## Complex Polynomial with Certain Properties

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### Complex Polynomial with Certain Properties

Hello, I have recently been wondering if there exists a polynomial with complex coefficients, p(x) , such that the only x values where the imaginary part of p(x) is equal to zero lie on the unit circle in the complex plane.

brenok
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### Re: Complex Polynomial with Certain Properties

Maybe i*x^4 - 2*i*x^2 + i ?

I don't know if this is exactly what you want...

Dark Avorian
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### Re: Complex Polynomial with Certain Properties

that definitely doesn't work as the complex zeroes of the imaginary part certainly don't all lie in the unit circle
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### Re: Complex Polynomial with Certain Properties

That doesn't work, as expanding i(a+bi)^4-2i(a+bi)^2+i gives ia^4-4a^3*b-6i*a^2*b^2-2ia^2+4a*b^3+4a*b+ib^4+2ib^2+i, which has an imaginary part of
a^4-6a^2*b^2-2a^2+b^4+2b^2+1=0 , which is certainly not a circle. Hopefully that also explains the process to go through... if you could get an imaginary part of a^2+b^2-1, then a^2+b^2 =1 and you would have the unit circle.

brenok
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### Re: Complex Polynomial with Certain Properties

Oh, I thought that the roots needed to be in the unit circle, not form a unit circle... well, I knew that this was too easy

Now I will need to try another way.

Edit:I just thought of something now - for the solutions form a circle, wouldn't the polynomial need infinite roots? And a polynomial with infinite roots wouldn't have infinite degree?

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### Re: Complex Polynomial with Certain Properties

I certainly don't want every point in the unit circle to be a root; I want every point on the unit circle to be real valued, and real valued only on the unit circle. So, all the roots must be on the unit circle, but not every point on the unit circle must be a root; every point must be real.

brenok
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### Re: Complex Polynomial with Certain Properties

Now I noticed something: looks like the imaginary roots of the first equation I wrote are, in fact, in the unit circle. Could you please check that again?

This is the expanded imaginary part: http://www.wolframalpha.com/input/?i=a% ... %5E2+%2B+1

Wolfram says that the roots are +-1/sqt(2) +- i/sqrt(2)

Arent these complex numbers with arguments 45º, 135º, 225º and 315º and absolute value 1?

jestingrabbit
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### Re: Complex Polynomial with Certain Properties

Pretty sure that the answer is "No".

Notice that a complex polynomial must take on every possible value, or else you violate the fundamental theorem of algebra. The polynomial is also continuous, and so is its imaginary and real parts. Therefore, to satisfy your condition, the polynomial would have to take on every real value on the circle, but the circle is compact. The continuous image of a compact set is compact, and the reals aren't. Contradiction.
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### Re: Complex Polynomial with Certain Properties

Thanks! That's some very cool reasoning there, certainly rigorous enough for me! And as to the other poster, while the roots may be all on the unit circle, I also had wanted every real value on the unit circle as well; it turns out that doesn't work.

Qaanol
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### Re: Complex Polynomial with Certain Properties

There is, however, a complex function that is analytic except for a simple pole at -1, which sends the rest of the unit circle to the real line, the interior of the unit disk to the lower half of the complex plane, and everything outside the unit circle to the upper half of the complex plane. The function can be made so it hits every complex value except i.

Can you construct such a function?
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### Re: Complex Polynomial with Certain Properties

Hmmmm, sounds like quite an interesting function! I'll think about, maybe I'll think up a solution, perhaps it has some sort of polynomial expansion I could use, but it seems unlikely.

edit: it seems i/(ln(-x)) maps the unit circle to the real line, has a pole at x=-1, and sends the interior of the circle to the lower half of the plane, and the exterior to the top half... but it hits every value but 0, not every value but i... I might have messed up though, didn't really write anything out but in text.

Qaanol
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### Re: Complex Polynomial with Certain Properties

Natural log needs a branch cut, it is not continuous.

My approach involves
Spoiler:
Rotating the Riemann sphere.

If you are still having fun calculating on your own, don’t read the following spoiler because I give away one such function.
Spoiler:
f(z) = i·(z-1)/(z+1)

Which means,
f(x+iy) = ( -2y + i·(x2+y2-1) ) / ( (x+1)2 + y2 )
wee free kings

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### Re: Complex Polynomial with Certain Properties

I can visualize what is going on with rotating the sphere, and see how your solution works in that sense, but I am unaware as to how I can find a function for this.

