If x is complex, as in this case, then yes, ln(x) is a multi-valued function.
If one has e[sup]lnz[/sup]=z, then because of the periodicity of the function e[sup]lnz + 2n(pi)i[/sup], the logarithm function in the complex plane is multi-valued, which explains your strange results - ln is not uniquely defined.
The reason is that exponentiation in the complex plane is defined as follows: a[sup]b[/sup] = exp(b ln(a)). But as you’ve noted, ln() is a multi-valued function when applied to complex numbers. Therefore, exponentiation is also multi-valued in the complex domain.
Now consider 1[sup]ln(5)/2πi[/sup]. By definition, this must equal exp( (ln(5)/2πi) ln(1)). If we let ln(1)=0, then we get exp(0) which equals 1, which we expect. But since we’re considering complex numbers, ln(1) also equals 0+2πi, in which case 1[sup]ln(5)/2πi[/sup] = exp(ln(5)) = 5. In the same fashion, we can show that 1[sup]ln(5)/2πi[/sup] = exp(2 ln(5))=25, and so on.
If we restrict ourselves to the “primary branch” of the logarithm function…that is, if we arbitrarily require that the imaginary part of ln(z) always lies between -πi and πi…then some of these problems go away. But that in turn causes other problems, like breaking the rule that ln(ab) = ln(a)+ln(b). In short, complex numbers are weird.
i has two square roots, (1+i)/sqrt(2) and (-1-i)/sqrt(2). Those are also the two possible numbers you get when evaluating i[sup]1/2[/sup] via the formula exp(ln(i)/2).
Might be. It could be a new theory of mine (all mine), and it works perfectly well if you don’t care for all the roots. So nyah!
Actually if one of the mods could correct that it’d be appreciated – okay it’ll make MikeS look pretty silly, but he’s been here two years and only posted forty-six times (pulls out calculator), that’s less than one post per fortnight! Why do have to pick on mine?