09-240/Classnotes for Thursday September 17
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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{ x:(a+bi)(c+di)= }[/math]...you know what
• 0 = you know what
• 1 = you know what
• Thm:
• 1. [math]\displaystyle{ \mathbb C }[/math] is a field
• 2. [math]\displaystyle{ (0+1i)^2=(0,1)^2=i^2=-1_{C}=(-1,0) }[/math]
• 3. [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)
