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LN400

Maths question

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I'll get straight down to it:

I have a polynomial with a complex variable z and complex coeff's (real parts all zero):

z3 + 8i z2 - z + 42i, i being the imaginary unit.

 

I am wondering for which values of z does this polynomial give real numbers, that is numbers with no imaginary part.

I know I can write the equation

z3 + 8i z2 - z + 42i - r = 0 where r is a real number.

However, I am looking for keywords to look up, techniques to read up on and (hopefully) learn, different kind of plots, to be able to plot r as z moves across the complex plane.

The more suggestions the merrier.

 

I know that there can be more than one z in the original polynomial that produces the same r. You might say I want to know at which points does the polynomial turn all real.

Edited by LN400
typo

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Simplify it into

z (z2 + 8i z) - (z - 42i + r) = 0 

Dont know exactly how to help from there, but that simplifies the polynomial. 

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Thanks. Not sure how it simplifies the polynomial. I'm still stuck with a cubic imaginary mess. I know the roots of the original polynomial btw, they are

2i, -3i and -7i

so factorized it's

(z - 2i)(z + 3i)(z + 7i)

so I know that for these roots, I get the real number zero.

I have looked at various plots on wolfram alpha, some using

z = a + bi

 

The fact that all roots are imaginary thows me off quite a bit making it extra tricky to think how the polynomial "pops in and out of reality" so to speak, and where apart from zero that might happen.

 

Edited by LN400

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[edited tired + chit chat] 

Edited by WinkAllKerb''

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You basically want to find z numbers such that the complex part of the polynomial is zero.

Define z = a + ib with a,b real.

Your polynomial becomes:
P(z) = z3 + 8iz2 - z + 42i = (a + ib)3 + 8i(a + ib)2 - (a + ib) + 42i

Develop the cube and square and rearrange a bit to obtain the polynomial with separated real and imaginary parts:
P(z) = (a3 - 3ab2 - 16ab - a) + i(-b3 + 3a2b + 8a2 - 8b2 - b + 42)

If you want your polynomial to return real numbers, you need to find a number z = a + ib such that the imaginary part of P is zero:
Im[P(z)] = -b3 + 3a2b + 8a2 - 8b2 - b + 42 = 0

This is basically all you need to solve this kind of problem, there are methods that are more efficient but this one is reliable and fairly simple as it gives the imaginary (real) part of z as a function of the real (imaginary) part, so you won't need to find roots (except maybe for the domain of b).

For this particular polynomial, the solution is z = a + ib with:
b ∈ ( -∞, -7] ∪ [-3, -8/3) ∪ [2, ∞), and
a = ± √[(b3 + 8b2 + b - 42) / (3b + 8)]

 

You could also use the variable r as you did, and solve z3 + 8i z2 - z + 42i - r = 0 for z directly. This amounts to solving a third order polynomial for z, keeping in mind that z may be complex. If you seek a numerical solution, this is probably easiest/fastest method since root-finding algorithms are pretty efficient.

If you want a function that maps z to r, you will need to solve it analytically. 3rd order polynomials are doable but surprisingly annoying for what they're worth. The solution for a and b is quite complicated in itself (a being the square root of a rational function), so I don't expect the math to be trivial.

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Math and me are natural enemies. But i once had to attend the lectures in analysis and algebra.

So there was the professor writing an argument on the board (yes, there still existed blackboards). He turned around to the auditorium and said with a firm voice:

"You surely can intuitively accept that this is trivial !"

350 students and you could hear a needle drop.

He turned back to the blackboard, rubbing his chin. Then he rushed to the door "Give me a minute !".

It took him two minutes and he returned with a grin on his face:

"I was right, it IS trivial !".

 

 

ok, this is an old one ...

:-)

 

Edited by Green Baron

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12 hours ago, Gaarst said:

[for a] function that maps z to r, you will need to solve it analytically. 3rd order polynomials are doable but surprisingly annoying for what they're worth. The solution for a and b is quite complicated in itself (a being the square root of a rational function), so I don't expect the math to be trivial.

This is where things get wonky for real, no pun intended. If I try that, then I get that the real part a is complex with a non-zero imaginary part and the imaginary part b is also complex. Then again, it is possible to factorize the Q(z) = P(z) - r since Q(z) is just another polynomial with the same number of roots as P(z).

Edited by LN400
Blooper concerning b being complex and not pure Re or Im

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