# I E^pi*i ?

#### Quarky nerd

fig 1

Given:

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

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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#### .Scott

Homework Helper
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

Homework Helper
Gold Member
2018 Award
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.

#### 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

Homework Helper
Gold Member
2018 Award

#### Quarky nerd

might this be the same reason that extraneous solutions exist?

#### PeroK

Homework Helper
Gold Member
2018 Award
might this be the same reason that extraneous solutions exist?
Extraneous solutions to what?

"E^pi*i ?"

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