What makes Euler's Identity puzzling in complex number calculations?

[Quarky nerd]

fig 1

Given:

e^iΘ= cosΘ + i sinΘ (radians)

e^πi=-1

Deduced

e^2πi=(-1)²

e^2πi=1

e^(2/3)πi=1^1/3

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(3^1/2/2)=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

1ⁿ=1

∴

(-1/2+i(3/2))ⁿ=1However this is absurd and I
have no idea what's wrong
(sorry about the formating it was better in docs)

 

[.Scott]

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, $e^{(2/3)πi)} = -1/2 + (\sqrt{3}/2)i$

 

[PeroK]

fig 1

Given:View attachment 238257

e^iΘ= cosΘ + i sinΘ (radians)

e^πi=-1

Deduced

e^2πi=(-1)²

e^2πi=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(3^1/2/2)=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=1However 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.

 

[Quarky nerd]

why does this happen?

 

[Quarky nerd]

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

 

[PeroK]

why does this happen?

Why does what happen?

 

[Quarky nerd]

might this be the same reason that extraneous solutions exist?

 

[PeroK]

might this be the same reason that extraneous solutions exist?

Extraneous solutions to what?

 

[DrClaude]

why does this happen?

Have a look at https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/