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- Feb 21st 2010, 05:51 PMamberxinyabranch cut problem
- Feb 21st 2010, 06:25 PMdavismj
What do you have so far?

- Feb 21st 2010, 07:19 PMamberxinyare
$\displaystyle \Delta_{C}{arg f}=1,-1,0,0 $ for (i)(ii)(iii)(iv) respectively.

how about the behaviour - Feb 22nd 2010, 05:52 AMshawsend
I'll try (i) for $\displaystyle \alpha=1/2$:

You can let $\displaystyle z=1+se^{i\phi}$ and $\displaystyle z=re^{i\theta}$ then $\displaystyle \sqrt{\frac{re^{i\theta}}{se^{i\phi}}}=\left(\frac {r}{s}\right)^{1/2}e^{i/2(\theta-\phi+2k\pi)}$ then $\displaystyle \arg f=\frac{\theta-\phi}{2}+2k\pi$. Now consider (i) at various points around the circle $\displaystyle 1/2 e^{it}$. At 1/2, $\displaystyle \theta=0$ and $\displaystyle \phi=\pi$ so $\displaystyle \arg f(1/2)=-\pi/2$. Now start going around that circle and show how the term $\displaystyle \frac{\theta-\phi}{2}$ changes smoothly between the values of $\displaystyle -\pi/2$ and $\displaystyle 3\pi/2$ except when you cross the line segment between zero and one and therefore the function $\displaystyle f(z)=\sqrt{z/(z-1)}$ has a single-valued analytic component over this circle except for the line segment (branch-cut) between zero and one.