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.