Expand $z$ in power series of $w$ for each branches, where $w=2z+z^2$.

Question

Define $$w=z^2+2z$$ Just heuristics: $$z=\pm\exp\left(\frac{\log(w+1)}2\right)-1$$ If we expand $\log (1+w)$ into a power series for $|w|<1$, and further exponentiate it, how would the series look like?

Answer 1

I assume we have to expand around $0$.

Consider the given dependence as an algebraic equation of second degree in $z$ with parameter $w$, $$ z^2 + 2z - w =0\ ,$$ then the (two branches) solution is $$ \begin{aligned} z_{1,2}=z_{1,2}(w) &=-1\pm\sqrt{1+w} =-1\pm\sum_{n\ge 0}\binom{1/2}n w^n \\ z_1=z_1(w) &=\sum_{n\ge 1}\binom{1/2}n w^n \\ z_2=z_2(w) &= -2-\sum_{n\ge 1}\binom{1/2}n w^n \ . \end{aligned} $$

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Computer check modulo $O(w^8)$, here sage:

sage: R.<w> = PowerSeriesRing( QQ, default_prec=8 )
sage: z1 = -1 + sqrt(1+w)
sage: z2 = -1 - sqrt(1+w)
sage: z1
1/2*w - 1/8*w^2 + 1/16*w^3 - 5/128*w^4 + 7/256*w^5 - 21/1024*w^6 + 33/2048*w^7 + O(w^8)
sage: z2
-2 - 1/2*w + 1/8*w^2 - 1/16*w^3 + 5/128*w^4 - 7/256*w^5 + 21/1024*w^6 - 33/2048*w^7 + O(w^8)

sage: z1^2 + 2*z1
w + O(w^8)
sage: z2^2 + 2*z2
w + O(w^8)