Originally posted by Ramiri15
Why does the natural log of -1 equal Pi(i)? Neither my calculus nor my physics teacher knew the answer to this (I don't know it either, that's why I'm posting the question).
The real exponential function is a map from the set of real numbers to the set of positive real numbers
x maps to e^x
and it has an inverse, called the logarithm
x maps to log(x).
The exponential function is defined by a power series
e^x = sum from k=0 to infinity of x^k/k!
and this also makes sense for complex numbers z instead of x. So we have z going to e^z.
The question now is does this new "complex" exponential function admit an inverse that extends the real logarithm mentioned above.
Well, actually it does. Let's call this new complex logarithm log(z). One has to be careful with what z values we put in (the "domain" of the function -- it has to be restricted in some way -- see below). For us we shall simply define
log(z) = log(|z|) + i*arg(z)
for non-zero complex numbers z, and insist that, say, the argument arg(z) lies between -pi/2 and 3*pi/2. This is what I meant by restricting the domain -- we're excluding the negative imaginary axis. We have chosen this definition simply because it *is* an inverse for e^z, and extends the original real log(x). The reason for cutting out the negative imaginary axis is because if we didn't the log(z) function would jump suddenly as we went across it (the imaginary part suddenly changes from -pi/2 to 3*pi/2) and this is a bad thing -- we like our functions to be continuous.
There are of course many other inverses -- add on fixed multiples of 2*pi*i to the above formula. These other inverses are called "branches" of the logarithm.
If we plug in z=-1 into the definition for log(z) above we get
log(-1) = log(1) + i*pi
so that's i*pi, as you wanted. Another eg. would be z=1+i with
log(1+i) = log(root(2)) + i*pi/4 = log(2)/2 + i*pi/4.