OT: mathematical conundrum

[pamela]

I'm sure I follow what you say about a circle of roots.

Isn't the circle being used to illustrate the roots of -1? On the other hand, Cursitor wrote "The number of roots is the same as the highest power in any polynomial".

What Cursitor says is true enough without identifying any complex roots. Presumably if there are any complex roots to a polynomial then they would be in addition to the number of non-complex roots.

That would make his statement about the number of roots incorrect. If you see what I mean.

[Clive George]

No, roots of +1 is normal.

And he's right.

This is incorrect. Consider the cube roots of 1. How many are there? Three. How many aren't complex? One. I listed them above.

Ditto on 4th roots of 1 - four roots, two are complex (i,-i). Ditto on 5th - five roots, four are complex.

6th - six roots, four are complex. etc

Nope

It would, but they aren't in addition to it.

[Fredxxx]

A second order polynomial with real coefficients if it has complex roots, will have roots in conjugate pairs.

[pamela]

I'm not sure I follow your logic. In your examples you use the roots of

1 but Cursitor wasn't talking about how many roots exist of a given number. He is talking about the number of roots of a polynomial. Have I overlooked something here?

My understanding is that his statement is correct for all *Real* roots of a polynomial.

I would guess (forgive me if I'm wrong and this may be where I am missing something) that his observation about the number of roots was first made without realising the possible existence of Complex roots.

If that's so then any Complex roots which exist would be in addition to the number of Real roots and this would presumably make the total greater than the highest power in the polynomial.

Don't beat me up for not understanding because I'm trying my best!

[pamela]

Er, I'm lost. I don't know what a conjugate pair is and when I looked it up in Google I got some definition to do with acid bases in chemistry.

I'm afraid I've probably run out of the necessary steam to fully grasp mathematical conjugate pairs.

From what litle I can deduce from your link, doesn't it say that for every Complex root of a polynomial there is a second "conjugate" root?

In turn does this mean there can sometimes be more roots of a polynomial than the highest power it contains?

[Tim Streater]

Well the other coefficients are zero:

ax^4 + bx^3 + cx^2 + dx + e = 0 with a = 1, b = c = d = 0 and e = -1

so it's a polynomial.

[Graham Nye]

A Smith chart is a graphical tool for performing complex impedance [1] calculations in a constant impedance transmission line environment (such as a co-axial cable).

[1] Tautology alert: impedance is inherently a complex concept.

You don't need a Smith chart just to plot complex numbers. A simpler way of plotting them is an Argand diagram. This represents a complex number as a point on a 2-dimensional surface known as the complex plane. The real part of the number is plotted along the horizontal (x) axis and the imaginary part of the number is plotted along the vertical (y) axis.

An example of a couple of Argand diagrams with various complex numbers plotted on them can be seen at:

This is an engineering text so it uses j for the imaginary unit. Here's a mathematician's page, using i:

[Graham Nye]

A complex conjugate pair is a pair of complex numbers which have the same real part but have imaginary parts with opposite signs.

Try the page at:

[Clive George]

Yes.

Nth roots is just another way of saying a special case of a polynomial of power N. So eg "what's the cube root of 1" can be expressed as

x cubed -1 = 0

What is x?

That's a polynomial whose highest power is 3. Therefore we expect three roots, and indeed three is what we get. One real, two complex.

And that's not correct.

Here's an even simpler example

x squared + 1 = 0

How many *Real* roots of that polynomial are there?

The answer is zero, but there's two complex ones (x=i, x=-i).

This guess is also almost certainly wrong.

Which makes this statement wrong too.

Yes, I understand what you're saying, but it's simply wrong. You need to let go of the idea that anybody has said there are N real roots - nobody who knows what they're talking about has said that.

[pamela]

No more, please. My head is starting to spin. Arghhhh!

[pamela]

You wrote "let go of the idea that anybody has said there are N real roots" and maybe that's where I am going wrong. Google has found for me the statement: "A polynomial P(x) of degree n has exactly n roots, real or complex". As you said, I had incorrectly assumed the roots were always all real.

Can I ask you about some details of something else you wrote. I understand your example of "x squared + 1 = 0" having two roots (x=i, x=-i).

Does it mean that "(x to the power of 4) + 1 = 0" has four roots? Does it mean that "(x to the power of 9) + 1 = 0" has nine roots?

I can't easily see what those 4 or 9 roots would be even assuming they are complex numbers. Yet the statement says they are there.

[Clive George]

Well done :-)

This is how I remember it, but it might not help. Later on I'll do the specific cases.

Earlier I mentioned the Nth roots of 1 are on a circle.

Somebody else mentioned an Argand diagram, which is a normal graph. Except X is the axis of "real numbers" and Y the axis of "imaginary numbers".

The following ascii diagrams will only work in a fixed pitch font

| | |

------------+------------ | | |

The basic graph paper with no points on it. Let's mark a point corresponding to 1. Note I'm not going to make any attempt to scale this.

| | |

------------+----*------- | | |

Now one corresponding to i

• | |

------------+------------ | | |

One corresponding to -1

| | |

-------*----+------------ | | |

and -i

| | |

------------+------------ | | *

And finally a more complex number, 1+i

|

• | |

------------+------------ | | |

That should give you enough to be able to draw an argand diagram on a piece of paper and plot points on it.

