Ok so we know e^(pi*i)=-1 what about this?

Question

e^(pi*i)=-1 using calculus we know this is true but square both sides and you get:
e^(2pi*i)=1 which is also true but if you natural log both sides you get:
2pi*i=ln1 which is obviously false since ln 1 is real
why can't you natural log e^(2pi*i)?

Answer 1

Logarithms, when applied to imaginary numbers can have many results. This is because exponentiation, when applied to imaginaries, is defined as a trig function:

e^(ix) = cos x + i sin x

So, the logarithm is an inverse trig function. And we know inverse trig functions can have many results. ln e^(2 pi i) can be 2 pi i, but it could also be 0, or -2 pi i, or basically any ix where (cos x + i sin x = 1). In this case 0 fits with the real logarithm on the right side, since ln(1) = 0.

Answer 2

In the same sense you could write:
e^(0)=e^(2*i*pi)=e^(4*i*pi)=...
and it would seem that 0=2=4=.....
which is false. The problem here is with the definition of the natural logarithm for complex numbers.
If a complex is given as z=A*e^(i*phi) Then ln (z)= ln(A) + i*phi
but when defining the function we must agree in which of the sets ([0, 2pi), [2pi, 4pi), and so on...[meaning inclusive, ")" - non-inclusive) we want the phi to fall in. These definitions are said to lie on different Riemann's planes (free translation). Thus we can't get two phis that differ by more than 2*pi.

For illustration compare with the arctg or arcsin functions. Here we also have to agree, what values we allow the function to take, when we define them.

Answer 3

Because you have to keep 1 as (-1)^2.
If you take ln of (-1)^2 you get 2 ln (-1) which is imaginary.
So you'd have:
2i*pi = 2 ln (-1)
i*pi = ln (-1) which by the way = the original equation

Answer 4

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