E^pi*i ?

  • #1
fig 1

Given:
5LgiS_EU09s8UP911Ko28KLQVMC-RVkNA8izH6Cyba7MX48oMODerFJoh7hyvl07rRE5lCGrbEqVOLOHWUGlyDzOYr7ZuhCT.png


e= cosΘ + i sinΘ (radians)

eπi=-1

Deduced

e2πi=(-1)2

e2πi=1

e(2/3)πi=11/3

e(2/3)iπ=1

e(2/3)iπ=cos(2i/3)+i sin(2i/3)

e(2/3)iπ=-1/2+i(3/2)

-1/2+i(31/2/2)=1

where n is greater than or equal to 1 or n=a/b where a is greater than or equal to 1 and b is odd
726a0xR7EIDHt-17t443594r2qWP8KCc98Da6ood-H5sIaZPKhu4-NSENIPLdcu4NPgQaZDCxbdChyb59mEmk3GAlWSpgBYz.png


1n=1



(-1/2+i(3/2))n=1


However this is absurd and I
have no idea what's wrong
(sorry about the formating it was better in docs)
 

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    5LgiS_EU09s8UP911Ko28KLQVMC-RVkNA8izH6Cyba7MX48oMODerFJoh7hyvl07rRE5lCGrbEqVOLOHWUGlyDzOYr7ZuhCT.png
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Last edited:

Answers and Replies

  • #2
.Scott
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As soon as you put a denominator into the exponent, you run into trouble.
For example, ##1^1=1##. But does ##1^{1/2}=1##? It could equal -1.

Also, Look at your third "deduced" line.
Where did it come from? ##e^{(2/3)πi}=31## ?

In any case, ##e^{(2/3)πi)} = -1/2 + (\sqrt{3}/2)i##
 
  • #3
PeroK
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fig 1

Given:View attachment 238257

e= cosΘ + i sinΘ (radians)

eπi=-1

Deduced

e2πi=(-1)2

e2πi=1

e(2/3)πi=31

e(2/3)iπ=1

e(2/3)iπ=cos(2i/3)+i sin(2i/3)

e(2/3)iπ=-1/2+i(3/2)

-1/2+i(31/2/2)=1

where n is greater than or equal to 1 or n=a/b where a is greater than or equal to 1 and b is oddView attachment 238258

1n=1



(-1/2+i(3/2))n=1


However this is absurd and I
have no idea what's wrong
(sorry about the formating it was better in docs)

What you have really shown is that:

##(e^{2\pi i/3})^3 = 1##

In other words:

##(\cos(2\pi /3) + i \sin(2\pi /3))^3 = 1##

And, in fact, there are three complex numbers ##z##, where ##z^3 = 1##.

But, this doesn't mean that in this case ##z = 1^{1/3} = 1##.

In the same way that ##-1 \ne \sqrt{1} = 1##, although we do have ##(-1)^2 = 1##

And, in general, we have:

##(z^n)^{1/n} \ne z##

Where ##\ne## here means is not necessarily equal to.
 
  • #5
As soon as you put a denominator into the exponent, you run into trouble.
For example, ##1^1=1##. But does ##1^{1/2}=1##? It could equal -1.

Also, Look at your third "deduced" line.
Where did it come from? ##e^{(2/3)πi}=31## ?

In any case, ##e^{(2/3)πi)} = -1/2 + (\sqrt{3}/2)i##
sorry that was meant to be one root three
 
  • #7
might this be the same reason that extraneous solutions exist?
 
  • #8
PeroK
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might this be the same reason that extraneous solutions exist?

Extraneous solutions to what?
 

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