# 09-240/Classnotes for Thursday September 17

## 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.

WARNING: The notes below, written for students and by students, are provided "as is", with absolutely no warranty. They can not be assumed to be complete, correct, reliable or relevant. If you don't like them, don't read them. It is a bad idea to stop taking your own notes thinking that these notes can be a total replacement - there's nothing like one's own handwriting! Visit this pages' history tab to see who added what and when.

## 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)