Study Guides
(238,408)

Canada
(115,131)

University of Toronto St. George
(7,978)

Mathematics
(538)

MAT224H1
(48)

Sean Uppal
(42)

Final

# Summary notes 1

Unlock Document

University of Toronto St. George

Mathematics

MAT224H1

Sean Uppal

Summer

Description

MAT224H1a.doc n Complex n-Space C , Complex Matrices, Spectral Theorem C OMPLEX N UMBERS Definition 1 = i is the imaginary unit. Definition a +bi,a,bR is a complex number (C). a is the real part, b is the imaginary part. Properties of a Complex Number 1) (a +bi = a +b i ) iff a = a and b = b. 2) (a +bi + a +b i = a + a + b+b i ) . 3) (a +bi a +b i = a a + b b i ) . 4) (a +bi a +b i = aa bb + ab +ba i ) . a bi aa + bb) (a b ab ) 5) a +b i = 2 2 + 2 2 i. a )+ b ( )( a )+ b ( ) 6) (a+bi ) is a real number iff b = 0 (R C ). Definition Let us denote z = a+bi . The conjugate of z is z = abi. The absolute value of z is z a +b . 2 Geometric Interpretation of a Complex Number z = a +bi R 2(a,b 2 2 The absolute value z = a +b is just the distance from imaginary (a,b to the origin. (a, b) z1 z2 is the distance from z1= a1 + b1i to z 2= a 2+ b2i . real Polar Coordinates a = rcos i b = rsin =z+ a bi = r cos + r sin i=)r cos(+ isin = r e . Note: r = z = a2 +b 2 . b Note: = arctan the argument of z. a Definition ei = cos + (sin)i . Page 1 of 17 www.notesolution.com MAT224H1a.doc Example Write z = 2 2i in polar form. z =re i Let . r = ( 2 2)+ 2 2( ) 8 = 2 2 . = arctan 2 =arctan = 3 2 4 . i3 z = 2+ 2i= 2 2 e 4 So . Theorem: Multiplication In Polar Coordinates i i i + If z1 = r1e 1 and z 2= r2e 21 are complex numbers in polar form, then z z1 2 = 1 2 e 1 2 . Proof: z z = r e i1 (r ei2 )= r r ei1e i2 . 1 2 1 2 1 2 Want: e i1ei2 = ei 1 +2 . ei1ei2= (cos 1+ isin 1)cos 2 + isin 2) = (cos 1cos 2 sin1 sin 2 + sin 1c) ( 2+ cos 1sin 2 i. ) = cos ( + +i)in + ( ) 1 2 1 2 = e (1+2 ) Theorem: De Moivres Theorem i n i n) If is any angle, then (e) = e holds for all integers n. Proof (sketch): If n 0 , use induction. n i n i If n < 0 , then n > 0 . So (e =) e ( . Since i 1 1 1 cos i sin i ) (e) = i = = cos isin = cos + )i sin (= e ) , so e cos +isin cos isin i n i )n (e) ( = e ) ,n < 0 . Example 3 Find ( + 3i . 2 i i3 Let 1 3i = re = 2 . 2 2 r = (1 )+ ( 3 = 2 . 2 = arctan() 3 = 3 . Page 2 of 17 www.notesolution.com

More
Less
Related notes for MAT224H1