Exercises involving contour integrals and trig integrals

Transcription

1 8::9::9:7 c M K Warby MA364 Complex variable methods applications Exercises involving contour integrals trig integrals Let = = { e it : π t π }, { e it π : t 3π } with the direction of both arcs corresponding to the parameter t increasing thus both arcs are circular start at i end at i Compute In the case of we have = In the case of we have t) = e it, π/ π/ t) = i e it π/ e it ) i e it dt = i π/ dt = i π t) = e it, t) = i e it = 3π/ π/ 3π/ e it ) i) e it dt = i dt = i π π/ This demonstrates that in this example the value of the integral depends on the path taken from i to +i Let be points such that the straight line segment from to does not contain the point Also let be the straight line segment from to Explain why Show that = { Log Log, Log ) Log ), = if does not cross the negative axis, otherwise where Log denotes the principal valued logarithm We are given that the line segment does not contain the origin thus the integr is bounded the integral exists Parametriations of the line segments can be as follows = { + t ) : t }, = { + t + ) : t } Hence = + t ) dt

2 8::9::9:7 c M K Warby MA364 Complex variable methods applications = Both integrals have the same value + + t + ) dt = + t ) dt If the straight line segment from to does not cross the negative real axis then / has the anti-derivative F ) = Log in a region containing the segment we have = F ) F ) = Log Log If to is part of the negative real axis then is part of the positive real axis If to crosses the negative real axis then the intersection of the segment the axis is just one point the corresponding segment from to crosses the positive real axis at the corresponding point on the positive real axis In all cases there is a region containing at which / has the anti-derivative F ) = Log we have = = Log ) Log ) 3 Let f) = g) = 9 let C denote the circle with centre at radius traversed once in the anti-clockwise sense Compute f) C g) C The rational function f) can be represented in the form f) = = polynomial) + A + B + both + are inside the circle C The loop integral of the polynomial part is hence C f) = πia + B) Similar reasoning applies to g) we have g) = 9 = polynomial) + C + C g) = πic + D) D +

3 8::9::9:7 c M K Warby MA364 Complex variable methods applications 3 For the coefficients A, B, C D we have ) ) A = lim = lim = +, ) + ) B = lim = lim =, ) 9 ) C = lim 9 = lim = +, ) + 9 ) D = lim 9 = lim = Thus C f) =, C g) = πi 4 Let let f) = n, n, n an integer C R = { Re it : t π } a circle of radius R with centre at traversed once in the anti-clockwise sense Also let I R = f) C R i) What is I R when R <? ii) If R > n = then what is I R? iii) If R > then explain why I R is independent of R iv) If R > n then use the ML inequality to show that I R =

4 8::9::9:7 c M K Warby MA364 Complex variable methods applications 4 i) When R < the integr f) is analytic inside C R thus by the Cauchy integral theorem I R = ii) In this case f) = / ) which has one simple pole at = which is inside C R I R = πi iii) If R > then f) is analytic between C R C R hence I R = f) = f) = I R C R iv) To bound f) on C R we just need to observe that n takes its smallest value when = thus on C R we have f) C R R n The length of C R is πr thus by the ML inequality we have I R πr R n = π/rn ) n as R /R) As I R is independent of R for R > we have I R = This was not part of the question but an alternative proof that I R = for R > is obtained if f) is written in partial fraction form n = ) ω) ω n ), ω = expπi/n) The partial fraction decomposition is with The value of the integral is f) = n k= A k ω k ω k A k = lim ω k n = nω k ) = n n πi I R = f) = πi A k = C R n k= As we have a finite geometric series we have n ω k = ωn ω = k= ωk nω k ) = ωk n n ) n ω k k=

5 8::9::9:7 c M K Warby MA364 Complex variable methods applications 5 5 The following were part of question on the May 7 exam a) Let f) be a function which is analytic in a domain D Explain what is meant by an anti-derivative F ) of f) Let F ) = Log) let F ) = Log ) where Log denotes the principal valued logarithm Give expressions for F ) F ) for values of where the functions are differentiable Hence, or otherwise, give an expression for an anti-derivative of f) = / which is valid in the half plane with positive real part, also give an expression for an anti-derivative of f) = / which is valid in the half plane with negative real part b) Suppose that f) the domain D are such that an anti-derivative F exists on D Let denote a simple arc in D starting at ending at It can be shown that f) = F ) F ) This result can be used in the questions below Let be the straight line segment from to +i let be the line segment from i to + i Evaluate the following integrals giving the answer in Cartesian form i ii iii iv a) An anti-derivative F of the function f is a function such that f) = F ) b) df =, df = = F ) is a an anti-derivative of f) throughout the half plane with positive real part F ) is a an anti-derivative of f) throughout the half plane with negative real part i With f) = an anti-derivative is F ) = / The value of the integral is ) ) F + i) F ) = + i) 4) = i 4) = + i

