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Find a branch of $\log(z)$ on domain $\mathbb{C}\setminus T$ where $T=\{x+iy:x\ge 0, y=x^2\}$

I know the branch cut will be a parabola, branch cuts are usually rays, and my prof. did explain it briefly but could someone explain it in more detail? Greatly appreciated, thanks!

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2 Answers 2

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Given a point $z=re^{i\theta}$, not on the parabola, with $0\le\theta\lt2\pi$, imagine starting at $(r,0)$ and going around the circle of radius $r$ counterclockwise until you get to $z$. If you get to $z$ without crossing the parabola, define $\log z=\log r+i\theta$. If you cross the parabola before you get to $z$, define $\log z=\log r+i(\theta-2\pi)$.

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DJ, you accepted this answer earlier, then unaccepted it. That is, of course, your prerogative, and you don't owe anyone an explanation --- but if there is something particular that you find unsatisfactory in the answer, maybe I can improve it, if you let me know. –  Gerry Myerson Feb 14 '13 at 2:15
    
Well, I'm just still a little confused about this whole parabola branch cut idea. So are you saying, depending on where my point z is, my log branch will differ? What if instead branch cut was y=(x)^1/2 what would be different then? –  DJ_ Feb 14 '13 at 2:46
    
I take it you are familiar with the principal branch of the logarithm, with branch cut the nonpositive reals. Think how my description works for that case. Start at $(r,0)$, go counterclockwise until you reach $z$. If you reach $z$ before crossing the branch cut, then $z$ is in the upper half plane, and you can take $\log z=\log r+i\theta$. If you cross the nonpositive reals before you reach $z$, then $z$ is in the lower half-plane, and you have to subtract $2\pi$ from the argument. It's the same for the parabola, or any other branch cut (unless the cut winds around the origin). –  Gerry Myerson Feb 14 '13 at 2:59
    
Oh I see, thanks Gerry! –  DJ_ Feb 14 '13 at 3:01

This is a simply connected domain and $i$ lies in the domain. Define $$F(z) = \int_{\gamma} {dz\over z},$$ for any curve $\gamma$ lying inside the domain. You can use a line since the domain is convex.

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I haven't done complex integrals yet. Is there another way of explaining it? –  DJ_ Feb 13 '13 at 2:25

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