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We know that

e^i(pi)=-1

this gives,

e^2i(pi)=1

taking log to base e on both sides ,

we get,

2i(pi)=0

which means i=0. Can u explain me what went wrong in the above calculation ?(From what i know,taking log for e^2i(pi)is defined, since e^2i(pi)=1 gives [e^2i(pi)] is a real number and thus logarithm is defined) .

There are two different ways of asking this question:

1. Why is e^(2π i) = 1 ?

This is because e^(2π i) is like a rotation by 2π. The number e^(θ i) can be described by Euler's formula, which is a fundamental part of the theory of complex numbers:

e^(θ i) = cos(θ) + i sin(θ)

This is periodic -- it will repeat when θ=2π, precisely as you observed.

2. Why is it 2π i = 0 when I take the log of both sides?

Logarithmic functions are multivalued. You can read more here. Basically, the original real-number notion of logarithm is that the function f(x)=e^x is a one-to-one function. It should have an inverse, and that inverse is the logarithm. On the other hand, f(z)=e^z for complex values of z is not one-to-one, as you have noted e^0 = e^(2π i) = e^(4π i) = ...

So indeed, the logarithm is a multivalued function. Because of this, you can draw funky pictures like the one attached (from here). The graph on the left shows that the logarithm can take many values, and what you do is sort of cut this spiral and choose a "standard" set of values (usually 0 to 2π or -π to π). This gives you an inverse for e^x as log(x) in the same sense that arcsin(x) is an inverse to sin(x) by restricting the domain.

By creating this "cut" in the surface, if you do it from -π to π you create a discontinuity for numbers with the angle π or -π in them. In other words, you declare that those numbers have no logarithm. But those numbers are precisely the negative real numbers, for which there is no classical logarithm anyway.

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