※This image is by the painter ChatGPT and is named “A surreal imaginary world with a cyberpunk aesthetic.”
Problem Formulation
Regarding the problem of finding \(\small z\) for a complex quadratic equation:
\[ \small \alpha z z^\ast+\beta z+\gamma z^\ast+\delta = 0, \]
it seems like it can be solved analytically, so let us try that. Let us assume that \(\small \alpha,\beta,\gamma,\delta\) can be complex numbers. Including \(\small z^2,z^{\ast2}\) makes things complicated, so for now we will limit the quadratic terms to only the terms above.
Calculation
Each constant coefficient is represented as:
\[ \small \begin{align*} &\alpha = \alpha_1 + \alpha_2i \\ &\beta = \beta_1 + \beta_2i \\ &\gamma = \gamma_1 + \gamma_2i \\ &\delta = \delta_1 + \delta_2i. \end{align*} \]
Let denote \(\small z=x+yi\), and calculate the real and imaginary parts of the equation to get
\[ \small \begin{align*} &\alpha_1(x^2+y^2)+(\beta_1+\gamma_1)x+(\gamma_2-\beta_2)y+\delta_1 = 0 \\ &\alpha_2(x^2+y^2)+(\beta_2+\gamma_2)x+(\beta_1-\gamma_1)y+\delta_2 = 0. \end{align*} \]
Multiplying the second equation by \(\small \alpha_1/\alpha_2\) and subtracting it from the first equation gives us
\[ \small \left(\beta_1+\gamma_1-\frac{\alpha_1}{\alpha_2}(\beta_2+\gamma_2) \right)x+\left(\gamma_2-\beta_2-\frac{\alpha_1}{\alpha_2}(\beta_1-\gamma_1) \right)y+\left(\delta_1-\frac{\alpha_1}{\alpha_2}\delta_2\right) = 0. \]
Therefore,
\[ \small \begin{align*} y &= -\frac{\beta_1+\gamma_1-\frac{\alpha_1}{\alpha_2}(\beta_2+\gamma_2) }{\gamma_2-\beta_2-\frac{\alpha_1}{\alpha_2}(\beta_1-\gamma_1)}x-\frac{\delta_1-\frac{\alpha_1}{\alpha_2}\delta_2}{\gamma_2-\beta_2-\frac{\alpha_1}{\alpha_2}(\beta_1-\gamma_1)} \\ &= px+q \end{align*} \]
can be obtained. By substituting this into the first equation, we obtain a quadratic equation for \(\small x\), which means we can find \(\small x\) using the formula for solving quadratic equations. Note that just like with any quadratic equation, there are two solutions. If we rewrite the equation using \(\small y=px+q\), it can be expressed as:
\[ \small \begin{align*} &ax^2+bx+c = 0 \\ &a = \alpha_1\left(1 + p^2\right) \\ &b = 2\alpha_1pq + (\beta_1+\gamma_1)+(\gamma_2-\beta_2)p \\ &c = \alpha_1q^2+(\gamma_2-\beta_2)q+\delta_1. \end{align*} \]
Since we are assuming that \(\small x\) is real, if this quadratic equation has no real solutions, then there is no solution.
Extension
If we think about it in the above way, it seems that we can also analytically solve the problem of finding \(\small z\) for
\[ \small \alpha z^2+\beta z z^\ast+\gamma z^{\ast2}+\delta z+\epsilon z^\ast+\zeta = 0\]
by extending complex quadratic equations, so let us try that. Let us assume that \(\small \alpha,\beta,\gamma,\delta,\epsilon,\zeta\) can be complex numbers.
