Why does 2*pi*i equal 0 in euler's formula?

[tom92373]

I was playing around with euler's formula the other day and found this odd proof which says 2ipi=0. I know this is obviously wrong, but what was the step that went wrong? The "proof" goes like this:

start with e^(i*pi)+1=0
so -e^(i*pi)=1
multiply both sides by e e(-e^(i*pi))=e
factor a (-1) (-1)(e)(e^(i*pi))=e
add exponents (-1)(e^(i*pi+1))=e
take ln of both sides ln((-1)(e^(i*pi+1)))=1
use properties of logs ln(-1) + ln(e^(i*pi+1))=1
take ln of -1 i*pi + ln(e^(i*pi+1))=1
properties of logs i*pi + (i*pi+1)ln(e)=1
take ln(e) i*pi + i*pi+1=1
subtract 1 i*pi + i*pi=0
combine like terms 2*pi*i=0 ...

this even works in the original formula:
e^(i*2*pi)=cos(2*pi)+isin(2*pi)
e^(i*2*pi)=1
e^(i*2*pi)=e^0
i*2*pi=0 ...

 

[NoMoreExams]

Well what's the period of sine and cosine? What's e^(4pi*i)?

 

[abelxing]

in complex analysis, log function is a Multi-valued function，ln(z) = |z| + arg(z), you should choose a single value branch for log, or use Riemann surface

 

[arildno]

e^(i*2*pi)=e^0
i*2*pi=0

That inference, from line 1 to line 2 is invalid.

 

[HallsofIvy]

In "polar form", $z=re^{i \theta}$, $\theta$ represents the angle the line from 0 to z makes with the positive real axis. In that sense, yes, $2\pi$ is the same angle as 0. That's why, as arildno suggested, in complex numbers, the exponential function is no longer single valued. f(x)= f(y) implies x= y only if f is single valued.

 

[tom92373]

Ok I get it; so it's because complex exponentials have more than one solution.

 

[adriank]

That's why, as arildno suggested, in complex numbers, the exponential function is no longer single valued. f(x)= f(y) implies x= y only if f is single valued.

I think you mean it's not one-to-one! It's the logarithm that's not single-valued. :)