6 8::9::9:7 c M K Warby MA364 Complex variable methods applications 6 ii With f) = / with the segment being in the half plane with positive real part an anti-derivative is F ) The value of the integral is F + i) F ) = Log ) + i π 4 Log) = Log ) Log)) + i π 4 = Log) + i π 4 iii With f) = / with the segment being in the half plane with negative real part an anti-derivative is F ) The value of the integral is F + i) F i) = Log i) Log + i) = i π 4 iπ 4 = iπ iv With f) = / an anti-derivative is F ) = / The value of the integral is F + i) F i) = = i + i + i) i) = i 6 The following were part of question on the May 4 exam i) ii) + ) where = {e it : t π/} the direction of integration corresponds to t increasing + where = { + i)t : t } the direction of integration corresponds to t increasing i) With f) = + the anti-derivative is F ) = The start end of the path are i respectively thus the value of the integral is F i) F ) = + i 3 = 4 3 i 3 3 marks ii) With f) =, F ) = Log + ) +

7 8::9::9:7 c M K Warby MA364 Complex variable methods applications 7 The line segment does not cross the branch cut of F ) which is {x + iy : x, y = } The start end points are = + i) = + i) F ) = Log i) = iarg i) = i π, F ) = Log + i) = Log 5) + i tan /) hence the value of the integral is F ) F ) = Log 5) + i tan /) + π ) 3 marks 7 The following were part of question on the April 5 exam Suppose that f) the domain D are such that an anti-derivative F exists on D Let denote a simple arc in D starting at ending at Given that f) = F ) F ) evaluate the following integrals i) + ), where is a line segment joining + i ii), where is the part of the unit circle = from i to i in the anti-clockwise direction iii), where is the part of the unit circle = from i to i in the anti-clockwise direction as in the previous part) i) With f) = + we have the anti-derivative F ) = + F ) =, F + i) = + i) i) = + i) + 8i = + i Thus + ) = F + i) F ) = i marks

8 8::9::9:7 c M K Warby MA364 Complex variable methods applications 8 ii) With f) = / we have the anti-derivative F ) = F i) = i = i = i, F i) = i = F i) F i) = i i) = i marks iii) With f) = / we have the anti-derivative throughout the path being considered F ) = Log F i) = iarg i) = iπ, F i) = iargi) = iπ = F i) F i) = iπ 3 marks 8 The following were part of question on the May 6 exam Let denote the straight line segment from to i let denote the straight line segment from i to Evaluate the following integrals justifying your answer in each case You need to express your answer in cartesian form a) b) c) d) ) a) With f) = an anti-derivative is F ) = The value of the integral is F i) F ) = i + 4i + 8i 3 ) + + ) = 4 6i) 3 = 7 6i

9 8::9::9:7 c M K Warby MA364 Complex variable methods applications 9 b) With f) = / ) an anti-derivative involves the principal valued logarithm is F ) = Log ) The value of the integral is F i) F ) = Log i) = Log 8) + i π 4 c) As in the last part the value of the integral is F ) F i) = Log3) + Log 8) i π 4 d) = + = Log3) 9 The following was a part of question on the May 3 exam With Log denoting the principal valued logarithm describe the location of the branch cut of f) = Log give f ) in a region where f) is analytic Hence, or otherwise, evaluate the integral Log where denotes the straight line segment from = i to = Log has a jump discontinuity on the negative real axis thus f) also has a jump discontinuity on the negative real axis this is the location of the branch cut f ) = + Log = + Log An anti-derivative of Log is thus F ) = Log Log = F ) F ) Log i) = iπ, Log) = F ) = =, F ) = i) iπ/) i) = π + i The value is + π ) i

10 8::9::9:7 c M K Warby MA364 Complex variable methods applications This was question 4a of the 7 MA364 paper was worth marks By first using the substitution = e iθ determine the value of the integral π 3 + cos θ By using the value just obtained, or otherwise, determine the value of the following integral π cos θ 3 + cos θ = e iθ, = ieiθ = i, = i, cos θ = + The interval θ π maps to the unit circle C traversed once in the anticlockwise direction we have that the integral is I = F ), i where C F ) = ) = By inspection the quadratic in the denominator of F ) can be factored as = 3 + ) + 3) F ) has one simple pole inside C at = /3 The value of the integral is The residue of F at the pole is I = πresf, ) Hence ResF, ) = = lim )F ) = lim = + 3 = 5 I = π 5 cos θ = 3 + cos θ) 3 Thus cos θ 3 + cos θ = cos θ By using the answer to the first part we hence have π ) cos θ π = π 3 = 6π 3 + cos θ 5 5

