[Complex Analysis] Branch cuts of the logarithm

1. Oct 2, 2011

ZeroSum

1. The problem statement, all variables and given/known data

Consider a branch of $\log{z}$ analytic in the domain created with the branch cut $x=−y, x≥0.$ If, for this branch, $\log{1}=-2\pi i$, find the following.

$$\log⁡{(\sqrt{3}+i)}$$
2. Relevant equations

$$\log{z} = \ln{r} + i(\theta + 2k\pi)$$
3. The attempt at a solution

This one is actually given in the textbook (odd numbered problem), but I'm having trouble understanding how the answer was arrived at.

The answer given: $0.693 - i\frac{11\pi}{6}$

I can see easily that $\log{\sqrt{(1)^2 + (\sqrt{3})^2}}=\ln{2} = 0.693...$ The real part here makes sense since it's the (real) log of the modulus of the given complex number $\sqrt{3}+i$.

I can also understand that the branch cut is made along $x=-y$. Where I'm getting confused is how the cut actually affects this log. So $r = 2, \theta=\frac{\pi}{6}$. Winding around counterclockwise from 0, we reach $\frac{\pi}{6}$ easily, since it does not cross the branch cut at all.

Does the restriction $\log{1}=-2\pi i$ actually restrict this to moving around the circle clockwise from $-\frac{\pi}{4}$ such that $-\frac{9\pi}{4} < \theta \le -\frac{\pi}{4}$? When using this log with principal values and restricted to a domain of analyticity of $-\pi < \theta \le \pi$ we traditionally wind around counterclockwise toward $\pi$ and clockwise toward $-\pi$. This one, if I understand it correctly, winds around $-2\pi$ from the cut so that it's restricted to one set of values for an otherwise multi-valued log.

Why do both of these ways of figuring a single-valued log (the traditional principal valued log cut on the negative x axis and the one used in this problem) seem to involve winding around the axis different ways? What should I be understanding here that I'm not?

2. Oct 2, 2011

Bacle

The issue is that the angle between the real axis and your point is π/6 , so the total angle from the ray is the angle from the ray to 1--which you're told is 2π , minus the angle from the ray to 31/2+i , which is π/6.

3. Oct 2, 2011

ZeroSum

It's actually $-2\pi$, not $2\pi$ where the log of 1 is defined. That is what is throwing me. I'm trying to get a clear picture in my head instead of just a plug and chug with the single-valued (analytic) definition of the log in complex, which works but doesn't lead me to using or understanding the nature of the branch cut involved.

4. Oct 2, 2011

Bacle

I think then it has to see with which is the positive direction, i.e., the direction in which angles increase; I don't know if there are standard rules for this.