jjmontero9 said:
Those equations are obtained by applying the "[URL identity[/URL] of the complex numbers:
\ln{(-2)}
= \ln{(2\cdot{}e^{(2k+1)\pi{}i})}
= \ln{(2)} + \ln{(e^{(2k+1)\pi{}i})}
= \ln{(2)} + (2k+1)\pi{}i
Where k = 0, 1, 2, ...The second one is:
\ln{(16)}
= \ln{(16\cdot{}e^{2\pi{}mi})}
= \ln{(16)} + \ln{(e^{2\pi{}mi})}
= \ln{(16)} + 2\pi{}mi
Where m = 0, 1, 2, ...
To clarify this a bit, when dealing with complex numbers the logarithm is a multivalued function, so you must choose a "branch" of the function that is single valued. However, all branches are related to each other by multiples of 2\pi. If \log(z), for complex numbers z, is the complex logarithm function (which is always base e), you can relate it to the real valued logarithm, ln(x), and the phase of the number z. In polar coordinates, z = r\exp(i\theta), where r is the modulus of the complex number (and is purely real) and \theta is the phase. Then, we may write
\log(z) = \log(re^{i\theta}) = \ln r + \log(e^{i\theta}) = \ln r + i(\theta + 2n\pi)
where n is an integer.
The 2\pi n term is there for the same reasons as when solving an equation like y = \sin(\theta). You get \theta = \arcsin y + 2n\pi, because sin is 2\pi periodic. Similarly, the complex exponential function is 2i\pi periodic, so when taking the inverse you get this 2\pi i n term added on:
e^{i\theta} = e^{i\theta + 2n\pi i}.
In the example cited, \theta was zero, but no branch of the complex logarithm was chosen, so there are infinitely many solutions, one for each integer n.
