Specific Calculation of the Germs of a Holomorphic Function As a specific example of this question and a follow up to this one does anyone know a nice way to calculate the germs at $z=1$ of
$$f(z)=\sqrt{1+\sqrt{z}}$$
My attempts have been messy at best, and I'd rather avoid trying to wade through Taylor series if I can! Any ideas would be most welcome!
 A: I'm not sure what you're looking for since you asked two distinct questions in the other posts, so I'll answer both.
Power series of the principal branch
For all $|t|<1$:
$$\left(\sqrt{1+t}+\sqrt{1-t}\right)^2=2\left(1+\sqrt{1-t^2}\right)$$
$$\frac{\sqrt{1+t}+\sqrt{1-t}}{2}=\frac{1}{\sqrt 2}\sqrt{1+\sqrt{1-t^2}}$$
The left hand side is simply the even part of $\sqrt{1-t}$, thus if $\sqrt{1-t}=\sum_{n=0}^\infty a_n t^n$ we have:
$$\frac{1}{\sqrt 2}\sqrt{1+\sqrt{1-t^2}}=\sum_{n=0}^\infty a_{2n} t^{2n}$$
$$\sqrt{1+\sqrt z}=\sum_{n=0}^\infty \sqrt 2 a_{2n} (1-z)^n$$
(Note: $a_0=1$ and $a_n=-1/2 \frac{(2n-2)!}{n!(n-1)!4^{n-1}}$ for $n>0$.)
Call this germ $g_0$.
Other germs
If we interpret the square root as a multi-valued function, there are in principle 3 other branches obtained by flipping the sign of one or both radicals. But $\sqrt{1-\sqrt{z}}~\sim \sqrt{(1-z)/2}$ is not analytic around $z=1$, so around $z=1$ there are only two (distinct) germs $g_0$ and $-g_0$.
A: We have
$$ \sqrt{1+\sqrt{z}} = \sqrt{2}\left(1+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{16^n n}\binom{4n - 2}{2n-1} (z-1)^n\right) $$
Here is a very general method, in the particular case of an algebraic function, to prove the identity.
I — Notations
Let $f$ be the function $z \mapsto \sqrt{1+\sqrt{1+z}}/\sqrt 2$, which is holomorphic in a neighbourhood of zero. We take a determination of the square root holomorphic around 1 and such that $\sqrt 1 = 1$. Note that in order to simply the computation, I shifted the variable and normalized the value at zero.
II — Algebraic equation
The function $f$ obviously satisfies the algebraic equation
$$ (f(z)^2 - 1/2)^2 = \frac{1+z}{4}, $$
or, equivalently, $P(f(z), z) = 0$, where
$$ P(Y, z) = 4Y^4 - 4Y^2 -z. $$ 
III — Differential equation
The function $f$ satisfies the following linear differential equation :
$$ 16 z (z+1) f''(z) + 8(1+2z) f'(z) - f(z) = 0 $$
This is a general fact that an algebraic function satisfies a linear differential equation with polynomial coefficient.
IV — Recurrence
Write $f(z) = \sum_{n\geqslant 0}  u_n z^n$. Thus
$$ 16 z (z+1) f''(z) + 8(1+2z) f'(z) - f(z) = 16n(n-1)u_n + 16 (n+1)n u_{n+1} + 8 (n+1) u_{n+1} + 16 n u_n - u_n $$
which implies that
$$ (4 n - 1)(4n+1)u_n + 8(n+1)(2n+1)u_{n+1} = 0. $$
V — Resolution
We check easily that the sequence defined by
$$v_n = \frac{(-1)^{n-1}}{16^n n}\binom{4n - 2}{2n-1}$$
if $n>0$ and $v_0 = 1$
satisfies the first order recurrence above. Since $u_0 = 1 = v_0$, we can conclude that
$$u_n = v_n$$
VI — Automation
Here is a Maple session showing how to automate the proof of steps II to V.

> with(gfun):
> f := sqrt(1+sqrt(1+z))/sqrt(2);
                                                              1/2 1/2  1/2
                                                  (1 + (1 + z)   )    2
                                             f := ------------------------
                                                             2

> holexprtodiffeq(f, y(z));      
                                                                         / 2      \
                                              /d      \        2         |d       |
                          {-y(z) + (8 + 16 z) |-- y(z)| + (16 z  + 16 z) |--- y(z)|, y(0) = 1}
                                              \dz     /                  |  2     |
                                                                         \dz      /

> diffeqtorec(%, y(z), u(n));    
                                         2                         2
                              {(-1 + 16 n ) u(n) + (24 n + 8 + 16 n ) u(n + 1), u(0) = 1}
> rsolve(%, u(n));
                                                                        n
                                                   GAMMA(2 n - 1/2) (-1)
                                              -1/2 ----------------------
                                                      1/2
                                                    Pi    GAMMA(2 n + 1)

You have to use the duplication formula for $\Gamma$ to retrieve the binomial coefficient.
A: Let $w = \sqrt{1 + \sqrt{z}}$. Since taking square roots of complex numbers is unpleasant, let's get rid of them to get a nicer equation:
$$(w^2 - 1)^2 = z$$
Since we want to expand around $z = 1$, we should change variables: set $u = z - 1$. Our problem now is to find $a_0, a_1, a_2, \ldots$ so that $w = a_0 + a_1 u + a_2 u^2 + \cdots$ satisfies
$$(w^2 - 1)^2 = 1 + u$$
First, by neglecting terms of order $u$ or higher, we find that $({a_0}^2 - 1)^2 = 1$, so
$$a_0 \in \{ \pm \sqrt{2}, 0 \}$$
(Officially, what we are doing is working in the power series ring $\mathbb{C} [[ u ]]$ modulo $u$.) Now, we divide into cases.


*

*We ignore the case $a_0 = 0$, because this corresponds to a ramification point in the associated Riemann surface and so $w$ has no power series here. (If you try to do the computation by brute force, you will find that all the coefficients are $0$!)

*Take $a_0 = \sqrt{2}$. Neglecting terms of order $u^2$ or higher, we get
$$(w^2 - 1)^2 = ((\sqrt{2} + a_1 u)^2 - 1)^2 = (1 + 2 \sqrt{2} a_1 u)^2 = 1 + 4 \sqrt{2} a_1 u = 1 + u \pmod{u^2}$$
so $a_1 = \frac{1}{8} \sqrt{2}$. Now, neglecting terms of order $u^3$ or higher, we get
$$(w^2 - 1)^2 = 1 + u + \left( \frac{5}{16} + 4 \sqrt{2} a_2 \right) u^2 = 1 + u \pmod{u^3}$$
so $a_2 = - \frac{5}{128} \sqrt{2}$. And so on.

*Take $a_0 = -\sqrt{2}$. Observe that nothing changes if we substitute $-w$ for $w$, so the coefficients for this expansion must be the same as the ones computed in (2), except with the opposite sign.
