# University College Cork: MA2008 Complex Numbers and Functions Exercises Prove:

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1 University College Cork: MA8 Complex Numbers and Functions 5 Exercises. Show that (a) i, i i, i, i 5 i,... i i, i, i i,.... Let + i and 5i. Find in Cartesian form: (a) ( + ) (c) (d) (e) Im + i 7 i ( i) (f) Re + i. For x + iy, find in Cartesian form: (a) Im (Im ). Prove: (c) Im(/) (a) Any complex number is equal to the conjugate of its conjugate. is real if and only if. (c) i i (d) Re(i) Im (e) Im(i) Re (f) If then at least one of and must be ero. 5. Find (a) 5i (c) + (d) + i + i (e) (f) (g) ( + i) 6 i ( + i) (h) + i (i) cos θ + i sin θ 6. Represent in modulus argument form: 7. Determine the principal argument of (a) 7 i (c) + i (d) i 8. Show that multiplication by i corresponds to an anti-clockwise rotation about the origin through the angle π/. (a) + i + i (c) 8 (d) i 9. Verify the triangle inequality for i, + i.. Identify each of the following loci, and represent them on the Argand diagram: (a) Re (c) Re( ) (d) arg < π (e) π < Im π (f) + + (g) + (h) + i i (i) < Im <

2 University College Cork: MA8 Complex Numbers and Functions 5 Exercises. Determine all solutions to the following equations, and plot them on the Argand diagram: (a) i i (c) i (d) i (e) 5 (f) 8 (g) + i (h) + i (i) (j) 7 8 (k) 6 (l) i (m) 6 (n) + 8 (o) (p) For x + iy, demonstrate that the two square roots of are ( ) + x x ± + i(sign y) where sign y and the real square roots have positive sign. {, y, y < (Hint: Set w u + iv with w, and separate into two real equations to express u and v in terms of x and y.). Use Question to find the square roots of (a) i 8i (c) 5 + i (d) i. Use Question to find the solutions to the following equations: (a) (5 + i) i ( + i) 8 6i

3 University College Cork: MA8 Complex Numbers and Functions 5. Find f( + i), f( i) and f( + i) for each of the following functions: (a) f() f() Exercises, f() Re otherwise. Find the derivative with respect to of (c) f(). Find the real and imaginary parts of the following functions: (a) f() + f() (c) f(). Determine whether the following functions are continuous at the origin:, (a) f() Re otherwise (a) ( ) (c) + ( ) 5. Evaluate the derivative of the function at the given point: (a) ( i), i, i 6. Find the derivative of the following functions: (a) f() a + b f() + (c) f() ( + ) (d) f() (e) f() + (f) f() + 7. Are the following functions analytic? (a) f() Im f() + (c) f() Re( ) (d) f() + (e) f() (f) f() 8. Find the most general analytic function f(x + iy) u(x, y) + iv(x, y) for which (a) u xy v xy (c) u e x cos y 9. Show that the following functions are harmonic, and find a corresponding analytic function f(x + iy) u(x, y) + iv(x, y) : (a) u x v xy (c) u xy (d) u x xy (e) v e x sin y

4 University College Cork: MA8 Complex Numbers and Functions 5 Exercises. Find a representation (t) of the line segment with endpoints (a), i + i, + 5i (c), 5 + i (d) + i, + 5i (e) i, 9 5i (f) i, 7 + 8i. Identify the curves represented by the following functions: (a) it : t ( + i)t : t (c) + i + e it : t < π (d) i e it (e) cos t + i sin t : π < t < π (f) t + it : t. Find a function (t) representing the following loci: (a) i + i (c) y x between (, ) and (, ) (d) y x between (, ) and (, ) (e) x + y (f) y + between (, ) x and (, 5 ). Integrate the function f() along the line segment running: (a) from to + i from to i (c) from + i to + 5i (d) from to i 5. Integrate the following functions: (a) along the parabola y x from (, ) to (, ) around the unit circle in the clockwise direction (c) from vertically to i and then horiontally to + i (d) along the straight line segment from to + i (e) a + b along the straight line segment from to + i. 6. Evaluate: d (a) clockwise around the C circle d anti-clockwise around the C circle (c) Im d anti-clockwise around C the circle r (d) d along the following curves C running from i to i : i. the straight line segment ii. the unit circle in the left half-plane iii. the unit circle in the right half-plane

