How to solve for the roots of a 4th degree polynomial with complex coefficients? Given the following equation
$c_4Z^4+c_3Z^3+c_2Z^2+(c_1+\frac{i}{\beta})Z^1-\frac{i}{\beta} =0$
where $Z$ lies in the complex domain. Each of the coefficients, $c_n$ for $n=1:4$, are real and can be either positive or negative. They are known, but left as variables here for the sake of generality. My objective is to determine expressions for each of the 4 roots and plot them versus $\beta$. 
I have done this numerically using Matlab's roots function and got the correct results, but in order to plot each of the roots as a function of $\beta$ in a meaningful way, I have to sort them to appear as continuous curves or plot them as points in a scatter plot. Either way, it takes way too long to explore the full complex domain for $\beta$. This is a whole other issue, so don't lose focus here.
That being said, I'm wondering if I can solve for each of the roots in the form of an analytic expression as a function of $\beta$? If so, I could program each of the expressions as anonymous functions and save a bunch of time exploring the complex domain for $\beta$.
Unfortunately, my subpar math skills have left me scratching my head on how to proceed. I've looked at a bunch of different examples/tricks on determining the roots for 4th degree polynomials, but they all seem to apply to real polynomials with real coefficients. In the polynomial I've expressed above, $\beta$ can be complex. So again, my question is, can I solve for each of the roots in the form of an analytic expression as a function of $\beta$ and if so, how do I proceed given the complex nature of the polynomial?
 A: Here is a solution to the general quartic compactified. Given,
$$Ax^4+Bx^3+Cx^2+Dx+E = 0$$
divide by $A$ to get the simpler,
$$x^4+ax^3+bx^2+cx+d=0$$
Then the four roots are,
$$x_{1,2} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}+\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag1$$
$$x_{3,4} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}-\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag2$$
where,
$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(v_1^{1/3}\zeta_3+\frac{b^2 - 3 a c + 12 d}{v_1^{1/3}\zeta_3}\right)$$
with $v_1$ any non-zero root of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
and a chosen cube root of unity $\zeta_3^3 = 1$ such that $u$ is also non-zero. (Normally, just use $\zeta_3=1$, but not when $a^3-4ab+8c = 0$.)
P.S. This is essentially the method used by Mathematica, though much simplified for aesthetics.
A: First depress the equation by translating the variable $x\to x-\dfrac a4$, giving
$$x^4+px^2+qx+r=0.$$
Now factor this in two quadratic polynomials in such a way that the cubic term is missing (mind the reuse of the coefficient names)
$$(x^2+ax+b)(x^2-ax+c)=x^4+(-a^2+b+c)x^2+a(c-b)x+bc.$$
By identification,
$$\begin{cases}-a^2+b+c=p\implies ab+ac=a^3+pa,\\ac-ab=q,\\bc=r\end{cases}.$$
From the first two equations, you draw $ab,ac$ in terms of $a$ and plug in the third.
$$\left(a^3+pa-q\right)\left(a^3+pa+q\right)=4ra^2,$$
giving by magic a cubic equation in $a^2$,
$$(a^2)^3+2p(a^2)^2+(p^2-4r)a^2-q^2=0.$$
Once solved for $a$, you get $b,c$ and solve the two quadrics.

For the cubic $x^3+ax^2+bx+c$, first depress by the change $x\to x-\dfrac a3$ to the form
$$x^3+px+q=0.$$
Then set 
$$x=u-\frac{p}{3u}$$ so that by substitution
$$u^3+q-\frac{p^3}{27u^3}=0$$ or
$$(u^3)^2+qu^3-\frac{p^3}{27}=0.$$
Solve the quadratic for $u^3$ and you are... done.
