[Complex Analysis] Branch cuts of the logarithm

[ZeroSum]

Homework Statement

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)}

Homework Equations

\log{z} = \ln{r} + i(\theta + 2k\pi)

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?

 

[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 3^1/2+i , which is π/6.

 

[ZeroSum]

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 3^1/2+i , which is π/6.

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.

 

[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.