Now, one of the nice things about the Nth roots of 1 is that they all lie on a circle drawn on this diagram _____ / | \ / | \ | | |

-------|----+----|-------- | | | \ | / \__|__/

(yes, I know it looks like an octagon, you try doing a circle in ascii :-) )

One of the roots is always +1

_____ / | \ / | \ | | |

-------|----+----*-------- | | | \ | / \__|__/

And the others are evenly spread round the circle.

So for the square roots we have +1 and -1

_____ / | \ / | \ | | |

-------*----+----*-------- | | | \ | / \__|__/

180 degrees apart

For the cube roots, we have +1, -0.5 + 0.866i, -0.5 -0.866i as mentioned before (root 3 divided by 2 = 0.866)

_____ * | \ / | \ | | |

-------|----+----*-------- | | | \ | / *__|__/

For the fourth roots, I'll draw the picture first

_____ / * \ / | \ | | |

-------*----+----*-------- | | | \ | / \__*__/

Those points are 1, i, -1, -i.

And that's the answer to your question - those are the fourth roots of 1.

Let's check if we're right

1*1*1*1 = 1 obviously, so that one's ok i*i*i*i = -1 * -1 = 1

-1 * -1 * -1 * -1 = 1 * 1 = 1

-i*-i*-i*-i = -1*-1*-1*-1 * i * i * i * i = 1 * 1 = 1

To get the 9 roots, you have to distribute the 9 points evenly round the circle. And it's simple trigonometry to work out the numbers.

The coordinates of a circle are real = cos (angle), imaginary = sin (angle) Angle is between the horizontal (real) axis on the right and the point - so increasing angle moves anticlockwise round the circle.

First 9th root of 1 is at angle zero, so real = 1, imaginary = 0, ie 1 + 0i.

Second is at angle 40 degrees. Real = cos(40), imaginary = sin(40), ie

0.766 + 0.643i Third at 80 degrees, 0.174 + 0.985i Fourth 120 degrees, -0.5 + 0.866i Fifth 160 degrees, -0.940 + 0.342i Sixth 200 degrees, -0.940 - 0.342i Seventh 240 degrees, -0.5 - 0.866i Eighth 280 degrees, 0.174 - 0.985i Ninth 320 degrees, 0.766 - 0.643i

If you bother to calculate the 9th power of all those numbers, you'll find they're all 1.

Another thing is if you take the first, 0.766 + 0.643i, and multiply it by itself, you get to the second - 0.174 + 0.985i. Multiply by

0.766+0.643i again and you get to the third, -0.5 + 0.866i. And so on around the circle.

If you multiply by the third, you go in steps of 80 degrees. Etc.

The general case is that if you multiply any point on the circle with any other point on the circle, the result is still on the circle - it'll be at the angle formed by the sum of the other two. I think that's quite a nice result, which is why I've spent quite so much time writing it up.

Now, who wants to introduce radians? :-)

[pamela]

Now that's a very neat representation.

I haven't tried to check any of the 9 roots you list are correct by substituting one back in the original polynomial (x to the power of 9 + 1 =

0) but I'm happy to accept that it would solve the equation.

It's a lovely property that you mentioned in which multiplying a given root by the first root gives the next root (nicer to my thinking than the property of summing the angles to multiply two roots). You don't get that with real roots. In fact I would guess you don't get that for the complex roots of any other polynomial of the 9th degree except the specific one we're talking about which is perhaps a special case.

Thank you for the explanation.

[Martin Bonner]

And (although this may be what you meant), writing complex numbers as r * e^(i theta)

[Clive George]

FWIW I've just realised I was actually answering x to the power of N -1 = 0, ie the Nth roots of +1 rather than -1, but it's the more interesting special case IMO.

[pamela]

I hadn't noticed that. As I said, I didn't test the results. I suspect it doesn't make much of a difference because the way you described it even the roots of +1 can be complex numbers.

Perhaps it is far too simplistic to assume that the ninth roots of -1 (as opposed to the ninth roots of +1) shown in your diagrammatic way would have the real and imaginary parts transposed. It probably is!

[pamela]

Well, I guess Clive's comments about radians went right above my head!

However the equaton you wrote reminds me of a wonderful result in which e^(i pi) + 1 is zero. You must already know this is the Euler Identity. Personally I find it very lovely for at least two reasons:

Two irrational numbers (each of great mathematical importance), one imaginary number, unity and zero are all linked in one equation.

Also that raising a real positive number to a power can actually result in the value -1.

Hey, I've got real world bathrooms to think about rather than this elegant but abstract stuff. :-)

[Clive George]

I was leading up towards that :-)

pi is 180 degrees, so you can see where that fits on the circle I've been talking about.

[Clive George]

Sqrt -1 = 1, -i

4th root -1 = +- root2/2 +- root2/2i (all four combinations of it)

Which is the same as the four eighth roots of +1 which aren't also the

4th roots of +1.

(Cube roots of -1 are the three sixth roots of +1 which aren't also the cube roots of +1.)

So if you took the 18th roots of +1, then removed the 9th roots of +1 from there, you'd get the circle showing the 9th roots of -1.

It's a rotation by half the angle between the roots. For odd roots, that comes out as the same as a mirror.

[pamela]

How very neat.

I wouldn't have guessed a rotation. To go from solving for -1 as opposed to

+1, my wild guess was that you might *swap* the real and imaginary axes whilst retaining the location of the points representing the roots.

How comes you know the alebgra of imaginary numbers so well?