Is 2 pi i equal to 0, contradicting the fact that pi and i cannot equal 0?

[nicktacik]

This one has me stumped.

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

Or is 2 pi i actually 0, and this does not actually imply that either pi = 0 or i = 0?

 

[Integral]

Since $2 \pi i$ is an imaginary number you must use the definition of LN in the complex plane. You are using the definition for the real number line.

in the complex plane we have:

$$logz = log(r) + i \theta$$

edit (changed my definition of z)
$$r = |z|$$
and
$$\theta = arg(z)$$

 

[matt grime]

More than that - logs of complex numbers are only defined up to multiples of 2pi. One can ought to take the principal branch - this is just a slightly more complicated variation on the square root 'fallacies'.

 

[Kummer]

The mistake is here:
$$\ln e^z = z$$
This is not true for complex numbers.

Note: the other way around:
$$\exp (\ln z) = z, \ z\not =0$$
Is true.

 

[nicktacik]

Ok thanks. As you can see, I haven't taken my complex variables class yet.

 

[mathwonk]

its sort of like saying, (2^2 =4 and (-2)^2 = 4 so 2 = -2.)