I am asked to find the principle branch of the complex function $$f(z) = \sqrt{1-z}$$

I know that the principle branch of $z^{1/2}$ is given by $\exp(\frac{1}{2} \log(z))$ where $$\log(z) = \log(|z|)+i\arg(z)$$ is the principle branch of the logarithm. So then to find the principle branch of $\sqrt{1-z}$, is it just

$$\exp\bigg(\frac{1}{2} \log(1-z)\bigg)$$ with $\log(1-z)$ being the principle branch of the logarithm i.e. $$\log(1-z) = \log(|1-z|) + i\arg(1-z).$$

I think I am confusing myself. Is the principle branch correct?


You need to specify what you do with the arg(1-z) when $z$ winds around 1 (and the arg increases with $2\pi$). You will have to introduce a cut (where arg, whence $\sqrt{1-z}$ is not defined), a typical choice here being $[1,+\infty[$ along the positive real axis. This is compatible with the principal branch of Log(w) which is usually defined by a cut along $R_-$.

  • $\begingroup$ What about the radius of convergence of the Taylor series for this function around z = 0. Is the same ? |z|<1 ? $\endgroup$ – Richard Clare Aug 23 '17 at 18:00
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    $\begingroup$ If the cut does not enter the unit disk then the Taylor series converges for $|z|<1$ as you say. There are less natural choices of the cut for which the radius of convergence may be smaller than one, even zero (if the cut goes through the origin). $\endgroup$ – H. H. Rugh Aug 23 '17 at 22:17
  • $\begingroup$ So It if they ask me in an exam to explain detailed the set of all complex numbers in which this series converges, did you recommend me to define that branch cut $[1,\infty)$ and then say that it is $|z| < 1$ ? $\endgroup$ – Richard Clare Aug 24 '17 at 0:10
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    $\begingroup$ More like: The Taylor series for $\sqrt{1-z}$ around $z=0$ is $1-\frac{z}{2} + ...$ and converges for $|z|<1$ regardless of other things. It converges to a function which is analytic for $|z|<1$ and which coincide with the function $\sqrt{1-z}$ when defined in the complex plane with a branch cut going from $1$ to $\infty$ without intersecting the unit disk (so in particular $[1,+\infty)$ along the real line will do). $\endgroup$ – H. H. Rugh Aug 24 '17 at 5:23

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