The key to finding the natural logarith of a complex number is in an equation given by Euler.

     e^iy = cos(y) + i*sin(y) (proof)

     We'll need to reshape this a bit before it will be really useful.

     e^{x + iy} = e^x*(cos(y) + i*sin(y))

     We simply find the natural logarith of both side and we have something promising.

     x + iy = ln(e^x*(cos(y) + i*sin(y)))

     or

     x + iy = ln(e^x*cos(y) + i*e^x*sin(y)))

     Now we have something that looks like it might help us. On one side we have a natural logarithm of a complex number. On the other side we have a complex number result. The result is simple to understand. There are two parts, x and y.
     When we find the natural logarithm of a complex number we start with ln(m + in). We can find m and n in the above equation.

     x + iy = ln(m + in)
     m = e^x*cos(y)
     n = e^x*sin(y)

     Using these identities we can start solving for x and y in terms of m and n.

     Note: o, a, and h in the following are respectively opposite, adjacent, and hypotenuse sides relative to the angle y.
     n/m = e^x*sin(y)/e^x*cos(y)
     n/m = sin(y)/cos(y)
     n/m = (o/h)/(a/h)
     n/m = (o/a)/(h/h)
     n/m = (o/a)
     n/m = tan(y)
     arctan(n/m) = y

     Now we can just plug those numbers in for y.

     m = e^x*cos(arctan(n/m))

     Since arctan take the opposit over adjacent and converts it into an angle, we can think of n as an opposite side and we can think of m as and adjacent side. This also means that we can define the hypotenuse as the square root of n² + m². So:

     m = e^x*(m/((n² + m²)^1/2))
     m(n² + m²)^1/2 = me^x
     (n² + m²)^1/2 = e^x
     ln((n² + m²)^1/2) = x

     This has a natural logarithm in it, but it's a natural logarith of real numbers, so it really isn't a problem, but it does mean that the real number method of doing logarithms isn't made obsolete by the more general method. However, if that is displeasing, one can replace the ln statement by writting out a process for finding the ln of a real number.
     Now that we know both x and y in terms of m and n, all that remains is to write it out.

     x + iy = ln(m + in)
     ln((n² + m²)^1/2) + i*arctan(n/m) = ln(m + in)

     Note: If n = 0 (the number isn't complex), then this equation can quickly simplify to a normal ln.

     ln((0² + m²)^1/2) + i*arctan(0/m) = ln(m + i0)
     ln((m²)^1/2) + i*arctan(0) = ln(m + i0)
     ln(m) + i*0 = ln(m + i0)
     ln(m) = ln(m + i0)

What's It Good For?

ln(i)

Calculating xi

More Resources

Ln(x + iy) Explained By Dr. Math