# Is ei=1e^i =1 a wrong belief in mathematics or it is true as shown below? [on hold]

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

i will try to write eie^i by other way :

Short Proof:

ei=[(ei)2π)]12π=[e2iπ]12π=112π=1e^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 ei=1e^i =1 is not true by using the above proof ?

Note: ii is the unit imaginary part

Thank you for any help

=================

=================

3

=================

Same proof method shows that e=1e = 1:

e=(e2iπ)12iπ=112iπ=1e = (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.

1

Very nice explanation. +1 With some minor changes we can prove using this that any number equals one…
– DonAntonio
Oct 20 at 22:29

1

True, but it doesn’t answer why either proof fails.
– fleablood
Oct 20 at 22:37

2

@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.
– DonAntonio
Oct 20 at 22:40

@xyzzyz , thanks for the answer , but i asked why is not true and i never said it’s true !!!!
– user51189
Oct 20 at 22:46

You should use my answer to try to pinpoint the exact spot where the proof goes wrong.
– xyzzyz
Oct 20 at 22:51

What your argument shows, is that it is not true for complex numbers in general that (za)b=zab.\tag{*}(z^a)^b=z^{ab}.

For another example,
1=(−1)2=(i1/2)2=i.
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,bz,a,b are real).

(bn)m=(bn)m=bnm(b^n)^m = (b^n)^m = b^{nm} doesn’t always hold.

In particular it won’t hold if mm or nn contain pertainent information that gets “cancelled out”.

They hold for integer n,mn,m by basic definition ((b3)5=(bbb)(bbb)(bbb)(bbb)(bbb)(bbb)=(bbbbbbbbbbbbbbb)=b15(b^3)^5 = (bbb)(bbb)(bbb)(bbb)(bbb)(bbb) = (bbbbbbbbbbbbbbb) = b^{15}) but they break down for more precise definitions.

If c>0c > 0 then there exists one and only one distinct positive real number, cc, where cn=bc^n = b so it “makes sense” to define b1/n=cb^{1/n} = c (that way we can have (b1/n)n=cn=b=b1=b1nn(b^{1/n})^n = c^n = b = b^1 = b^{\frac 1n n}). A little handwaving but defining bm/n=n√bmb^{m/n} = \sqrt[n]{b}^m and (br)s=brs(b^r)^s=b^{rs} clearly holds for b>0b > 0 and r,s∈Qr,s \in \mathbb Q.

But note we immediately have problems with b<0b < 0. There is no distinct c2n=bc^{2n} = b so we can't define b1/2nb^{1/2n} and as (−b)2n=b2n(-b)^{2n}= b^{2n} we get ambiguities. So for oddm,nm,n, ((−b)2m)n/2=(b2m)n/2=bmn≠(−b)mn((-b)^{2m})^{n/2}=(b^{2m})^{n/2} = b^{mn}\ne (-b)^{mn}. So whats going on? Well, I view it as (−b)2(-b)^2 has important information (−b-b is negative) that is lost when rising to an even power. so although ((−b)n)m)=(−b)nm((-b)^n)^m) = (-b)^{nm} for odd n,mn,m if nn is even and its "evenness" is "cancelled out" out if m=k/2m = k/2 we lose −b-b being negative. It's similar for ea+bi=ea(cosa+isinb)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" (ea+bi)1/(a+bi)=(ea(cosa+isinb))1/(a+bi)(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)∗1a+bi=e1e^{(a+bi)*\frac{1}{a+bi}} = e^1 simply isn't compatible with the definition of complex powers.