $z,a,b$ are real).

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$(b^n)^m = (b^n)^m = b^{nm}$ doesn't always hold.

In particular it won't hold if $m$ or $n$ contain pertainent information that gets "cancelled out".

They hold for integer $n,m$ by basic definition ($(b^3)^5 = (bbb)(bbb)(bbb)(bbb)(bbb)(bbb) = (bbbbbbbbbbbbbbb) = b^{15}$) but they break down for more precise definitions.

If $c > 0$ then there exists one and only one distinct positive real number, $c$, where $c^n = b$ so it "makes sense" to define $b^{1/n} = c$ (that way we can have $(b^{1/n})^n = c^n = b = b^1 = b^{\frac 1n n}$). A little handwaving but defining $b^{m/n} = \sqrt[n]{b}^m$ and $(b^r)^s=b^{rs}$ clearly holds for $b > 0$ and $r,s \in \mathbb Q$.

But note we immediately have problems with $b < 0$. There is no distinct $c^{2n} = b$ so we can't define $b^{1/2n}$ and as $(-b)^{2n}= b^{2n}$ we get ambiguities. So for odd$m,n$, $((-b)^{2m})^{n/2}=(b^{2m})^{n/2} = b^{mn}\ne (-b)^{mn}$.

So whats going on? Well, I view it as $(-b)^2$ has important information ($-b$ is negative) that is lost when rising to an even power. so although $((-b)^n)^m) = (-b)^{nm}$ for odd $n,m$ if $n$ is even and its "evenness" is "cancelled out" out if $m = k/2$ we lose $-b$ being negative.

It's similar for $e^{a+bi} = e^a(\cos a + i \sin b)$. A key component is that we are raising to a non-real complex power. If we raise to an inverse power and "flatten it out" $(e^{a+bi})^{1/(a+bi)}= (e^a(\cos a + i \sin b))^{1/(a+bi)}$ will lose all complex information if we cancel out first. $e^{(a+bi)*\frac{1}{a+bi}} = e^1$ simply isn't compatible with the definition of complex powers.

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