Polynomials with degree \(n\) always have \(n\) roots, no matter if it's real, complex, or multiple roots.

Obviously, the real roots represent the intersection of x-axis.

But how about complex roots?! Does it say anything about the graph or not? If yes, can we know anything about it? Thank you! ^_^

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TopNewestYou could probably think of a polynomial \(f: \mathbb{C} \rightarrow \mathbb{C}\) as a mapping from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{2}\) (since \(\mathbb{R}^{2}\) and \(\mathbb{C}\) can be taken to be equivalent). If \((z, w) = f(x, y)\), then the zeroes are then \(f\)'s intersections with the plane \((z, w) = (0, 0)\).

Since this can be hard to visualise, you could perhaps also graph \(|f(x, y)|\) in 3D. The zeroes would then of course intersect the plane \(|f(x, y)| = 0\). Since \(|f(x, y)|\) is always nonnegative, the zeroes are then also the global minima of the function. Probably.

I might just be spouting nonsense though.

Also, while \(f\) always has \(\deg(f)\) complex roots, without counting for multiplicity you have your good old Fundamental Theorem of Algebra.

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What if we think about simple xy-plane, does this still work?

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Take \(y = x^{2}+1\), which has roots \(\pm i\); you can see that the parabola doesn't intersect the x-axis, but all that tells you is that its roots are nonreal. Maybe the focus and directrix might tell you a bit more, but I doubt it.

If you have a polynomial \(f\) with even \(\deg(f)\) of constant term 0, it passes through the origin. If you compare it with a congruent graph translated in the ±ve y direction with no real roots, you can deduce that the product of the roots is ±ve if the graph is "smiling" and ∓ve if the graph is "frowning", even if all the roots are nonreal.

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well if z axis is thought as imaginary axis!!how about that!!!u should get a 3 d shape

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What do you mean? Every polynomial with complex coefficients should have complex (2D) inputs and complex (also 2D) outputs so you would need a 4D graph.

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no it is still 3d.the x-z plane is like an input plane!!!

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Good!

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f (x) = 0 is the thing you mentioned.

f (z) = f (x + j y) = F (x)+ j G (y) = 0 is for complex numbers.

When we say Z = 0, we mean F (x) + j G (y) = 0. It is an x y- plan of both F (x) = 0 and G (y) = 0.

With Taylor's expansion, we are able to solve a polynomial equation with complex coefficients for all roots. There are n roots for polynomial of power n. This method allows a 1-dimensional search for 2-dimensional complex roots. Without which a two dimensional search for complex roots is tedious; I would say it as incorrect approach.

I invented a theorem which can obtain all roots for n-polynomial equations with complex coefficients. Theoretically, there is no limit to n allowed. I attempted up to z^10 + .... = 0 for real coefficients and up to z^5 + ... = 0 for complex coefficients. n roots for n power polynomial equations is true. However, I only completed in hands with z^8 + ... = 0 and z^4 + ... = 0 respectively. No intermediate term or coefficient is missing and they are very general.

Tell you an interesting thing. For cubic x^3 + p x^2 + q x + r = 0 of complex coefficients, cubic formula shall give all the roots, provided proper combination of two surds times omega are chosen correctly for each case. Hard to tell unless you study and investigate your own. Have a pleasant trial!

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