Singularity on power series expansion of implicit function I am looking for help to understand the singular point coming up on my power series expansion of an implicit function:
$$x^2 z^4 (x^2-(z-k)^2)=(z-k)^4(z^2-1)$$
where $z$ is complex-valued function of $x$ and $k$ is a pre-defined constant.
When $x$ is very small, I assume $z(x)$ is a power series of $x$ and obtain one of the power series expansions as below:
$$z(x) = k + \sqrt{\frac{k^2 \left(k^2+\sqrt{k^4+4k^2-4}\right)}{2(1-k^2)}} x$$
The problem I have is when I calculate the value of $z(x)$ when $k$ is set close to $1$, say $k=0.9$ with small value of $x=0.005$, my value from my power series approximation of $z(x)$ diverges from the numerical values of $z(x)$ from the first equation. In fact, the solution from first equation stay finite for all $k$ but the power series blow up due to the singularity $k=1$ on the 2nd equation.
Is there anyone aware of the situation that the series expansion of a function introduces a singular point that is absent from the original function?
 A: Given three real variables $\,x,y,z\,$ define the polynomial
$$ P:=(z\!-\!y)^4 (z^2\!-\!1) \!-\! x^2 z^4 (x^2\!-\!(z\!-\!y)^2) \tag1 $$
of degree $6$ in $\,z.\,$ Factor $\,P\,$ into linear factors
$$ P=(1+x^2)\prod_{n=1}^6(z-z_n) \tag2 $$
where the $6$ real roots $\,z_1<\dots<z_6\,$ depend on $\,x\,$ and $\,y.\,$
Assuming $\,y\,$ is constant, then the $6$ roots power series expansions in
$\,x\,$ have coefficients which depend on $\,y.\,$
The first root has the expansion
$$ z_1(x,y) = -1 + \frac{x^2}{2(1+y)^2} -
  \frac{(7+11y)x^4}{8(1+y)^5} + O(x^6). \tag3 $$
The sixth root then is given by
$$ z_6(x,y) = -z_1(x,-y). \tag4 $$
It will be convenient to define
$$ Y := \sqrt{-4 + 4y^2 + y^4}. $$
The second root has the expansion
$$ z_2(x,y) \!=\! y \!-\!
 \frac{y\sqrt{y^2\!+\!Y}}{\sqrt{2(1-y^2)}}x \!+\!
 \frac{y(2\!-\!y^2)(y^4\!+\!y^2(Y\!+\!2)\!-\!2)}{2Y(1-y^2)^2}x^2
 \!+\!O(x^3). \tag5 $$
The fifth root then is given by
$$ z_5(x,y) = z_2(-x,y). \tag6 $$
The third root has the expansion
$$ z_3(x,y) = y - \frac{y\sqrt{y^2-Y}}{\sqrt{2(1-y^2)}}x
+ c_2 x^2 + O(x^3) \tag7 $$
where
$$ c_2 = \frac{y(2-y^2)(-3y^2+3y^4+y^6+(1-y^2-y^4)Y)}
{(1-y^2)^2Y(Y-y^2)}. $$
The fourth root then is given by
$$ z_4(x,y) = z_3(-x,y). $$
The radius of convergence of these series is not yet known to me.
Note that the value of $\,Y\,$ is imaginary unless
$\,|y| \ge \sqrt{\sqrt{8}-2} \approx 0.9101797.\,$

The cases where $\,y=\pm1\,$ causes some of the denominators of the
power series expansions to equal zero. This is clear for $\,z_1,z_6\,$
where the denominators are zero for $\,z_1\,$ if $\,y=-1\,$ and the
same for $\,z_6\,$ if $\,y=1.\,$ It would seem to be the case for
$\,z_3,z_4\,$ if $\,y=\pm1\,$ but this is not so since the limit as
$\,y\to\pm1\,$ of the coefficients is finite. In fact,
$$ z_3 = 1 - x - x^2 - 8x^3 - 71x^4 -\frac{1427}2x^5 - 7753x^6 + O(x)^7. $$
Thus, if $\,y=\pm1\,$ then three of the power series have finite
coefficients and the other three have coefficients which "blow up".
The resolution in these three cases is that the solutions are really
Puiseux series. Thus,
$$ z = 1 + t + \frac{11}6t^2 + \frac1{12}t^3 + \frac{5653}{648}t^4 
- \frac{62587}{19445}t^5 + \frac{6147}{32}t^6 + O(t)^7 $$
where $\,t\,$ is one of the three roots of $\,x^2 = -2t^3\,$ and if
$\,x\,$ is real, two of the roots are conjugate complex numbers.
The general case is given in the Wikipedia article:

Puiseux's theorem, sometimes also called the Newton–Puiseux
theorem, asserts that, given a polynomial equation $\,P(x,y)=0\,$ with complex coefficients, its solutions in $y$, viewed as functions of
$x$, may be expanded as Puiseux series in $x$ that are convergent in some neighbourhood of $0$. In other words, every branch of an algebraic curve
may be locally described by a Puiseux series in $x$ (or in $x − x_0$ when considering branches above a neighborhood of $x_0 \ne 0$).