I wonder if there is anything fundamental about mapping the unit circle to the real line that always results in the interior and exterior of the circle being mapped to the upper and lower half of the plane (in no particular order, as you could take the opposite) ; can this be proved for all continuous functions? For my almost solution, all I tried to do was map the unit circle to the reals, and the other properties (to my pleasure) naturally manifested.

jestingrabbit
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### Re: Complex Polynomial with Certain Properties

This movie will probably help.

Edit: and then I forgot the movie address...
Last edited by jestingrabbit on Fri Dec 28, 2012 2:55 am UTC, edited 2 times in total.
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Qaanol
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### Re: Complex Polynomial with Certain Properties

Heptadecagon wrote:I can visualize what is going on with rotating the sphere, and see how your solution works in that sense, but I am unaware as to how I can find a function for this.

I wonder if there is anything fundamental about mapping the unit circle to the real line that always results in the interior and exterior of the circle being mapped to the upper and lower half of the plane (in no particular order, as you could take the opposite) ; can this be proved for all continuous functions? For my almost solution, all I tried to do was map the unit circle to the reals, and the other properties (to my pleasure) naturally manifested.

If you can visualize what is going on, picture it on the Riemann sphere. The unit circle is the equator, and the real line is the prime meridian (minus the north pole).

You can think of the surface of the Riemann sphere as being deformable like rubber (but the sphere itself retains shape!) so you can push stuff around on the sphere smoothly. But no matter how you push it around, you can never make anything pass through other stuff.

Now it should be clear that any smooth bijective mapping that sends the equator to the prime meridian, must also send the northern hemisphere to either the eastern or the western hemisphere, and the southern hemisphere to the other one.

You can find such a function by using the projection that sends a point on the complex plane to the Riemann sphere, the rotation that turns the sphere how you want, and the projection that sends a point on the sphere to the plane. Compose those three operations, and you’ll get a single function that does what you want.

The maths are a little complex, but if you draw triangles you should be all right. Notably, a point at distance r from the origin in the complex plane, gets sent to the point at height (r2-1)/(r2+1) on the Riemann sphere, in the same radial direction. You can invert that relation to project the other way.

Also, I edited the function I was thinking of into my previous post here, probably while you were posting. So if you want to calculate the function on your own, don’t open the second spoiler there.
wee free kings

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### Re: Complex Polynomial with Certain Properties

Thanks! I understand most of that except that I cannot yet work with the equations to project the sphere onto the plane and such, I'll have to learn that at some point.

Qaanol
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### Re: Complex Polynomial with Certain Properties

Heptadecagon wrote:Thanks! I understand most of that except that I cannot yet work with the equations to project the sphere onto the plane and such, I'll have to learn that at some point.

It’s a just taking a straight line from the top of the sphere to the point on the plane, and seeing which other point of the sphere that line goes through.
wee free kings

mr-mitch
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### Re: Complex Polynomial with Certain Properties

You could have a constant polynomial (which has no zeroes). Since they have no zeroes, all of the zeroes lie on the unit circle in the complex plane.

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### Re: Complex Polynomial with Certain Properties

Then indeed, every point on the unit circle would be real valued, but I want the unit circle to be the only place where the polynomial is real valued; a constant would leave the function real everywhere.
--------------- Now I'll try and extend my work.... (recall Im(a+bi)=b)
Suppose I wanted a polynomial, p(x), where Im(p(a+bi))=0 if an only if q(a,b)=0 (q is a designated polynomial in two variables)
if q(a,b)=b than potential options for p(x) are C*x, where C is any real (non-zero) constant.
if q(a,b)=a^2+b^2-1 , we have just proved no polynomial p(x) can satisfy these conditions.
if q(a,b)=ab+1 , a solution is p(x)=x^2+2i
if q(a,b)=a^2-b^2-1 , a solution is ix^2-i
------------------------------------------------------------------ This all seems rather interesting to me. So, how can all this be studied?
How can we tell what polynomials q(a,b) create solutions for p(x) ? This all seems to me to be a neat way to relate polynomials in two variables to polynomials in one variable.
Any thoughts/ideas on this new little theory?