5 University College Cork: MA8 Complex Numbers and Functions 5 Exercises 5. Integrate the following functions around the anti-clockwise unit circle, and in each case indicate whether Cauchy s Theorem may be applied: (a) f() f() e (c) f() (d) f() 5 (e) f() Im (f) f() Re (g) f() (h) f() (i) f() (j) f() + (k) f() tanh (l) f() sec. Given Im d π about the C anti-clockwise unit circle, use Cauchy s Theorem applied to the function f() to deduce the value of Re d about the same circle. C. Integrate the function f() along the following curves from to + i : (a) C runs vertically to i then horiontally to + i C is the straight line segment from to + i (c) C runs horiontally to then vertically to + i. (a) Show that +. Use the principle of deformation of path to show that d πi C where C is an anti-clockwise circle enclosing both the points and. 5. Integrate f() about the anti-clockwise circles (a) Could you obtain the second result from the first by the principle of deformation of path? 6. Evaluate the following integrals along the given curves: + (a) C d where C is the + anti-clockwise circle i. ii. iii. + C d where C is the clockwise boundary of the rectangle with vertices at ± i and ± i. 5

6 University College Cork: MA8 Complex Numbers and Functions 5 Exercises 6: Definite Integration; Cauchy s Integral Formula; Cauchy s Integral Formula for derivatives.. Evaluate the following integrals: (a) (c) (d) (e) +i i i i i i πi d ( + ) d ( + ) d ( ) d e d (f) (g) (h) (i) (j) πi π πi πi i i i i e d e / d e d sinh π d sin d (k) (l) (m) (n) (o) πi π πi πi i +i πi sin d cos d sinh π d e d cos d. Integrate the function f() + about the following anti-clockwise circles: (a) + i (c) i (d) i. Integrate the function f() about the following anti-clockwise circles: (a) + + (c) i (d). Integrate the following functions anti-clockwise about the unit circle: (a) (c) (d) cos + e e (e) i (f) ( π) cos (g) sin (h) e i 5. Integrate the following functions about the anti-clockwise circle : (a) (c) ( i) e π ( i) (d) (e) (f) (g) cos ( + ) sin π cos (h) (i) (j) e ( ) e e sin 6

7 University College Cork: MA8 Complex Numbers and Functions 5 Exercises 7: The Exponential Function; Trigonometric and Hyperbolic Functions; Logarithms.. Find (a) e πi/ e πi/ +i (c) e (d) e +5i. Express the following complex numbers in the form re iθ for some r and θ : (a) + i i. Find the real and imaginary parts of (a) e e (c) e (d) e e (c) The square roots of i and of i (d) The square roots of re iθ (e) The n th roots of re iθ. Find (a) cos sin (c) tan 5. Calculate (a) cos i cosh i (c) sin i (d) sinh( + i) 6. Show that (a) cosh cosh x cos y + i sinh x sin y (cosh ) sinh (c) cosh sinh (d) cos cosh i (e) sin i sinh i 7. Calculate the principal value of Ln for (a) + i i (c) 5 (d) i 7

8 University College Cork: MA8 Complex Numbers and Functions 5 Exercises 8: Taylor Series; Laurent Series.. Find the Taylor Series expansion about the given point a of each of the following functions and in each case determine the radius of convergence: (a) e, a, a (c) e, a i (d) cos, a π/ (e) sin, a π/ (f) sin π, a (g) cos( ), a (h) cos (), a (i) (j), a, a i. Expand the following functions as Laurent Series about the origin, and determine the precise region < < R of convergence: (a) sin ( ) (c) (d) e / 6 ( + ) (e) (f) 5 +. Find all Taylor Series and Laurent Series expansions of the following functions about the given point a, and determine the precise regions of convergence: (a) (c) +, a i +, a i, a (d) (e) (f), a ( ), a e ( ), a (g) (h) sin ( π ), a π, a i 8

