# Branch cut/ points for square roots inside cubic roots- incorrect branching by mathematica or my mistake?

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

$(z+1)^{1/3}$ 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?

• 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 Mar 29, 2023 at 18:01
• 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 $~S = \{t \times \xi_{k,3} ~: ~k = 0,1,2 ~\}.$ Mar 29, 2023 at 18:08