English Courses

Description

• cours - matière potentielle : the literature
• cours - matière potentielle : consent
• leçon - matière potentielle : 25 -40 pages per night
• cours - matière potentielle : social studies
• exposé - matière potentielle : work
• expression écrite
• cours - matière : literature
• leçon - matière potentielle : 20 -35 pages per night
• expression écrite - matière potentielle : quality of student
• leçon - matière potentielle : critical analysis
• cours - matière potentielle : at language
• expression écrite - matière potentielle : processes
• expression écrite - matière potentielle : westerners
• revision - matière potentielle : process
• expression écrite - matière potentielle : practice
• expression écrite - matière potentielle : by seutonius
• cours - matière potentielle : load
• expression écrite - matière potentielle : assignments
• expression écrite - matière potentielle : process
• cours - matière potentielle : 1 unit
• cours - matière : history
• cours - matière : history - matière potentielle : history
• exposé - matière potentielle : to the class
• expression écrite - matière potentielle : from india

1 English Courses English I: Ancient Literature Full year – 1 unit This course is taught in conjunction with History I: Ancient Cultures. Please see the history course descriptions for the write- up of the class. This course is a requirement for ninth grade students. English II: American Studies Prerequisite: sophomore standing Full year course - 1 unit Designed to correlate closely with History II: American Studies English II is a survey course of American literature of all genres and various perspectives.

• requirement for ninth grade students
• literature prerequisite
• economic development policies
• literature
• world
• history
• language
• students
• course

Sujets

• Consent
• Social studies
• Literature

Informations

• Publié par: viol0
• Nombre de lectures: 19
• Langue: English

Extrait

Lecture 14: Cauchy-Riemann Equations:Polar Form

Dan Sloughter FurmanUniversity Mathematics39

March 31, 2004

14.1 Polarform of the Cauchy-Riemann Equations Theorem 14.1.Supposefis defined on anneighborhoodUof a point iθ 0 z 0 =r 0 e, iθ f(re) =u(r, θ) +iv(r, θ), andu r ,u θ ,v r , andv θ exist onUand are continuous at (r 0 , θ 0 ). Iffis differentiable atz 0 , then r 0 u r (r 0 , θ 0 ) =v θ (r 0 , θ 0 ) andu θ (r 0 , θ 0 ) =−r 0 v r (r 0 , θ 0 ) and 0 −iθ 0 f(z 0 ) =e(u r (r 0 , θ 0 ) +iv r (r 0 , θ 0 )). Proof.By the chain rule from multi-variable calculus, ∂u ∂u∂x ∂u∂y = + ∂r ∂x∂r ∂y∂r and ∂u ∂u∂x ∂u∂y = +. ∂θ ∂x∂θ ∂y∂θ Sincex=rcos(θ) andy=rsin(θ), we have u r =u x cos(θ) +u y sin(θ)

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and u θ =−u x rsin(θ) +u y rcos(θ). Similarly, v r =v x cos(θ) +v y sin(θ) and v θ =−v x rsin(θ) +v y rcos(θ). Nowfis differentiable atz 0 , and so satisfies the Cauchy-Riemann equations atz 0 . Thatis, u x (r 0 , θ 0 ) =v y (r 0 , θ 0 ) andu y (r 0 , θ 0 ) =−v x (r 0 , θ 0 ). Hence v r (r 0 , θ 0 ) =−u y (r 0 , θ 0 ) cos(θ) +u x (r 0 , θ 0 ) sin(θ) and v θ (r 0 , θ 0 ) =u y (r 0 , θ 0 )rsin(θ) +u x (r 0 , θ 0 )rcos(θ). It follows that r 0 u r (r 0 , θ 0 ) =v θ (r 0 , θ 0 ) andu θ (r 0 , θ 0 ) =−r 0 v r (r 0 , θ 0 ) The final statement of the theorem is left as an exercise. Theorem 14.2.Supposefis defined on anneighborhoodUof a point iθ 0 z 0 =r 0 e, iθ f(re) =u(r, θ) +iv(r, θ), andu r ,u θ ,v r , andv θ exist onUand are continuous at (r 0 , θ 0 ). If r 0 u r (r 0 , θ 0 ) =v θ (r 0 , θ 0 ) andu θ (r 0 , θ 0 ) =−r 0 v r (r 0 , θ 0 ), thenfis differentiable atz 0 . Proof.The proof is left as a homework exercise. Example 14.1.Forz6= 0, let 1 f(z) =. 2 z iθ If we writez=re, then 1 1 f(z= (cos(2) =θ)−isin(2θ)). 2 2iθ2 r er 2

Hence, in the notation of the above theorems, 1 u(r, θcos(2) =θ) 2 r and 1 v(r, θ) =−sin(2θ). 2 r It follows that 2 2 u r (r, θ) =−cos(2θ) andu θ (r, θ) =−sin(2θ) 3 2 r r and 2 2 v r (r, θsin(2) =θ) andv θ (r, θ) =−cos(2θ). 3 2 r r Thus ru r (r, θ) =v θ (r, θ) andu θ (r, θ) =−rv r (r, θ), and sofis differentiable at allz6= 0.Moreover,   2 2 0 −iθ f(z) =e−cos(2θ) +isin(2θ) 3 3 r r 2 −iθ−2iθ =−e e 3 r 2 −3iθ =−e 3 r 2 =−. 3 z

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