Arturo O. answered • 10/31/16
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You can get it if you work with complex numbers. You cannot get it working only with real numbers. In polar form, a complex number
z = x + iy
may be expressed in the form
z = reiθ
where r = √(x2 + y2) and θ = tan-1(y/x)
Furthermore,
x = r cosθ and y = r sinθ
Then
-1 = -1 + i(0)
r = √[(-1)2 + 02] = 1
θ = tan-1(0/-1) = π
Therefore,
-1 = 1(cos π + i sin π) = 1eiπ
-1 = eiπ
Now take ln() on both sides and get
ln(-1) = iπ
Note the solution is not real.
Arturo O.
Kenneth,
I have seen ln(z), where z is complex, defined in mathematical physics textbooks as
ln(z) = ln(r) + iθ
In this case, -1 is just 1eiπ in polar form, so ln(-1) = ln(1) + iπ = 0 + iπ = iπ
But there is the matter of the multivalued nature of ln(z) when z is complex, which I did not get into in my solution, where we should really say
ln(z) = ln|z| + i(θ ± 2πn), n = integer
If I recall correctly, the infinite possible values of n are the Riemann branches. (Now that is something I have not worked with in 20+ years!)
Of course, the mathematical physics treatments will not be as formal as in textbooks in pure math. A formal proof of the validity of taking ln(z), where z is something other than a positive real number, is probably found in a textbook on complex analysis, written for the pure mathematician. My exposure to the subject has focused on applications. But this is a widely used form, especially in problems involving oscillations.
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11/01/16
Kenneth S.
11/01/16