1. ## the i*pi=ln(-1) identity

Famously, we have...

$\displaystyle e^{i \pi (2n-1)}+1=0$

where n is any positive integer. This is of course because $\displaystyle \cos{(\pi (2n-1))}=-1$ and $\displaystyle i\sin{(\pi (2n-1))}=0$

My question is, from this relation that Euler discovered doesn't it also follow that...

$\displaystyle e^{i \pi 2n}-1=0$ ?

Since $\displaystyle \cos{(\pi 2n)}=1$ and $\displaystyle i\sin{(\pi 2n)}=0$ ?

And if this is true, that means that $\displaystyle x^{i\pi 2n}=x^{\ln{1}}=x^0=1$, right?

Thanks

2. Originally Posted by rainer
Famously, we have...

$\displaystyle e^{i \pi (2n-1)}+1=0$

where n is any positive integer. This is of course because $\displaystyle \cos{(\pi (2n-1))}=-1$ and $\displaystyle i\sin{(\pi (2n-1))}=0$

My question is, from this relation that Euler discovered doesn't it also follow that...

$\displaystyle e^{i \pi 2n}-1=0$ ?

Since $\displaystyle \cos{(\pi 2n)}=1$ and $\displaystyle i\sin{(\pi 2n)}=0$ ?

And if this is true, that means that $\displaystyle x^{i\pi 2n}=x^{\ln{1}}=x^0=1$, right?

Thanks
You are correct that $\displaystyle (e^{i\pi})^{2n} - 1 = 0$.

I don't see how you turned $\displaystyle i\pi 2n$ into $\displaystyle \ln{1}$ though... What is $\displaystyle x$?

3. Originally Posted by Prove It
You are correct that $\displaystyle (e^{i\pi})^{2n} - 1 = 0$.

I don't see how you turned $\displaystyle i\pi 2n$ into $\displaystyle \ln{1}$ though... What is $\displaystyle x$?
$\displaystyle i\pi 2n=\ln1$ follows from euler's equation: $\displaystyle e^{i\pi 2n}-1=0$ Add 1 to both sides and then take the natural log of both sides and you get $\displaystyle i\pi 2n=\ln1$

Right?

As for x, it is any real number.

4. You cannot go from $\displaystyle e^{2ni\pi}= 1$ to $\displaystyle 2ni\pi= ln(1)= 0$ so quickly. In the complex numbers the logarithm is multi-valued. Specifically, if, in polar form, $\displaystyle x= re^{i\theta}$ then it is also true that $\displaystyle x= re^{i(\theta+ 2n\pi}$. $\displaystyle ln(x)= ln(r)+ 2n\pi$ where n can be any integer. In particular, since 1, in polar form, has r= 1, $\displaystyle \theta= 0$, ln(1) has many values, one of which is 0. The others are, just as you say, multiples of $\displaystyle 2\pi i$. It does not follow that, just because as a real function, ln(1)= 0 that $\displaystyle 2n\pi i= 0$.

5. Originally Posted by HallsofIvy
You cannot go from $\displaystyle e^{2ni\pi}= 1$ to $\displaystyle 2ni\pi= ln(1)= 0$ so quickly. In the complex numbers the logarithm is multi-valued. Specifically, if, in polar form, $\displaystyle x= re^{i\theta}$ then it is also true that $\displaystyle x= re^{i(\theta+ 2n\pi}$. $\displaystyle ln(x)= ln(r)+ 2n\pi$ where n can be any integer. In particular, since 1, in polar form, has r= 1, $\theta= 0[/tex], ln(1) has many values, one of which is 0. The others are, just as you say, multiples of \displaystyle 2\pi i. It does not follow that, just because as a real function, ln(1)= 0 that \displaystyle 2n\pi i= 0. Hmmm, so if it's not \displaystyle i\pi 2n=\ln1 what equation would capture this multi-valued character of \displaystyle \ln1 in the complex numbers? Also, I do not quite follow your math. If \displaystyle x= re^{i(\theta+ 2n\pi)}, then \displaystyle ln(x)= ln(r)+ i(\theta+2n\pi), not \displaystyle ln(x)= ln(r)+ 2n\pi as you wrote. Or is what you wrote somehow the same? What am I missing here? 6. Sorry, I shifted too quickly to the specific case with \displaystyle \theta= 0. In general, if \displaystyle x= re^{i\theta} then it is also true that \displaystyle x= re^{i(\theta+ 2n\pi} so \displaystyle ln(x)= ln(r)+ i(\theta+ 2n\pi) In particular, if x= 1, then r= 1, \displaystyle \theta= 0 so \displaystyle ln(x)= 2n\pi for all n. 7. Originally Posted by HallsofIvy Sorry, I shifted too quickly to the specific case with \displaystyle \theta= 0. In general, if \displaystyle x= re^{i\theta} then it is also true that \displaystyle x= re^{i(\theta+ 2n\pi} so \displaystyle ln(x)= ln(r)+ i(\theta+ 2n\pi) In particular, if x= 1, then r= 1, \displaystyle \theta= 0 so \displaystyle ln(x)= 2n\pi for all n. Wait, if x=1, and consequently r=1 and theta=0, then what you get is... \displaystyle \ln(1)= \ln(1)+ i(0+ 2n\pi)= i\pi 2n not \displaystyle \ln(1)= \pi 2n as HallsofIvy wrote... In either case this seems to confirm rather than repudiate the original absurd result I started the thread with via my variation of Euler's identity, viz.: \displaystyle e^{i\pi2n}=1 \displaystyle i\pi2n=\ln(1)=0 Correct me if I'm wrong, but as far as I can see we're back to square one again. Any other suggestions as to where the fallacy lies in this absurd result? 8. Yes, you are wrong for exactly the reasons stated before: ln(z) is NOT a single valued function. You cannot just assert that ln(1)= 0 in the complex numbers. \displaystyle ln(1)= 2m\pi i for any integer m. Yes, m= 0 gives ln(1)= 0 as one of the values of ln(1) but you cannot, in the complex numbers, start from a= ln(1) and declare that a= 0. If a= ln(1) then a is equal to one of the infinite number of possible values of ln(1). Saying that "[itex]2n\pi i= ln(x)$" only says that $2n\pi i= 2m\pi i$ for some integer m- which is trivially true.