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


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Class notes for today

• Convention for today: x,y,a,b,c,d,... will be real numbers; z,w,u,v,... will be complex numbers

Dream: Find a field \mathbb C that contains \mathbb R and also contains an element i such that i^2=-1

Implications:

b \in \mathbb R \Rightarrow bi \in \mathbb C

a \in \mathbb R \Rightarrow a+bi \in \mathbb C

c,d \in \mathbb R \Rightarrow c+di \in \mathbb C

\Rightarrow (a+bi)+(c+di) must be in \mathbb C

=(a+c)+(bi+di)
=(a+c)+(b+d)i
=e+fi

(a+bi)(c+di)=(a+c)+(b+d)i

=a(c+di)+bi(c+di)
=ac+adi+bic+bidi
=ac+bdi^2 + adi+bci
=(ac-bd)+(ad+bc)i
=e+fi
0_C=0+0i
1_C=1+0i
(a+bi)+(c+di)=0+0i
-(a+bi)=(-a)+(-b)i
a+bi \neq 0 \Rightarrow (a,b) \neq 0

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

(a+bi)(x+yi)=(ax-by)+(ay+bx)i=1+0i
ax-by=1 (1)
bx+ay=0 (2)
a,b are given
x,y unknowns

b \times (1) abx-b^2y=b

a \times (2) abx+a^2y=0

\Rightarrow a^{2}y+b^{2}y=-b
y=\frac{-b}{a^{2}+b^{2}}
x=\frac{a}{a^{2}+b^{2}}

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

(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 }}

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

with +: (a,b)+(c,d)=(a+c,b+d)

(a+bi)+(c+di)=(a+c)+(b+d)i

\times :(a+bi)(c+di)=...you know what

• 0 = you know what

• 1 = you know what

Theorem:

  1. \mathbb C is a field
  1. (0+1i)^2=(0,1)^2=i^2=-1_{C}=(-1,0)
  1. \mathbb R \rightarrow \mathbb C by a \rightarrow a+0i

Proof: F_{1},F_{2},F_{3},...

Example: F_{5} (distributivity)

• Show that z(u+v)=zu+zv

Let z=(a+bi)

u=(c+di)
v=(e+fi)

When a,b,c,d,e,f \in \mathbb R

(a+bi)[(c+di)+(e+fi)]=(a+bi)(c+di)+(a+bi)(e+fi)=(ac-bd)+\ldots

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