the i*pi=ln(-1) identity

**rainer** · July 30th 2010, 12:51 AM

Famously, we have...

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

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

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

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

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

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

Thanks

**Prove It** · July 30th 2010, 01:08 AM

Originally Posted by rainer

Famously, we have...

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

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

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

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

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

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

Thanks

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

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

**rainer** · July 30th 2010, 01:22 AM

Originally Posted by Prove It

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

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

$i\pi 2n=\ln1$ follows from euler's equation: $e^{i\pi 2n}-1=0$ Add 1 to both sides and then take the natural log of both sides and you get $i\pi 2n=\ln1$

Right?

As for x, it is any real number.

**HallsofIvy** · July 30th 2010, 01:37 AM

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

**rainer** · July 30th 2010, 09:18 AM

Originally Posted by HallsofIvy

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

Hmmm, so if it's not $i\pi 2n=\ln1$ what equation would capture this multi-valued character of $\ln1$ in the complex numbers?

Also, I do not quite follow your math.

If $x= re^{i(\theta+ 2n\pi)}$ , then $ln(x)= ln(r)+ i(\theta+2n\pi)$ , not $ln(x)= ln(r)+ 2n\pi$ as you wrote. Or is what you wrote somehow the same? What am I missing here?

**HallsofIvy** · July 31st 2010, 01:21 AM

Sorry, I shifted too quickly to the specific case with $\theta= 0$ .

In general, if $x= re^{i\theta}$ then it is also true that $x= re^{i(\theta+ 2n\pi}$ so $ln(x)= ln(r)+ i(\theta+ 2n\pi)$

In particular, if x= 1, then r= 1, $\theta= 0$ so $ln(x)= 2n\pi$ for all n.

**rainer** · July 31st 2010, 11:32 PM

Originally Posted by HallsofIvy

Sorry, I shifted too quickly to the specific case with $\theta= 0$ .

In general, if $x= re^{i\theta}$ then it is also true that $x= re^{i(\theta+ 2n\pi}$ so $ln(x)= ln(r)+ i(\theta+ 2n\pi)$

In particular, if x= 1, then r= 1, $\theta= 0$ so $ln(x)= 2n\pi$ for all n.

Wait, if x=1, and consequently r=1 and theta=0, then what you get is...

$\ln(1)= \ln(1)+ i(0+ 2n\pi)= i\pi 2n$ not $\ln(1)= \pi 2n$ 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.:

$e^{i\pi2n}=1$

$i\pi2n=\ln(1)=0$

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?

**HallsofIvy** · August 1st 2010, 04:26 AM

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. $ln(1)= 2m\pi i$ 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)[/itex]" only says that [itex]2n\pi i= 2m\pi i[/itex] for some integer m- which is trivially true.

Thread: the i*pi=ln(-1) identity

LinkBack

Thread Tools

Display

the i*pi=ln(-1) identity

Similar Math Help Forum Discussions

identity

identity 2

identity

Sin/Cos Identity

Identity

Search Tags