Analyticity of the complex logarithm on the negative real axis

[Fosheimdet]

A theorem in my textbook states the following:

For every n=0,±1, ±2, --- the formula ln z=Ln z ± 2nπi defines a function, which is analytic, except at 0 and on the negative real axis, and has the derivative (ln z)'=1/z.

I don't understand why the logarithm isn't analytic for negative real values. It is after all defined for negative values, contrary to the real logarithm. Is it because it is not differentiable for negative real values? And if so, why?

 

[UVCatastrophe]

Analytic, in a nutshell, means differentiable around a point. On the negative real axis, complex log has a discontinuity across the real axis, which means it's not differentiable. Actually, that's purely a matter of convention; the so-called branch cut can be defined along any ray starting at the origin (the branch point). But it's common to use the negative real axis.

Using polar representation, we find that $\log z = \log \left(r e^{i \theta} \right) = \log r + i \theta$. Suppose we approach the negative real axis from $\theta = 0$ and go around counter-clockwise into the second quadrant. Close to the negative real axis, we find that $\text{Im}\left[ \log z\right] = \pi - \epsilon$. On the other hand, if we go clockwise through the fourth quadrant into the third quadrant, we have $\text{Im}\left[ \log z\right] = - \pi + \epsilon$. The discontinuity in the imaginary part is thus $\text{Disc} = 2 \pi$.

The reason we need a branch cut is because otherwise, complex logarithm would be multivalued (imagine you keep going around across the branch cut; every time you traverse the circle and end up at the same point, you'd add another factor of $2 \pi$ to the imaginary component). We'd like it to be single valued, so we need a convention.

 

[Fosheimdet]

Thank you so much! Great explanation, made things allot clearer.

 

[mathwonk]

the log ios the inverse of the exponential, so the way to see what is going on is to look at the expopnential and try to see how to invert it. I.e. one wants to choose a subset of the domain of the exponential on which it is ionjective.

a simpler case, as indicated above, is the map taking t on the real axis to (cos(t), sin(t)) on the unit circle. this wraps around the unit circle infinitely many times and is thus highly non injective, but it becomes injective if restricted to any open interval of length 2π. The omitted endpoint can be chosen to be anywhere on the real axis, and the resulting omitted point of the circle can thus also be anywhere.

 

[WWGD]

To add to what, Mathwonk and Catastrophe said, a branch of logz is a local inverse of the exponential; expz maps onto $\mathbb C-{0}$ , and it is periodic with period $2\pi$ . What you do is you take a subset of $\mathbb C-{0}$ satisfying $y< Imz \leq y+ 2\pi$, (while the Real part is in $(-\infty, \infty)$), to avoid hitting any value more than once. In this subset, $expz$ is 1-1, and, by the inverse function theorem it has an inverse, which is "usually" (meaning in my experience ) called log_y , so that in this set, you have exp(log_y (z))= log_y(exp(z))=z .

Just to add a bit more, you can also define Riemann surfaces to avoid multivaluedness, so that, informally, once a function is about to repeat itself, you go up
on a "parking ramp". You use as many up-ramps as the number of times you need to go back to your original value when winding around a branch point. Think of the case of $z^{1/2}$, which is 2-valued. Once can create a Riemann surface to avoid hitting any value twice; in this case, the Riemann surface would have just 1 ramp, since if you wind around the origin ( the branch point here ) twice , you go back to the original value. Consider a point w in $ {#z: |z|=1#} $ .
Then we have that $$w=e^{i\theta}; e^{\frac {i(\theta+ 2\pi)}{2}}=e^{i\theta/2}e^{i\pi}=-w $$, so if you wind around the branch point once you do not go back to the wriginl value, but if you go around twice, you do return to your original value. This defines a group called the monodromy group, which in this case is isomorphic to $\mathbb Z_2$, but in the case of logz is isomorphic to $\mathbb Z$.