wi4243AP: Complex Analysis week 1, Friday K. P. Hart Faculty EEMCS - PowerPoint PPT Presentation

Section 1.1 Section 1.2 Section 1.3 wi4243AP: Complex Analysis week 1, Friday K. P. Hart Faculty EEMCS TU Delft Delft, 5 September, 2014 K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Section 1.2 Section 1.3 Outline Section 1.1 1 Complex numbers Section 1.2 2 Algebra Algebra and geometry Section 1.3 3 More geometry K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Section 1.2 Complex numbers Section 1.3 Definition A complex number is an ordered pair ( x , y ) of real numbers. We write z = ( x , y ) = x (1 , 0) + y (0 , 1) = x + yi Note: we abbreviate (1 , 0) = 1 and (0 , 1) = i and we shall define multiplication in such a way that (0 , 1) 2 = − (1 , 0), i.e., i 2 = − 1. Notation: Re z = x — real part Im z = y — imaginary part K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Section 1.2 Complex numbers Section 1.3 Representation Every complex number, x + yi , corresponds to a point, ( x , y ), in the plane. z = x + yi | z | θ x 2 + y 2 ; the modulus � | z | = θ ; angle, the argument K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Section 1.2 Complex numbers Section 1.3 Representation We introduce some geometry via polar coordinates. Properties x = | z | cos θ y = | z | sin θ There are infinitely many values for θ . Principal value: Arg z , chosen in interval ( − π, π ] Other values: arg z = Arg z + 2 k π K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Addition and multiplication Addition: coordinate-wise z + w = x + yi + u + vi = ( x + u ) + ( y + v ) i Multiplication (use distributive law and i 2 = − 1): z · w = ( x + yi )( u + vi ) = ( xu − yv ) + ( xv + yu ) i for example: (2 + i )(3 + 4 i ) = (6 − 4) + (8 + 3) i = 2 + 11 i K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Division The formula ( x + iy )( x − iy ) = x 2 + y 2 helps with division: 1 ( x + iy )( x − iy ) = x − yi x − iy x + iy = x 2 + y 2 For example 16 + 63 i = 16 + 63 i · 3 − 4 i 3 + 4 i 3 + 4 i 3 − 4 i = (48 + 252) + ( − 64 + 189) i 9 + 16 = 300 + 125 i = 12 + 5 i 25 K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Complex conjugate We write x + yi = x − yi , the complex conjugate . Geometrically: reflection in real axis. | z | = | z | and z · z = | z | 2 Arg z = − Arg z (except when z is real and negative) z + z = 2 x = 2 Re z z − z = 2 iy = 2 i Im z z ± w = z ± w , z · w = z · w , z / w = z / w K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Multiplication Write z = r (cos ϕ + i sin ϕ ) and w = s (cos ψ + i sin ψ ), we get zw = rs (cos ϕ + i sin ϕ )(cos ψ + i sin ψ ) � � = rs (cos ϕ cos ψ − sin ϕ sin ψ ) + i (sin ϕ cos ψ + cos ϕ sin ψ ) � � = rs cos( ϕ + ψ ) + i sin( ϕ + ψ ) . So multiply moduli and add angles . K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Multiplication: example √ Set z = − 1 + i en w = 1 + 3 i . Then √ 2(cos 3 4 π + i sin 3 z = 4 π ) w = 2(cos 1 3 π + i sin 1 3 π ) √ √ zw = ( − 1 − 3) + (1 − 3) i And so . . . K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Multiplication, example . . . we find √ √ √ � cos 13 12 π + i sin 13 � ( − 1 − 3) + i (1 − 3) = 2 2 12 π √ � � cos − 11 12 π + i sin − 11 = 2 2 12 π so we write arg zw = arg z + arg w but not Arg zw = Arg z + Arg w K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Multiplication, example w z 3 4 π π 3 zw − 11 12 π K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Division Write z = r (cos ϕ + i sin ϕ ) and w = s (cos ψ + i sin ψ ), we get w = r (cos ϕ + i sin ϕ ) z s (cos ψ + i sin ψ ) = r s · cos ϕ + i sin ϕ cos ψ + i sin ψ · cos ψ − i sin ψ cos ψ − i sin ψ = r s · (cos ϕ + i sin ϕ )(cos ψ − i sin ψ ) cos 2 ψ + sin 2 ψ = r � � cos( ϕ − ψ ) + i sin( φ − ψ ) . s So divide moduli and subtract angles . K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 De Moivre’s formula For any angle θ and any integer n we have (cos θ + i sin θ ) n = cos( n θ ) + i sin( n θ ) Application: easy formulas for cos n θ and sin n θ : cos 3 θ + i sin 3 θ = cos 3 θ + 3 cos 2 θ i sin θ + 3 cos θ i 2 sin 2 θ + i 3 sin 3 θ = (cos 3 θ − 3 cos θ sin 2 θ ) + i (3 cos 2 θ sin θ − sin 3 θ ) thanks to the binomial formula K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Chebyshev polynomials Note cos 3 θ = cos 3 θ − 3 cos θ sin 2 θ = cos 3 θ − 3 cos θ (1 − cos 2 θ ) = 4 cos 3 θ − 3 cos θ So cos 3 θ = T 3 (cos θ ), where T 3 ( x ) = 4 x 3 − 3 x . T 3 is a Chebyshev polynomial ; these are used in interpolation and approximation theory. K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Chebyshev polynomials General definition: cos n θ = T n (cos θ ). Using De Moivre’s identity and the binomial formula we get � n ⌊ n / 2 ⌋ � � x n − 2 k (1 − x 2 ) k ( − 1) k T n ( x ) = 2 k k =0 See Problem 1.18 for an other formula for T n . K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Fractional exponents What is (cos θ + i sin θ ) q if q is a (proper) fraction, say q = m n ? Remember: for real positive x , by definition, n is that real positive number with y n = x m . m y = x In complex numbers: no such choice available, no positive/negative numbers. K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Fractional exponents We solve z n = (cos θ + i sin θ ) m . write z = r (cos ϕ + i sin ϕ ) ( r = | z | and ϕ = Arg z ) we get r n (cos n ϕ + i sin n ϕ ) = cos m θ + i sin m θ so r = 1 and cos n ϕ = cos m θ and sin n ϕ = sin m θ we get n ϕ = m θ + 2 k π ( k an integer) as − π < ϕ � π there are n suitable values for k m n has n values so: (cos θ + i sin θ ) K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Fractional exponents, example √ √ 2 What are the values of ( 1 2 + 1 3 ? 2 i ) 2 2 We have θ = 1 4 π , so we get 3 ϕ = 1 2 π + 2 k π or ϕ = 1 6 π + 2 3 k π with k = − 1, 0, 1, to get − π < ϕ � π . √ The values are − i ( ϕ = − 1 2 π ), 1 3 + 1 2 i ( ϕ = 1 6 π ) and √ 2 − 1 3 + 1 2 i ( ϕ = 5 6 π ) 2 K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Fractional exponents Difference between real and complex analysis: many-valued functions 1 2 a two-valued function z �→ z 2 3 a three-valued function z �→ z z �→ z − 2 5 a five-valued function K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Euler’s formula (one of the many) We write (for now as an abbreviation): e i θ = cos θ + i sin θ thanks to angle-adding we know e i θ + i ϕ = e i θ · e i ϕ (as it should be). Later we’ll see further justification for this formula. K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Euler’s formula, use If | z | = r and Arg z = θ we write z = re i θ . m n as We write the values of z m n e i ( m n θ +2 km n π ) r ( k = 0 , . . . , n − 1) m n is unambiguous. Note: r � 0 so r n θ as its m n e i m Also note: the values form a regular n -gon with r ‘first’ vertex. K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Square roots Using Euler’s formula (and because e i π = − 1) the values of √ z are √ re i 1 2 θ and − √ re i 1 2 θ Direct calculation: look at ( u + iv ) 2 = x + iy . We get u 2 − v 2 = x (real parts) 2 uv = y (imaginary parts) u 2 + v 2 = x 2 + y 2 (moduli) � Now solve . . . K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Algebra Section 1.2 Algebra and geometry Section 1.3 Square roots . . . the first and third to get � u 2 = 1 x 2 + y 2 + x ) or u = ± x 2 + y 2 + x ) � 1 � 2 ( 2 ( � v 2 = 1 x 2 + y 2 − x ) or v = ± x 2 + y 2 − x ) 1 � � 2 ( 2 ( Of the four combinations only two survive because of 2 uv = y . Example: ( u + iv ) 2 = 5 + 12 i gives u = ± 3 and v = ± 2; we get 3 + 2 i and − 3 − 2 i because 2 uv = 12 > 0. Don’t remember the formula but do remember the method. K. P. Hart wi4243AP: Complex Analysis

Section 1.1 Section 1.2 More geometry Section 1.3 Modulus again Useful formulas: with z = x + iy we have | z | 2 = x 2 + y 2 = z · z x 2 + y 2 = | z | or � | x | , | y | � | Re z | , | Im z | � | z | K. P. Hart wi4243AP: Complex Analysis