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There's a lot of great information here about understanding the branch cuts and branch points of functions of the form ( for example ) $(z^3+1)^{1/2}$, sums of simple roots and products thereof.

However, I am not entirely sure how to deal with more complicated examples, such as

$g(z)=(z+ \sqrt{z^2 - 3})^{1/3}$, or even worse iterations (for example, some rational function f[g(z)]).

With the simpler examples that I mentioned at the top, it is easy to get intuition for example by plotting in mathematica for example for $(z+1)^{1/3}$ we can see nicely that one needs three different sheets to obtain a smooth function over a larger domain than the complex plane in the plot

plot of the gluing of three sheets for the multivalued function <span class=$(z+1)^{1/3}$ for $z=r Exp(I y)$ while varying $r$ from $0$ to $10$ and $y$ from $0$ to $6\pi$" />

When one tries a similar thing for $g(z)=(z+ \sqrt{z^2 - 3})^{1/3}$, I cannot understand the output by mathematica, seemingly not following the correct root branches. How would one do the proper analysis 'by hand' so as to predict this structure?

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  • $\begingroup$ I like the way that it is handled in Chapter 1 of "An Introduction to Complex Function Theory" (Bruce Palka). Let $~\xi_{k,n} ~$ denote exp$[i(2k\pi)/n]~$ (i.e. one of the roots to $~z^n = 1).~$ For $~z = re^{i\theta},~$ Palka arbitrarily defines the principal $~n$-th root $~z^{(1/n)}~$ as $~r^{(1/n)}e^{i\theta/n},~$ so that all roots are given by $~z^{(1/n)} \times \xi_{k,n} ~: k \in \{0,1,2,\cdots,n-1\}.~$ Then, everything falls into place, and you have easy notation for describing all roots, even with a complicated expression such as yours. ...see next comment $\endgroup$ Mar 29, 2023 at 18:01
  • $\begingroup$ So, you would let $~w = ~$ the principal root of $~(z^2 - 3)^{(1/2)},~$ and then let the set $~V = \{~w \times \xi_{k,2} ~: ~k=0,1 ~\}.~$ Then, for each value $~v \in V,~$ you would let $~t = ~$ the principal root of $~(z + v)^{(1/3)}, ~$ and then let $~S = \{t \times \xi_{k,3} ~: ~k = 0,1,2 ~\}.$ $\endgroup$ Mar 29, 2023 at 18:08

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