09-240/Classnotes for Thursday September 17: Difference between revisions

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• <math>\Rightarrow (a+bi)+(c+di)</math> must be in <math>\mathbb C</math>
• <math>\Rightarrow (a+bi)+(c+di)</math> must be in <math>\mathbb C</math>


• <math>= (a+c)+(bi+di)</math>
• <math>=(a+c)+(bi+di)</math>


• <math>= (a+c)+(b+d)i</math>
• <math>=(a+c)+(b+d)i</math>


• <math>=e+fi</math>
• <math>=e+fi</math>
Line 43: Line 43:
• <math>(a+bi)(c+di)=(a+c)+(b+d)i</math>
• <math>(a+bi)(c+di)=(a+c)+(b+d)i</math>


• <math>= a(c+di)+bi(c+di)</math>
• <math>=a(c+di)+bi(c+di)</math>


• <math>=ac+adi+bic+bidi</math>
• <math>=ac+adi+bic+bidi</math>
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• <math>(a+bi)+(c+di)=0+0i</math>
• <math>(a+bi)+(c+di)=0+0i</math>


• <math>-(a+bi)=(-a)+(-b)i</math>
(rest of notes will be added in 1/2-hour)

• <math>a+bi \neq 0 \Rightarrow (a,b) \neq 0</math>

• Find another element of <math>\mathbb C</math>, <math>x+yi</math> such that <math>(a+bi)(x+yi)=(1+0i)</math>

• <math>(a+bi)(x+yi)=(ax-by)+(ay+bx)i=1+0i</math>

• <math>ax-by=1</math> (1)

• <math>bx+ay=0</math> (2)

• <math>a,b</math> are given

• <math>x,y</math> unknowns

• <math>b \times (1)</math> <math>abx-b^2y=b</math>

• <math>a \times (2)</math> <math>abx+a^2y=0</math>

• <math>\Rightarrow a^{2}y+b^{2}y=-b</math>

• <math>y=\frac{-b}{a^{2}+b^{2}}</math>

• <math>x=\frac{a}{a^{2}+b^{2}}</math>

• (Note: We can divide since we assumed that <math>(a,b) \neq 0</math>

• <math>(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>\mathbb C</math> be the set of all pairs of real numbers <math>{(a,b)}={a+bi}</math>

• with <math>+: (a,b)+(c,d)=(a+c,b+d)</math>

• <math>(a+bi)+(c+di)=(a+c)+(b+d)i</math>

• <math>x:(a+bi)(c+di)=</math>...you know what

• 0 = you know what

• 1 = you know what

• Thm:

• 1. <math>\mathbb C</math> is a field

• 2. <math>(0+1i)^2=(0,1)^2=i^2=-1_{C}=(-1,0)</math>

• 3. <math>\mathbb R \rightarrow \mathbb C</math> by <math>a \rightarrow a+0i</math>

• Proof: <math>F_{1},F_{2},F_{3},...</math>

• Example: <math>F_{5}</math> (distributivity)

• Show that <math>z(u+v)=zu+zv</math>

• Let <math>z=(a+bi)</math>

• <math>u=(c+di)</math>

• <math>v=(e+fi)</math>

• When <math>a,b,c,d,e,f \in \mathbb R</math>

• <math>(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)

Revision as of 18:18, 17 September 2009

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

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)