We know that:

e ^(iπ) = -1

then we square both sides

e ^(2iπ) = 1

then we raise to the 1/2π both sides

e^i = 1^(1/2π) = 1 (A)

On the other hand, we know that:

e^i = e^(1*i) = cos(1)+i*sin(1) = 0.540302… + i* 0.8414709…. (B)

(A) contradicts (B)

.... or I have made a mistake somewhere in the process

The problem here is that in (A) you considered 1 as cos(0)+isin(0) while in (B) you considered 1 as cos(2pi)+isin(2pi).

OMG, OMG. I woke up this morning and it hit me like a brick on the head. By squaring both sides we are picking up extraneous roots.

x = -1

x^2 = 1

therefore x = +/-1

But x=1 in an extraneous root due to squaring both sides.

A rookie high school error - what an idiot I am for not seeing it.George: that's where the contradiction came in

Squaring [e^2pi] = [-1]^2 Giving e^i = [1]^(i/2pi)

and that's where the the extraneous roots show up.

cos(0) + isin(0), cos(2) + isin(2), cos(4) + isin(4), ......

[1]^(1/2pi] includes both the even and odd angles.

Let pi =pe^ip = -1

-1 in polar form has R=1 and t = p

so -1 = cos(p+2kp) + isin(p+2kp) where k is an integer

Therefore

e^i

= (-1)^(1/p) --->the pi'th roots of -1 - not the 2pi'th roots of 1

= [cos(p+2kp) isin(p+2kp)]^[1/p]

= [cos[p+2kp)/p + isin(p+2kp)/p] --->by DeMoivre

= cos(1+2k) + isin(1+2k)

e^i = cos(1+2k) + isin(1+2k) where k is an integerSo we get:

k=0) cos(1) +isin(1) --> Principal Value and the one George stated

k=1) cos(3)+isin(3)

k=2 cos(5)+isin(5)

etc, etc ....

k=-1) cos(1) - isin(1)

k=-2) cos(3) - isin(3)

etc. etc ...

The even angles are the extraneous roots !!

My sincere and humble apologies to

anbownag12who had the right intuition all along.And my apologies to George for making such a dumb-ass error. But at least we resolved the contradiction. Still a good video idea and he can even show my error.

BTW George T: I think we have an

excellentcase for a BPRP video. With your permission I will pass this over to the video request section and plead our case. Thanks for the problem.I was unable to find an error but the result does not make sense. I was not suggesting he was wrong.

@ianfowler if there are infinitely man answers correct, then George T has just proven e^i=1 which is clearly wrong!