9 University College Cork: MA8 Complex Numbers and Functions 5 Exercises 9: Zeroes and Singularities; Residues; The Residue Theorem.. Find the location and order of the eroes of the following functions: (a) ( ) (c) (9 + ) ( ) (d) (e) + i + ( + ) ( + ). Suppose that f() has a ero of order n at ζ, with n >. (a) Prove that f () has a ero of order n at ζ. Prove that has a pole of f() order n at ζ.. Find the location and type of each singularity of the following functions: (a) + e (c) 7 ( + ) (d) e /( ) (e) e /. Find the residues at the singular points of the following functions: (a) (c) e (d) e (e) 5 cos (f) ( ) (g) ( + πi) 6 e 5. Find the residue at each singular point which lies inside the circle : (a) (e) ( + )( + 6) (c) (d) 5 + ( ) 6. Evaluate the following integrals about the anti-clockwise unit circle: d d (a) C + (c) sin C (e) + 6i C d (g) d ( + ) + 9 (d) + d e (f) sin d (h) C C C C C e cos π d ( + ) d 7. Integrate + about the following anti-clockwise circles: (a) i i (c) i 8. Integrate (a) + ( )( ) about the following clockwise circles: (c) (d) 9

10 University College Cork: MA8 Complex Numbers and Functions 5 Exercises : Contour Integrals. Evaluate the following (real) integrals using methods of contour integration: (a) (c) (d) (e) dx x + dx ( + x ) + x + x dx dx ( + x ) x + x 8 dx (f) (g) (h) (i) (j) x (x x + ) dx dx x + x + 9 sin x x + x + dx cos x x + dx cos x (x + )(x + ) dx (k) dx (x + )(x + )(x + 9). Use elementary methods to find. Find a value for dx (x + )(x + ) dx x +. by integrating around the following contour: Im i R R Re. Integrate e around the boundary of the rectangle with vertices ±a, ±a + ib to show that π e b e x cos bx dx. Hint: Let a, and use e x dx π.

11 Exercises b i d 7 + i c xy x + y 5a (x + ) 5c + y (x ) + y 5e (x + y ) 5f (x + y ) 5h 5 6a (cos π/ + i sin π/) 6c 8(cos π + i sin π) 7a π 7c π/ c The region between the two branches of the hyperbola x y. e Horiontal strip of width π. g The circle (x 7 5 ) + y 6 5 i The left half-plane without the closed disk of radius and centre (, ). Exercises a ± + i c i, ± + i e ±5i + i g ± Exercises i ± + i, ± + i + i i k ±, ±i, ± m, ± i o ±, ±i a ±( + i) c ±( + i) a + i, i a + i, + i, i c 9 i i, i, 5 b (x xy ) x, (x y y ) y { b f() x y x + y, y, x. Hence not continuous. a 6( ) c 5b ( ) 7 7b Yes ( ) 7d No 7f Yes ( ) 8a i + ic, c real 8c e (cos y + i sin y) + ic, c real 9b 9d

12 Exercises a ( i)t : t c ( + i)t : t 5 e i + ( i)t : t a line segment from + i to i c Circle with centre + i, radius e Ellipse x + y Exercises 5 a i + e it : t π c t + it : t e cos t i sin t : t π a + i c 7 + i 5a 88 6i 5c 5 + 6i 6a πi 6b πi 6(d)i i 6(d)ii i 6(d)iii i a, no d, yes g, no k, yes b, yes e π, no h πi, no l, yes c, no f iπ, no i, no 6b πi Exercises 6 a + i c i e l i sinh π n e e a (by Cauchy s Theorem) c π/ a πi c πi e π/8 5c 5e 6πi 5g πi g c π g 5i j a πi/ 5a π 5j πi Exercises 7 a i c e(cos + i sin ) a e x cos y, e x sin y c e x xy cos(x y y ), e x xy sin(x y y ) a 5e i tan b e πi/ b sin x + sinh y 5a 5d approx i 7a πi ln i 7c ln 5 + πi.69 +.i

