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 $W=Z^{2}$ 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: $x,y,a,b,c,d,...$ 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

• $b\times (1)$ $abx-b^{2}y=b$

• $a\times (2)$ $abx+a^{2}y=0$

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

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

$x={\frac {a}{a^{2}+b^{2}}}$

• (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)