Not getting the correct asymptotic behaviour when sending a small parameter to zero

Question

I want to solve this equation for $z$:

$-\frac{2\pi^2}{\beta^2}z^2+\frac{2i\pi^2u}{\beta^2}z^3+z^4=0$

$\beta$ is a positive real constant, $u$ is a real variable ranging from $0$ to $2\pi$.

The code is simply just the Solve function of Mathematica:

s1 = Solve[-((2*Pi^2)/(β^2))*z^2 + ((2*I*Pi^2*u)/(β^2))*z^3 + z^4 == 0,z]

Mathematica gives me these four solutions:

$z=0,z=0,z=\pi\left(\frac{-i\pi u}{\beta^2}\pm\frac{\sqrt{-\pi^2u^2+2\beta^2}}{\beta^2} \right),$

Now I am interested in adding a term $i\eta$ in the equation. The original motivation for this is that this polynomial is the denominator of a function which I have to integrate in the complex plane. Since there is a singularity in $z=0$ I apply the physicist method of adding a small imaginary term, which I will then send to zero at the end.

The Mathematica solutions of

$i\eta-\frac{2\pi^2}{\beta^2}z^2+\frac{2i\pi^2u}{\beta^2}z^3+z^4=0$

which correspond to the code

s2 = Solve[I*η - ((2*Pi^2)/(β^2))*z^2 + ((2*I*Pi^2*u)/(β^2))*z^3 + z^4 == 0,z]

now are very long and "bad looking". Of course I expect this, because it is applying the formula for the solution of an algebraic equation of degree four, which is well know to be very long and messy. However, I would expect, in the limit $\eta \longrightarrow 0$, to recover the previous solutions (of course I will send $\eta$ after using the residue theorem, here I just wanted to check the consistency of the solutions).

So for example, if $z_1$ is the first solution of the new equation, I would expect that $\lim_{\eta \longrightarrow 0}z_1=0$. This is however not the case, as you can check. The same happens for the other ones, I do not get back the old solutions. In terms of code I would write

pole1 = z/.s2[[1]]
Limit[pole1, η->0]

and the limit apparently does not match with the old solution obtained with the first equation.

I would appreciate if you could indeed solve the equations with your Mathematica and confirm the behaviour that I am talking about.

What is the reason behind this? I am missing something? I must recover the old solutions if I send $\eta$ to zero, right? So why does Mathematica not recover them?

Also note: I am interested in the symbolic form, not numerical values.

Answer 1

You could try using AsymptoticSolve instead:

AsymptoticSolve[
    I η - (2 Pi^2/β^2) z^2 + (2 I Pi^2 u/β^2) z^3 + z^4 == 0,
    z,
    η->0
] //Normal //TeXForm

$\left\{\left\{z\to -\frac{\sqrt{i \beta ^2} \sqrt{\eta }}{\sqrt{2} \pi }\right\},\left\{z\to \frac{\sqrt{i \beta ^2} \sqrt{\eta }}{\sqrt{2} \pi }\right\},\left\{z\to \frac{\pi \left(-\sqrt{2 \beta ^2-\pi ^2 u^2}-i \pi u\right)}{\beta ^2}-\frac{\beta ^6 \eta }{4 \pi ^3 \left(-i \pi ^2 u^2 \sqrt{2 \beta ^2-\pi ^2 u^2}+i \beta ^2 \sqrt{2 \beta ^2-\pi ^2 u^2}+\pi ^3 u^3-2 \pi \beta ^2 u\right)}\right\},\left\{z\to \frac{\pi \left(\sqrt{2 \beta ^2-\pi ^2 u^2}-i \pi u\right)}{\beta ^2}-\frac{\beta ^6 \eta }{4 \pi ^3 \left(i \pi ^2 u^2 \sqrt{2 \beta ^2-\pi ^2 u^2}-i \beta ^2 \sqrt{2 \beta ^2-\pi ^2 u^2}+\pi ^3 u^3-2 \pi \beta ^2 u\right)}\right\}\right\}$

Answer 2

Execute

s1 = Solve[-2*(Pi^2/\[Beta]^2)*z^2 + 2*I*((Pi^2*u)/\[Beta]^2)*z^3 + 
z^4 == 0, z]; s2 = 
Solve[I*\[Eta] - 2*(Pi^2/\[Beta]^2)*z^2 + 
2*I*((Pi^2*u)/\[Beta]^2)*z^3 + z^4 == 0, z]; 

and then compare

s1 /. {\[Beta] -> 0.7, u -> \[Pi]}

with

s2 /. {\[Eta] -> 0, \[Beta] -> 0.7, u -> \[Pi]}

to see that both equations numerically have the same solutions.