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

• Jul 30th 2010, 12:51 AM
rainer
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
• Jul 30th 2010, 01:08 AM
Prove It
Quote:

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 $\displaystyle x$?
• Jul 30th 2010, 01:22 AM
rainer
Quote:

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.
• Jul 30th 2010, 01:37 AM
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, $\displaystyle \theta= 0$, 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 $\displaystyle 2n\pi i= 0$.
• Jul 30th 2010, 09:18 AM
rainer
Quote:

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? • Jul 31st 2010, 01:21 AM 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. • Jul 31st 2010, 11:32 PM rainer Quote: 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? • Aug 1st 2010, 04:26 AM HallsofIvy 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.