<s>[INST] Proof similarity of complex power series Let be the functions
$$
f(x)=\sqrt{2}\frac{\cos(\frac{\pi}{4}-\frac{1}{2}\arctan(x))}{(1+x^2)^{1/4}} \qquad g(x)=\frac{1}{\sqrt{1-x}}
$$
Obviously, they are holomorphic functions in a neighborhood of $0$. The series of $g(z)$ is well known and using Mathematica we have
$$
f(x)= 1+\frac{x}{2}- \frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}-\frac{231x^6}{1024}-\frac{429x^7}{2048}+\dotsc
$$
$$
g(x)= 1+\frac{x}{2}+ \frac{3x^2}{8}+\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}+\frac{231x^6}{1024}+\frac{429x^7}{2048}+\dotsc
$$
Exercise. Proof the absolute value of the coefficients of the two series are the same.
Attemp. With trigonometric manipulation I get
$$
f(x)=\frac{1+x+\sqrt{1+x^2}}{\sqrt{2}(1+x^2)^{3/4}\sqrt{1+\frac{1}{\sqrt{1+x^2}}}}
$$
My idea is use the Cauchy Integral Theorem to calculate the coefficients 
$$
a_n = \frac{1}{2\pi i}\int_C \frac{f(z)}{z^{n+1}}dz
$$
and with changes of variables get something that seems to
$$
\frac{1}{2\pi i}\int_C \frac{g(z)}{z^{n+1}}dz
$$ 
But if I work in $\mathbb{C}$ I need to define $\log(1+z^2)$ and work with it in changes of variables doesn't look so friendly. 
Can you give me a hint to prove it?
 [/INST] Hint
Let $\phi(z)=\sum_k a_k z^k$.
The first step is to see how to express $a_0+a_1 z-a_2 z^2 -a_3 z^3 +a_4 z^4+\ldots$ using $\phi$.
The four roots of $X^4=1$ being $\lbrace \pm 1, \pm i \rbrace$. You have to solve the system:
$$\begin{pmatrix}1&1&1&1\\1&i&-i&-1\\1^2&i^2&(-i)^2&(-1)^2\\1^3&i^3&(-i)^3&(-1)^3 \end{pmatrix} \begin{pmatrix} a\\ b \\c \\d \end{pmatrix}=\begin{pmatrix} 1\\ 1 \\-1 \\-1 \end{pmatrix}$$
As the solutions are $a=d=0$, $b=\frac{1-i}{2}, c=\frac{1+i}{2}$ you obtain:
$$a_0+a_1 z-a_2 z^2 -a_3 z^3 +a_4 z^4+\ldots = \frac{1}{2} \left((1-i)\phi(ix)+(1+i)\phi(-ix) \right)$$

Here you then have to prove that near $0$:
$$f(x)=\frac{1}{2} \left(\frac{1-i}{\sqrt{1-ix}}+\frac{1+i}{\sqrt{1+ix}} \right)$$
Using your expression of $f$ this is the same to prove that near $0$:
$$(1+i) \sqrt{1-ix}+(1-i) \sqrt{1+ix}=\sqrt{2}\frac{1+x+\sqrt{1+x^2}}{\sqrt{1+\sqrt{1+x^2}}}=2\sqrt{x+\sqrt{1+x^2}}$$
which seems easier.
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