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I am not sure that I am understanding how to find branch points and cuts correctly. So for $\sqrt{z-1}$, I let $z=re^{i\theta+2nx}$ and substitute into $\sqrt{re^{i\theta+2nx}+e^{2kx}}$. But I don't know how to proceed from there.

And how should we find branch points and cuts for logarithm? Say take $log(z^2+z+1)$ for example, should we first substitute it into something like what we've done above? Thanks!

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  • $\begingroup$ For every $z \ne 0$, there is a branch such that $f(z)=(z-1)^{1/2}$ is holomorphic around $z$, since its derivative is $\frac{1}{2}(z-1)^{-1/2}$. At $z=0$ it is different, and its derivative diverges hence it is not holomorphic. Once you have the singularity at $z=a$, rotate around it for seing if it is an isolated singularity, or if $f(a+ze^{2i\pi}) \ne f(a+z)$, in that case it is a branch point. Rotating around $z=1$, you have $f(1+e^{it}) = (1+e^{it}-1)^{1/2} = e^{it/2}$ but $f(1+e^{it}e^{2i\pi}) = (e^{2i \pi}e^{it})^{1/2} = -e^{it/2}\ne f(1+e^{it})$ hence $z=1$ is a branch point. $\endgroup$
    – reuns
    Sep 29, 2016 at 23:06

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The branch points for $\sqrt z$ are at $z=0$ and $z=\infty $. Therefore, the branch points for $\sqrt{z-1}$ are at $z=1$ (if you like, let $z=1+re^{i\theta}$) and $z=\infty$. The plane can be cut by any contour that adjoins $1$ and the point at infinity. The Principal branch would be the ray along the negative real axis that starts at $z=-1$ and approaches $-\infty$.

The branch points for $\log(z)$ are also at $z=0$ and $z=\infty$. Therefore, the branch points for $\log(z^2+z+1)$ are at $z=\frac{-1\pm i\sqrt{3}}{2}$ and $z=\infty$. Branch cuts may be any contours that begin at the branch points $\frac{-1\pm i \sqrt 3}{2}$ and terminate at $\infty$. Alternatively, a suitable branch cut is any contour that adjoins $\frac{-1+i\sqrt3}{2}$ and ${-1-i\sqrt3}{2}$.

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  • $\begingroup$ Thank you! When do we need to substitute the parameters? $\endgroup$
    – J.doe
    Sep 29, 2016 at 22:36
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Sep 29, 2016 at 23:02
  • $\begingroup$ Help me understand what you mean by "substitute the parameters." $\endgroup$
    – Mark Viola
    Sep 29, 2016 at 23:03
  • $\begingroup$ Ah nothing, I get it now. Thanks! $\endgroup$
    – J.doe
    Sep 29, 2016 at 23:34

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