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**1. Homework Statement**

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

[tex]\log{(\sqrt{3}+i)}[/tex]

**2. Homework Equations**

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

**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: [itex]0.693 - i\frac{11\pi}{6}[/itex]

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

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

Does the restriction [itex]\log{1}=-2\pi i[/itex] actually restrict this to moving around the circle clockwise from [itex]-\frac{\pi}{4}[/itex] such that [itex]-\frac{9\pi}{4} < \theta \le -\frac{\pi}{4}[/itex]? When using this log with principal values and restricted to a domain of analyticity of [itex]-\pi < \theta \le \pi[/itex] we traditionally wind around counterclockwise toward [itex]\pi[/itex] and clockwise toward [itex]-\pi[/itex]. This one, if I understand it correctly, winds around [itex]-2\pi[/itex] 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?