### General Information

Student: Timothy E. Brown CoRE 434 Morgan State University tibro26@morgan.edu Experiments with Geometric Modulus Principle for Polynomials

### Project Description

Geometric Modulus Principle: Let p(z) be a nonconstant polynomial. If p(z0)=0, then every direction at z0 is an ascent direction for |p(z0)|. If p(z0)≠0 then the cones of ascent and descent direction at z0 partition the unit disc centered at z0 into alternating ascent and descent sectors of equal angle π ⁄ k where k≥1is the smallest index with pk(z0)≠0. (Bahman Kalantari 2011)

### Weekly Log

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 ∫ (zn - 1)k dz. This creates a polynomial p (z) s.t p'(z)= p''(z) =. ... pk(z) = 0 for roots of zn- 1. Example suppose n=3 and k=1. That is ∫ (z3 - 1)1 dz=z4/4 -z. This generates a polynomial p(z) s.t p'(z)=0 for roots of z3 - 1.
Week 6:
I'm writing a code for the polynomial I created last week. This code will;
1. 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) = √(u2 + v2)=|p(z)|. Example: p(z)=z2-1. Set z=x+iy. Then f(x,y)=√((x2-y2-1)2+(2xy)2).
2. Take a point z0 where p'(z)0)=0 but p''(z)0)≠0, e.g. if p(x)=z2-1, z0=0 is such a point.
3. Evaluate f(z0)=|p(z0)|, then in a box containing z0for many points evaluate f(x,y). Whenever it is larger than f(z0) color it red and if it is smaller than f(z0) color it black.Then give the image.