09-240/Classnotes for Thursday September 17

09-240/Navigation Panel   Additions to the MAT 240 web site no longer count towards good deed points # Week of... Notes and Links 1 Sep 7 Tue, About, Thu 2 Sep 14 Tue, HW1, HW1 Solution, Thu 3 Sep 21 Tue, HW2, HW2 Solution, Thu, Photo 4 Sep 28 Tue, HW3, HW3 Solution, Thu 5 Oct 5 Tue, HW4, HW4 Solution, Thu, 6 Oct 12 Tue, Thu 7 Oct 19 Tue, HW5, HW5 Solution, Term Test on Thu 8 Oct 26 Tue, Why LinAlg?, HW6, HW6 Solution, Thu 9 Nov 2 Tue, MIT LinAlg, Thu 10 Nov 9 Tue, HW7, HW7 Solution Thu 11 Nov 16 Tue, HW8, HW8 Solution, Thu 12 Nov 23 Tue, HW9, HW9 Solution, Thu 13 Nov 30 Tue, On the final, Thu S Dec 7 Office Hours F Dec 14 Final on Dec 16 To Do List The Algebra Song! Register of Good Deeds Misplaced Material Add your name / see who's in!

NSERC - CMS Math in Moscow Scholarships

The Natural Sciences and Engineering Research Council (NSERC) and the Canadian Mathematical Society (CMS) support scholarships at $9,000 each. Canadian students registered in a mathematics or computer science program are eligible.

The scholarships are to attend a semester at the small elite Moscow Independent University.

Math in Moscow Program http://www.mccme.ru/mathinmoscow/

Application details http://www.cms.math.ca/Scholarships/Moscow

For additional information please see your department or call the CMS at 613-733-2662.

Deadline September 30, 2009 to attend the Winter 2010 semester.

Some links

• Dori Eldar's work on "mechanical computations": Machines as Calculating Devices and Computing the function [math]\displaystyle{ W=Z^2 }[/math] the hard way.
• The "Dimensions" video on "Nombres complexes", is at http://dimensions-math.org/Dim_reg_AM.htm (and then go to "Dimensions_5".

Class notes for today

• Convention for today: [math]\displaystyle{ x,y,a,b,c,d,... }[/math] will be real numbers; [math]\displaystyle{ z,w,u,v,... }[/math] will be complex numbers

Dream: Find a field [math]\displaystyle{ \mathbb C }[/math] that contains [math]\displaystyle{ \mathbb R }[/math] and also contains an element [math]\displaystyle{ i }[/math] such that [math]\displaystyle{ i^2=-1 }[/math]

Implications:

• [math]\displaystyle{ b \in \mathbb R \Rightarrow bi \in \mathbb C }[/math]

• [math]\displaystyle{ a \in \mathbb R \Rightarrow a+bi \in \mathbb C }[/math]

• [math]\displaystyle{ c,d \in \mathbb R \Rightarrow c+di \in \mathbb C }[/math]

• [math]\displaystyle{ \Rightarrow (a+bi)+(c+di) }[/math] must be in [math]\displaystyle{ \mathbb C }[/math]

[math]\displaystyle{ =(a+c)+(bi+di) }[/math]

[math]\displaystyle{ =(a+c)+(b+d)i }[/math]

[math]\displaystyle{ =e+fi }[/math]

[math]\displaystyle{ (a+bi)(c+di)=(a+c)+(b+d)i }[/math]

[math]\displaystyle{ =a(c+di)+bi(c+di) }[/math]

[math]\displaystyle{ =ac+adi+bic+bidi }[/math]

[math]\displaystyle{ =ac+bdi^2 + adi+bci }[/math]

[math]\displaystyle{ =(ac-bd)+(ad+bc)i }[/math]

[math]\displaystyle{ =e+fi }[/math]

[math]\displaystyle{ 0_C=0+0i }[/math]

[math]\displaystyle{ 1_C=1+0i }[/math]

[math]\displaystyle{ (a+bi)+(c+di)=0+0i }[/math]

[math]\displaystyle{ -(a+bi)=(-a)+(-b)i }[/math]

[math]\displaystyle{ a+bi \neq 0 \Rightarrow (a,b) \neq 0 }[/math]

• Find another element of [math]\displaystyle{ \mathbb C }[/math], [math]\displaystyle{ x+yi }[/math] such that [math]\displaystyle{ (a+bi)(x+yi)=(1+0i) }[/math]

[math]\displaystyle{ (a+bi)(x+yi)=(ax-by)+(ay+bx)i=1+0i }[/math]

[math]\displaystyle{ ax-by=1 }[/math] (1)

[math]\displaystyle{ bx+ay=0 }[/math] (2)

[math]\displaystyle{ a,b }[/math] are given

[math]\displaystyle{ x,y }[/math] unknowns

• [math]\displaystyle{ b \times (1) }[/math] [math]\displaystyle{ abx-b^2y=b }[/math]

• [math]\displaystyle{ a \times (2) }[/math] [math]\displaystyle{ abx+a^2y=0 }[/math]

[math]\displaystyle{ \Rightarrow a^{2}y+b^{2}y=-b }[/math]

[math]\displaystyle{ y=\frac{-b}{a^{2}+b^{2}} }[/math]

[math]\displaystyle{ x=\frac{a}{a^{2}+b^{2}} }[/math]

• (Note: We can divide since we assumed that [math]\displaystyle{ (a,b) \neq 0 }[/math]

[math]\displaystyle{ (a+bi)^{-1}=\frac{a}{a^{2}+b^{2}}+\frac{-b}{a^{2}+b^{2}}i=\frac{a-bi}{a^{2}+b^{2}}=\frac{\overline{a+bi}}{|a+bi|^{2}}=\frac{\mbox{conjugate}}{\mbox{norm squared }} }[/math]

Def: Let [math]\displaystyle{ \mathbb C }[/math] be the set of all pairs of real numbers [math]\displaystyle{ {(a,b)}={a+bi} }[/math]

with [math]\displaystyle{ +: (a,b)+(c,d)=(a+c,b+d) }[/math]

[math]\displaystyle{ (a+bi)+(c+di)=(a+c)+(b+d)i }[/math]

[math]\displaystyle{ \times :(a+bi)(c+di)= }[/math]...you know what

• 0 = you know what

• 1 = you know what

Theorem:

1. [math]\displaystyle{ \mathbb C }[/math] is a field

1. [math]\displaystyle{ (0+1i)^2=(0,1)^2=i^2=-1_{C}=(-1,0) }[/math]

1. [math]\displaystyle{ \mathbb R \rightarrow \mathbb C }[/math] by [math]\displaystyle{ a \rightarrow a+0i }[/math]

Proof: [math]\displaystyle{ F_{1},F_{2},F_{3},... }[/math]

Example: [math]\displaystyle{ F_{5} }[/math] (distributivity)

• Show that [math]\displaystyle{ z(u+v)=zu+zv }[/math]

Let [math]\displaystyle{ z=(a+bi) }[/math]

[math]\displaystyle{ u=(c+di) }[/math]

[math]\displaystyle{ v=(e+fi) }[/math]

When [math]\displaystyle{ a,b,c,d,e,f \in \mathbb R }[/math]

[math]\displaystyle{ (a+bi)[(c+di)+(e+fi)]=(a+bi)(c+di)+(a+bi)(e+fi)=(ac-bd)+\ldots }[/math]

• NEXT WEEK: Complex numbers have geometric meaning, geometric interpretation (waves)