Each constant coefficient is represented as:
\[ \small \begin{align*} &\alpha = \alpha_1 + \alpha_2i \\ &\beta = \beta_1 + \beta_2i \\ &\gamma = \gamma_1 + \gamma_2i \\ &\delta = \delta_1 + \delta_2i \\ &\epsilon = \epsilon_1 + \epsilon_2i \\ &\zeta = \zeta_1 + \zeta_2i. \end{align*} \]
Let denote \(\small z=x+yi\), and by calculating the real and imaginary parts of the equation, we need to find \(\small x,y\) that satisfies
\[ \small \begin{align*} &(\alpha_1+\beta_1+\gamma_1)x^2+(\beta_1-\alpha_1-\gamma_1)y^2+2(\gamma_2-\alpha_2)xy+(\delta_1+\epsilon_1)x+(\epsilon_2-\delta_2)y+\zeta_1 = 0 \\ &(\alpha_2+\beta_2+\gamma_2)x^2+(\beta_2-\alpha_2-\gamma_2)y^2+2(\alpha_1-\gamma_1)xy+(\delta_2+\epsilon_2)x+(\delta_1-\epsilon_1)y+\zeta_2 = 0. \end{align*} \]
By subtracting the first equation multiplied by
\[ \small \frac{\beta_2-\alpha_2-\gamma_2}{\beta_1-\alpha_1-\gamma_1} \]
from the second equation, we can obtain the equation with \(\small y^2\) eliminated. Since this is in the form:
\[ \small ax^2+bxy+cx+dy+e = 0, \]
we can get the solution:
\[ \small y = -\frac{ax^2+cx+e}{bx+d}. \]
Just substitute this into the first equation and find the solution for \(\small x\). Since this equation is a fourth-order equation in \(\small x\), the solution can be found using Ferrari’s formula. Well, there may be some question as to whether this can be called an analytical solution… Since it is a solution to a fourth-order equation, there can be a maximum of four combinations of solutions.
What Is It Used for?
You may be wondering what the point is of solving such a problem that sounds like a college student’s homework, but I felt that when trying to numerically solve a partial differential equation like
\[ \small \frac{\partial \psi(x,t)}{\partial t}\frac{\partial \psi^\ast(x,t)}{\partial t} = \frac{m^2c^4}{\hbar^2}\psi(x,t)\psi^\ast(x,t)+c^2\frac{\partial \psi(x,t)}{\partial x}\frac{\partial \psi^\ast(x,t)}{\partial x}, \]
problems like this would appear even when using explicit methods. When discretized, it becomes
\[ \small \frac{\psi_{j,k+1}-\psi_{j,k}}{\Delta t}\frac{\psi^{\ast}_{j,k+1}-\psi^{\ast}_{j,k}}{\Delta t} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\ \small \qquad\qquad = \frac{m^2c^4}{\hbar^2}\psi_{j,k}\psi^{\ast}_{j,k}+c^2\frac{\psi_{j+1,k}-\psi_{j-1,k}}{2\Delta x}\frac{\psi^{\ast}_{j+1,k}-\psi^{\ast}_{j-1,k}}{2\Delta x}, \]
so if we set \(\small z=\psi_{j,k+1}\),
\[ \small zz^{\ast}-\psi^{\ast}_{j,k}z-\psi_{j,k}z^{\ast}+\psi_{j,k}\psi^\ast_{j,k}-\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\ \small \left(\frac{m^2c^4}{\hbar^2}\psi_{j,k}\psi^{\ast}_{j,k}+c^2\frac{\psi_{j+1,k}-\psi_{j-1,k}}{2\Delta x}\frac{\psi^{\ast}_{j+1,k}-\psi^{\ast}_{j-1,k}}{2\Delta x} \right)\Delta t=0 \]
will become the problem to be solved. Note that the time is calculated using forward differences, and the coordinates are calculated using central differences. In this case, since \(\small \alpha_2=0\), we can simply define it as:
\[ \small y = -\frac{\beta_2+\gamma_2 }{\beta_1-\gamma_1}x-\frac{\delta_2}{\beta_1-\gamma_1}. \]
However, since there are two solutions, you won’t know which one to adopt until you try it. It should be noted that if an appropriate solution is not chosen, an oscillatory solution may result. By the way, I thought that if we were to solve the Dirac equation numerically, we would solve this equation, but I’m not sure at the moment…A differential equation containing the square of a first derivative with respect to a complex function may branch into four solutions over time. In other words, the Dirac equation may be a forcible conversion of a quartic equation into a linear equation with four unknowns.
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