# Roots of the Following Series in the Complex Plane

1. Apr 9, 2014

### The Wanderer

1. The problem statement, all variables and given/known data

For the series x^n - x^(n-1) - x^(n-2) ... - x^(0) the roots seem to be x = 2 and the circle around the complex plane with radius i or 1 I'm not sure how you would say it as n approaches infinity. Here's an image of the roots where n = 15.

http://www.wolframalpha.com/input/?...x^11-x^10-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x-1

2. Relevant equations

3. The attempt at a solution

http://www.wolframalpha.com/input/?...x^11-x^10-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x-1

This is beyond me but it really intrigued me as to why the roots seem to create a circle in the complex plane.

I have almost complete Calculus AB and have read a lot into other advanced topics but I have no idea how I would go about finding all of the roots for this infinite series or why they seem to create a circle. I was hoping someone would as I thought it was pretty cool.

2. Apr 9, 2014

### electricspit

I can tell you that the Fundamental Theorem of Algebra states that each nth degree polynomial has exactly n roots.

If you think of the graph of a polynomial, the roots are where it crosses the x-axis (the "zeroes"). In the case where, let's say, a parabola doesn't cross at all, we still need to have two roots according to the F.T.A..

This is where complex analysis comes in, because the roots will no longer be real. For n=2 you can look at this case:

$x^2+1=0$

Will have solutions:

$x^2=-1$
$x=\pm i$

As you can see the solutions are complex conjugates of each other. This gives the symmetry in the complex plane. For an $n^{th}$ degree polynomial, if can have two cases:

$n\in \mathbb{Z} \text{ even}$
$n\in \mathbb{Z} \text{ odd}$

If n is even, there can be at most $\frac{n}{2}$ sets of imaginary roots. If n is odd there can be at most $\frac{n-1}{2}$ sets of imaginary roots because the last root will not have a conjugate, and therefore must be real.

Another good example of this is when:

$x^4+1=0$
$x=(-1)^{1/4}$

So there must be exactly 4 roots since it is a 4th degree polynomial. Using a bit of complex analysis:

$e^{i\pi}=-1$
$x=(e^{i\pi})^{1/4}$

Now in complex analysis you will learn eventually that you can't just take the 1/4 power of complex numbers without considering the fact that they have multi-value properties, so in this case:

$x=\exp{[i(\frac{\pi}{4}+\frac{2n\pi}{4})]}$

Since every $2n\pi$ gives the same value (a full rotation in the circle), we must make sure we divide this by 4 as well to get the correct last answer. Now n ranges from 0 to 3. If you plot these points in the complex plane they will form a circle of radius 1. You can use deMoivre's formula to plot them:

$\exp{i\theta}=\cos{\theta}+i\sin{\theta}$

Does this make sense? I hope I didn't go over your head, I'm not sure what Calculus AB is. To plot using deMoivre's formula, just consider $\cos{\theta}$ as the x-coordinate and $\sin{\theta}$ as the y-coordinate.

Last edited: Apr 9, 2014
3. Apr 9, 2014

### The Wanderer

Oh okay. So it is just giving the points on the circle using Euler's formula. That's pretty cool. And if there is an infinite number of points it just gives you a circle since every point is filled. Thank you very much.

4. Apr 9, 2014

### electricspit

It is very cool! I just finished a course on complex analysis yesterday (wrote the final!). It was one of the best classes I've taken, you learn a lot about crazy geometry. I recommend taking a course, or reading a book. The book I used was Brown and Churchill: Complex Variables and Applications, very readable and approachable.

5. Apr 9, 2014

### The Wanderer

Wow I actually will get that book. I was expecting a 70$or 100$+ textbook but that book only costs less than 30\$. Thank you very much.

6. Apr 9, 2014

### Dick

The negative terms in your expression form a geometric series. You should look up how to sum it. If you do that and do some algebra, you'll see that the roots of your expression solve (2-x)x^15-1. The root at 2 is only approximate, likewise the roots on the unit circle.

Last edited: Apr 10, 2014
7. Apr 9, 2014

### micromass

Staff Emeritus
Actually, before you get that book, I highly recommend: https://www.amazon.com/Visual-Complex-Analysis-Tristan-Needham/dp/0198534469
It's one of the best math books ever written. It really motivates complex analysis in a very beautiful way. Of course, it's not a textbook and you will need to get a textbook on complex analysis later to be more rigorous and fill in some gaps. But if you read visual complex analysis, you will have an incredibly intuition and love for complex analysis.

Last edited by a moderator: May 6, 2017