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

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 am very glad that you find this problem interesting.

I'll be waiting for the video...

Best regards

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.

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).

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

I was unable to find an error but the result does not make sense. I was not suggesting he was wrong.