Asymptotic expansion of an algebraic equation Consider $z^3 − z^2 = \lambda$ where $z ∈ \mathbb{C}$, $\lambda \geq 0$. Find the first 2 leading terms in the asymptotic expansion of all the roots of the equation near $\lambda = 0$.
My approach
For $\lambda =0$, the equation has a repeated root, $z=0$.
Let $z(\lambda) \sim z_0 + \sqrt\lambda z_1 + \lambda z_2 + \lambda^{3/2} z_2 + \cdot \cdot$
Can someone comment on my approach?
 A: Let us first solve $z^2-z^3=a^2$ for small complex $a$ and small complex $z$.
It is equivalent to
$$z(1-z)^{1/2}=\pm a.$$
It is well known how to solve $z(1-z)^{1/2}=b.$ Observe that for $|z|<1$, the function
$f(z)=z(1-z)^{1/2}$ is holomorphic and satisfies $f(0)=0,f'(0)=1$ Therefore, is has a holomorphic inverse function $f^{-1}$ in a certain neighborhood of $0$.
Then $z(1-z)^{1/2}=b$ is equivalent to $z=f^{-1}(b).$ The solutions of
$z^2-z^3=a^2$ are therefore $z=f^{-1}(\pm a)$.
This implies that the solutions of $z^3-z^2=\lambda$ or $z^2-z^3=-\lambda$ are $z=f^{-1}(\pm i\sqrt{\lambda})$, where $\sqrt{\lambda}$ denotes the principal value of the square root in the complex domain. Given that the holomorphic function $f^{-1}(y)$ in the neighborhood of $y=0$ has a convergent power series expansion, this justifies
the approach in the question for small $\lambda$ and small $z$, where we have actually $$f^{-1}(y)=c_1 y+c_2 y^2+...$$
Altogether the expansion for the two roots of  $z^3-z^2=\lambda$ close to $0$ is
$$z=\pm c_1i\sqrt\lambda-c_2\lambda+...,$$
where the coefficients can be found by inserting into $z^3-z^2=\lambda$. Of course $c_1$ is not uniquely determined since there are two roots, but once $c_1$ has been chosen, the other coefficients are determined. Except for this sign ambiguity, we find $c_1=1$ and $c_2=1/2$.
The equation $z^3-z^2=\lambda$, $\lambda$ small has a third root near $z=1$.
Here, the implicit function theorem can be applied directly because $g(z)=z^3-z^2$ satisfies $g(1)=0$ and $g'(1)=1$. Hence there exist a holomorphic function $h$ in the neighborhood of $0$ with $h(0)=1$ such that $z=h(\lambda)$. Here we have an expansion
$$z=h(\lambda)=1+h_1\lambda+h_2\lambda^2+...$$
for small $\lambda$. The coefficients can again be found by inserting into the equation, for example $h_1=1$.
A: For the only real root
Clearly, for $\lambda > 0$, the equation $z^3 - z^2 = \lambda$ always has one real root $z_1$ and two non-real conjugate complex roots $z_2, z_3$,
since the discriminant is $-\lambda(27\lambda+4) < 0$.
Let us analyze the only real root $z_1$.
By using the Lagrange inversion theorem [1], it is easy to get
\begin{align}
z_1 &= 1 + \sum_{n=1}^\infty \frac{(3n-2)!}{n!(2n-1)!}(-1)^{n-1} \lambda^n\\
&= 1+\lambda-2\lambda^2+7\lambda^3-30\lambda^4+143\lambda^5-728\lambda^6 + \cdots
\end{align}
(Note: If $0 < \lambda < \frac{4}{27}$, the series is convergent.)
Remark: Also see Interval of convergence of Lagrange's infinite series
Reference
[1] https://en.wikipedia.org/wiki/Lagrange_inversion_theorem