11 8::9::9:7 c M K Warby MA364 Complex variable methods applications This was question 4a of the 3 MA394 paper was worth 8 marks Let a be real with a > By using the substitution = e iθ show that π π a + cos θ = π a = i For the integr let Now f) = gives = i a + + ) = cos θ = eiθ + e iθ ) = ) i a + + = + a + = when = a ± a + ) i + a + The negative sign case gives a value which is less than hence outside of the unit disk hence we only need to consider With I denoting the integral we have = a + a I = πi)resf, ) Resf, ) = i) lim ) + a + = i) The required result then follows as i = ) + a = i + a = This was question 4a of the 4 MA364 paper was worth marks By using the substitution = e iθ determine I = π 6 4 sin θ + cos θ i a cos θ = eiθ + e iθ ) = = i gives = i + ), sin θ = i eiθ e iθ ) = ) i If C denotes the unit circle then the value of the integral is I = F ) i C

12 8::9::9:7 c M K Warby MA364 Complex variable methods applications with F ) = 6 = i ) + + ) + i) i) Let denotes the roots of the denominator The product of these is such that = i + i = Let denote the root with the smaller magnitude which is the position of the simple pole inside the unit circle By the residue theorem I = πresf, ) ResF, ) = lim + i) i) = + i) + 6 For the point we have by the quadratic formula that = ) + i) = i) = + i), ie thus + i) = ResF, ) = 4 I = π 3 This was question 4b of the 5 MA364 paper was worth 8 marks By using the substitution = e iθ determine π cos θ = e iθ, = ieiθ = i, = i, cos θ = + The interval θ π maps to the unit circle C traversed once in the anticlockwise direction we have that the integral is I = F ), i where F ) = 5 + C 4 + ) =

13 8::9::9:7 c M K Warby MA364 Complex variable methods applications 3 The poles of F ) are points where the quadratic in the denominator is By inspection the product of the roots of the quadratic is the root with magnitude less than is, by the quadratic formula, = 5 + To get the residue we have I = πresf, ) 4 ) ResF, ) = lim )F ) = lim = = 4 Hence I = 8π 4 This was part of question 4 of the May 6 exam was worth 8 of the marks on the question By using the substitution = e iθ determine π 3 cos θ sin θ cos θ = eiθ + e iθ ) = = i gives = i + ), sin θ = i eiθ e iθ ) = ) i If C denotes the unit circle then the value of the integral is I = F ) i with F ) = 3 + C ) i ) = 6 + ) + i ) = i ) i ) The denominator has roots at = 6 ± 36 4) i ) = 6 ± 7 + i)

14 8::9::9:7 c M K Warby MA364 Complex variable methods applications 4 F ) has one simple pole inside C one outside C with the pole inside C at The residue at is From the above = i) ResF, ) = lim )F ) = + i) i) + 6 = 7 hence ResF, ) = 7 = 7 The value of the integral is πresf, ) = π 7 5 Let a > b > Show that π a sin θ + b cos θ = π ab Let I = π a sin θ + b cos θ First note that when a = b the integr is the constant /a the result is correct thus we just need to consider the case when a b We can simplify the expression a bit before we make a substitution by using sin θ = cos θ so that a sin θ + b cos θ = b a ) cos θ + a As in the other questions let = e iθ note that With this substitution = i eiθ = i, = i ) + / cos θ = = + + / 4 I = C F ) i where C is the unit circle traversed once in the anti-clockwise sense where ) F ) = b a ) ) 4 + a 4 = b a ) ) + 4a, = 4 b a ) 4 + ) + b + a )

15 8::9::9:7 c M K Warby MA364 Complex variable methods applications 5 The denominator is a quadratic in is when The two values are i = b + a ) ± 4b + a ) 4b a ) b a ) = b + a ) ± 4a b b a ) = b + a ) ± ab b a ) b a) = b a b + a) = b a The product of the roots is is the smaller in magnitude as a > b > The function F ) has poles at ± inside C We can get the residue at the pole by using L Hopitals rule Similary ResF, ) = lim )F ), = 4 lim b a ) 4 + ) + b + a ), 4 = 4b a ) 3 + 4b + a ) = b a ) + b + a ) = b a) + b + a ) = ab ResF, ) = ResF, ) I = πresf, ) + ResF, )) = π ab