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Prove that the function $\log(z+ \sqrt{z^2-1})$ can be defined to be analytic on the domain $\mathbb{C} \setminus (-\infty,1]$

(Hint: start by defining an appropriate branch of $ \sqrt{z^2-1}$ on $\mathbb{C}\setminus (-\infty,1]$ )

It just seems typical language problem. I do not see point of being rigorous here. But I think I have difficulty understanding the branch cut. I don't know if someone will explain me in some easy way and try to explain the solution to the problem. Any help will be much appreciated.

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Is the term under the square root also included in the log? –  Host-website-on-iPage Dec 29 '12 at 3:56
    
The region is simply connected. See to it that $z+\sqrt{z^2-1}$ is nonzero in the region given to you. You can then define a branch of its logarithm. –  Host-website-on-iPage Dec 29 '12 at 4:08
    
Could you do in more rigorous way? –  Deepak Dec 29 '12 at 4:30
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$\sqrt{f(z)}=\exp(\frac12\log f(z))$. You choose a value for $\log f(z_0)$ and define $\log f(z)$ to be $\log f(z_0)+\int_{z_0}^z\frac{f'(z)dz}{f(z)}$. Here the path of integration does not matter because the region supplied is simply connected and $f(z)$ which here is $z^2-1$ is nonzero throughout it. So the integral is well-defined. Check that it is analytic. –  Host-website-on-iPage Dec 29 '12 at 4:32
    
I think you could make this as a nice solution. –  Deepak Dec 29 '12 at 6:20

2 Answers 2

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We know that $\sqrt{f(z)}=\exp(\frac12\log f(z))$. You choose a value for $\log f(z_0)$. Here the path of integration does not matter because the region supplied is simply connected and $f(z)$ which here is $z^2-1$ is non zero in the region. So the integral $I(z)=\int_{z_0}^z\frac{f'(z)dz}{f(z)}$ is well-defined. Check that it is analytic. You can thereafter define $\log f(z)=\log f(z_0)+I(z)$. This gives an analytic branch of $\log f$ and the choice of $\log f(z_0)$ determines the branch.

Having gotten a well defined branch for the square root. You can get one for $\log(z+\sqrt{z^2-1})$

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There is a more direct definition for a branch of $\sqrt{z^2-1}$. Consider $\sqrt{z-1}\sqrt{z+1}$ with the standard branch for $\sqrt{w}$ –  Thomas Andrews Dec 29 '12 at 14:13

Alternatively, you can just take the standard branch for $\sqrt{z}$ excluding $(-\infty,0]$ and then compute $\sqrt{z-1}\sqrt{z+1}$ which is defined for $z+1,z-1\notin(-\infty,0]$, that is, for $z\notin(-\infty,1]$

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