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This Week

We started the section complex dynamical systems. We introduced the complex numbers by looking at the following equations:

x+5=10.

x+10=5.

5x=6.

x^2=2.

x^2=-1.

We said that the ‘need’ for complex numbers is analogous to the fact that we can solve the first equation using natural numbers \mathbb{N} but we require

  • negative integers to solve the second equation (\mathbb{Z})
  • fractions to solve the third equation (\mathbb{Q})
  • real numbers to solve the third equation (\mathbb{R})
  • and finally, complex numbers to solve the fourth equation (\mathbb{C})

We then asked do we need more and more complicated number systems to solve more complicated equations and we mentioned in passing that the complex numbers are algebraically closed and hence we don’t need more complicated number systems to solve polynomial equations:

a_nx^n+\cdots+a_1x+a_0=0.

This is usually where people stop but we asked what about equations like xy=yx+1

We also showed that complex numbers multiply in the rotating manner: in the notation z=(\text{modulus},\text{argument});

(r,\theta)\times (s,\alpha)=(rs,\theta+\alpha).

On Monday we will talk about roots of unity and then we will start examining the iterates of the complex linear map.

In the tutorial we discussed some aspects of the homework.

Complex Number Exercises

One of the important things about this module is that you get some more exposure to complex numbers. For those of you who hope to go teaching we won’t be doing enough complex number algebra to get ye going great guns so you will either have to get comfortable with the complex number stuff you did in first year or else look at stuff like this.

Problems

2009 Autumn Q. 1(d)

Additional Exercises

  1. Show that 2\text{ Re}(z)=z+\bar{z} and 2\text{ Im}(z)=z-\bar{z} for all z\in\mathbb{C}.
  2. Show that \overline{(w+z)}=\bar{w}+\bar{z} and \overline{(wz)}=\bar{w}\bar{z} for all w,\,z\in\mathbb{C}. Hence prove the Conjugate Root Theorem — which exhibits the symmetry between i and -i with respect to the real numbers. In particular \sqrt{-1}=i is imprecise: i^2=-1 is better/
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