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I have a polynomial equation $-(a-ib)e^{(4\pi i/3)}(\sqrt{2}i+x^3/\sqrt{3})x- (a+ib) e^{(2\pi i/3)}(\sqrt{2}ix^3+1/\sqrt{3})=0$ with the conditions

$a^{2}+b^{2} \leq 1$,

$1/2 \geq a \geq -1$, and

$\sqrt{3}/2 \geq b \geq -\sqrt{3}/2$.

I want numerical solutions for this equation. I tried in Mathematica, but ended up with messy solutions. Also, I couldn't impose conditions on $a$ and $b$. Please help me in solving this.

Sunita
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  • The case $a=b=0$ satisfies your constraints and leads to all $x \in \mathbb{R}$ as a solution. Else, for numerical solutions you need to specify numerical values for $a,b$ and you will get two separate quartics (after separating the real and imaginary parts). It might have no solution. But, you can look for solutions by turning two quartics into a single cubic: $$ x^4 = f(x), x^4 = g(x) \implies f(x)=g(x)$$ with the understanding that a solution to your problem must satisfy $f(x)=g(x)$, but not all (or even any) points that satisfy $f(x)=g(x)$ solve your problem. – Michael Oct 09 '18 at 09:37
  • "I want numerical solutions for this equation. I tried in Mathematica, but ended up with messy solutions. Also, I couldn't impose conditions on a and b. Please help me in solving this." - You should probably learn more about Mathematica. It certainly allows you to search for roots numerically, AND impose conditions (they are called Assumptions there). Please ask a question on mathematica.stackexchange if using Mathematica for this task is convenient for you – Yuriy S Oct 09 '18 at 09:46
  • I am assuming you are looking for $x \in \mathbb{R}$ (you did not specify). – Michael Oct 09 '18 at 10:14
  • Thank you both. I am looking for $x \in \C $. – Sunita Oct 10 '18 at 17:38

1 Answers1

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It is not clear if you are seeking over $x \in \mathbb{R}$ or $x \in \mathbb{C}$. Nevertheless here are some particular cases.

  • Suppose $a=b=0$. Then any $x \in \mathbb{C}$ is a solution.

  • Suppose $(a+ib) = (\rho i)e^{-2 \pi i/3}$ for some $\rho \in \mathbb{R}$, $\rho \neq 0$. Then $$ (a+ib)e^{2\pi i /3} = \rho i \quad , \quad (a-ib)e^{4 \pi i /3} = -\rho i $$ So your equation reduces to $$ (ix\sqrt{2} + x^4/\sqrt{3}) = (ix^3\sqrt{2} + 1/\sqrt{3})$$ which is satisfied when $x=1$.

  • Suppose $(a+ib) = \rho e^{-2\pi i/3}$ for some $\rho \in \mathbb{R}$, $\rho \neq 0$. Then $$ (a+ib)e^{2\pi i /3} = \rho \quad , \quad (a-ib)e^{4 \pi i /3} = \rho $$ So your equation reduces to $$ (ix\sqrt{2} + x^4/\sqrt{3}) = -(ix^3\sqrt{2} + 1/\sqrt{3})$$ which has no solution over $x \in \mathbb{R}$ because it would require $x^4 = -1$.

Michael
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  • The cases with $(a+ib) = (\rho i) e^{-2\pi i/3}$ and $(a+ib) = \rho e^{-2\pi i/3}$ can have $\rho$ scaled as small as desired to satisfy the inequality constraints given in the question. – Michael Oct 09 '18 at 13:03
  • Hope this would help me! Thanks. – Sunita Oct 10 '18 at 17:39
  • Well, I gave an infinite number of solutions that satisfy your constraints, the simplest being $a=b=x=0$. I also showed an infinite number of examples of $a,b$ that satisfy the constraints but for which no real-valued $x$ exists for a complete solution. It is not clear if you are choosing $a,b$ or if you are given those. If you are given $a,b \in \mathbb{R}$, you can represent $(a-ib)e^{4\pi i/3}$ and $(a+ib)e^{2 \pi i/3}$ by $c$ and $\overline{c}$ (the complex conjugate) and then solve the quartic over $x\in\mathbb{C}$. The quartic is a 4th degree polynomial in $x$ with complex coefficients. – Michael Oct 10 '18 at 18:51