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. :)