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When creating a system of equations for complex cubic roots, I came across a paradoxical example of a cubic with 7 complex roots. Correct me if I'm wrong, but the fundamental theorem of algebra explicitly states that an nth degree polynomial has n roots, and they can either be real or complex. $$a(u+vi)^3+b(u+vi)^2+c(u+vi)+d=0,$$ so $$au^3+bu^2+(c-av^2)u+d-by^2=0,$$ and $$3au^2-ay^2+2bu+c=0.$$ This worked until I tried $a=-3.9$, $b=5.3$, $c=-1.6$, $d=0.3$, and this happens: https://www.desmos.com/calculator/xumlqtnam3

Arthur
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user1085395
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    The fundamental theorem is about polynomials in one variable. You have two variables. When you split $u$ from $v$, you're losing information, so your equations are weaker and so you get more solutions. – B. Goddard Aug 22 '22 at 14:38
  • I haven't really been able to interpret your desmos work yet but the fundamental theorem of algebra says there are $n$ roots counting multiplicity, for example $y=x^2$ has a single root in the complex plane at $x=0$ but it has multiplicity $2$. – Sidharth Ghoshal Aug 22 '22 at 14:39
  • So you wanna combine @B.Goddard's insight on multivariate polynomials which is formally stated as https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem# with the fact that that we count roots with multiplicity, and if you do that all then the universe will be back in order – Sidharth Ghoshal Aug 22 '22 at 14:41
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    TLDR: your 3 equation system has $3 \times 3 \times 2 = 18$ roots actually but for that specific choice of $a,b,c,d$ you're able to make it so there are only $7$ distinct roots with some of those roots having high multiplicity – Sidharth Ghoshal Aug 22 '22 at 14:46
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    I don't understand the other comments. Apart from the inconsitency where you switch from using v to using y, your main mistake is that you forgot one of the binomial coefficients when expanding the cube. Once you fix that, you should find at most 3 solutions. – Jaap Scherphuis Aug 22 '22 at 15:14
  • "the universe will be back in order" ? What do you mean with this statement ? – Peter Aug 22 '22 at 15:52
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    With those values of $a,b,c,d$, the solutions are $1.034205806 + i0$, $0.1623842763 + i0.2191123781$, $0.1623842763 - i0.2191123781$. Only three solutions. – GEdgar Aug 22 '22 at 15:54
  • Real and imaginary part both have to be $0$. I cannot see where the imaginary part went to. – Peter Aug 22 '22 at 15:56
  • For a broader response to what I'm doing here, I'm solving a system of equations by dividing the original function into It's real and imaginary parts, "graphing" each function as a 3d plane, looking at the parts where they equal 0, then seeing where the resulting plots intersect, which should give 2 pairs of u and v and 1 u value at v=0, which are the roots of the function. My question is why there are more than 3 intercepts when graphed. – user1085395 Aug 22 '22 at 16:10
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    In both your $au^3+...by^2=0$ and $3au^2-ay^2...=0$ equations you need to change $y$ back to $v$. Also the second equation needs to be multiplied by $v$. – Somos Aug 22 '22 at 17:51
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    OP: go read Jaap Scherphuis's comment above again. Your problem is an elementary error expanding the polynomials. – JBL Aug 22 '22 at 19:37

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