Student: | Timothy E. Brown |
---|---|

Office: | CoRE 434 |

School: | Morgan State University |

E-mail: | tibro26@morgan.edu |

Project: | Experiments with Geometric Modulus Principle for Polynomials |

Geometric Modulus Principle: Let p(z) be a nonconstant polynomial. If p(z_{0})=0, then every direction at z_{0} is an ascent direction for |p(z_{0})|. If p(z_{0})≠0 then the cones of ascent and descent direction at z_{0} partition the unit disc centered at z_{0} into alternating ascent and descent sectors of equal angle π ⁄ k where k≥1is the smallest index with p^{k}(z_{0})≠0. (Bahman Kalantari 2011)

- Week 1:
- I settled down on Sunday and began working on research in Complex Analysis. After orientation and talking with my advisor and roommates, I became excited to see the result of this program come mid July. On Monday I spoke with Dr. Kalantari and discussed research problems.Friday I gave my presentation.
- Week 2:
- After meeting my mentor I started look over Polynomial Interploation. To create a smooth uniform graph we agreed to use Hermite Interplotaion. If given zeros and critical points we should be able to construct a polynomial. Now we want to take this polynomial and see if it has the properties of Geometric Modulus Principle
- Week 3:
- I started to write the program for Hermite Interploation which will generate the polynomials I need for the Geometric Modulus Principle. My next goal is to create a program in Mathmatica which will display the modulus graph and the 2D contour which will highlight direction of ascent and descent.
- Week 4:
- I got my Hermite Interpolation code to work for complex numbers. Now I am attempting to code Newton's Method so I can visualize the roots that are in the convex hull. By putting all of these programs together I should get a nice 2D contour which highlight gradients and show roots in the convex hull.
- Week 5:
- This week I constructed a polynomial. In general you are given n and k ∫ (z
^{n}- 1)^{k}dz. This creates a polynomial p (z) s.t p^{'}(z)= p^{''}(z) =. ... p^{k}(z) = 0 for roots of z^{n}- 1. Example suppose n=3 and k=1. That is ∫ (z^{3}- 1)^{1}dz=z^{4}/4 -z. This generates a polynomial p(z) s.t p^{'}(z)=0 for roots of z^{3}- 1. - Week 6:
- I'm writing a code for the polynomial I created last week. This code will;
- Take the modulus of a complex polynomial p(z) (we write p(z) as u(x,y)+iv(x,y) then its modulus is f(x,y) = √(u
^{2}+ v^{2})=|p(z)|. Example: p(z)=z^{2}-1. Set z=x+iy. Then f(x,y)=√((x^{2}-y^{2}-1)^{2}+(2xy)^{2}). - Take a point z
_{0}where p^{'}(z)_{0})=0 but p^{''}(z)_{0})≠0, e.g. if p(x)=z^{2}-1, z_{0}=0 is such a point. - Evaluate f(z
_{0})=|p(z_{0})|, then in a box containing z_{0}for many points evaluate f(x,y). Whenever it is larger than f(z_{0}) color it red and if it is smaller than f(z_{0}) color it black.Then give the image.

- Take the modulus of a complex polynomial p(z) (we write p(z) as u(x,y)+iv(x,y) then its modulus is f(x,y) = √(u

- My Mentors