13 Exercises 8 a +!! +, R c e i ( + ( i) + ( i) + ),! R e! ( π ) +! ( π ) ), R f π π + π5 5, R! 5! + i j ( + + i ( ) + i ( i) + ( i) + ), R a c ( ) n n+ n, > (n + )! n n ( ) n, > n!n+6 e ( ) n n, < < n ( ) i n+ a ( i) n : n ( i) n < i < ; ( i) n+6 : i > c ( ) ( f n n : < ; n ) n n : > e( ) n : > n! n Exercises 9 a ±, ±i, simple d i,, simple e ±i, second order a (simple pole); ( nd order pole) c ±i ( rd order pole); (simple pole) a at c at e at f at ± 5a, i,, i at, i,, i 5c / at 5e 6a 7 6c π/ 6e at πi sin 6g i sinh 7a 6πi 7c 5πi 8b πi 8d Exercises b π/ d π/6 f π/ i πe / (sin + cos ) g π/ k π/6

14 Worked solutions to some of the exercise sheets Sheet 7 a b c d e pii/ cos π/ + i sin π/ i e πi/ cos π/ + i sin π/ i e +i e e i e(cos + i sin ) e cos + ie sin e +5i e e 5i e (cos 5 + i sin 5) e cos 5 + ie sin 5 a e e x iy e x e iy e x (cos y + i sin y) e x cos( y) + ie x sin( y) e x cos y i x sin y b e e (x+iy) e x y +ixy e x y e ixy e x y (cos xy + i sin xy) e x y cos xy + e x y sin xy c e e x +ix y xy iy e x xy ( cos (x y y ) + i sin (x y y ) ) d e e e ex+iy e ex (cos y+i sin y) e ex cos y+ie x sin y e ex cos y ( cos (e x sin y) + i sin (e x sin y) ) e ex cos y cos(e x sin y) + ie ex cos y sin(e x sin y) a + i + 5 arg + i tan (/) cos tan (/) + i sin tan (/) 5e i tan (/)

15 b i cos ( π/) + i sin ( π/) e iπ/ e iπ/ ; the square roots of i are e iπ/ and e πi/ d r e iθ/, r e i( θ +π) c The square roots of i are e iπ/ Sheet 9 a Simple eroes at ±, ±i and e n re i(θ+πk)/n for k,,..., n. a b Zero of order at c Double eroes when + 9, that is, when ±i. (Also triple poles at ±.) d Simple ero when i. (Also simple poles at ±i.) e Triple eroes when ±i. (Also a pole of order at.) a f() has a ero of order n at ζ if and only if f(ζ) f (ζ) f (n ) (ζ) f ( n)(ζ). Put g f and you see that g has a ero of order n. b f() has a ero of order n at ζ if and only if f(ζ) f (ζ) f (n ) (ζ) f (n) (ζ). This is the same as the condition for f to have a pole of order n at ζ. a Simple pole at b Double pole at c Triple poles at ±i d Isolated essential singularity at e Isolated essential singularity at f() Simple pole at ; res{f(); }. b Overt simple pole at ; res{f(); } (). c f() e g() with g() e. Pole of order at ; res{f(); } g ()!. d f() e h() k() has covert simple poles when e, that is, when πni, for n Z. res{f(); πni} h(πni) k (πni) 5

16 e Overt pole of order 5 at. [ ] d (cos ) d res{f(); }! f f() [cos ]! ( ) ( ) ( + )( ) ( + ) ( ) Double poles at ±. [ ] d res{f(); } d ( + ) res{f(); } e [ ( + ) ] [ ] d d ( ) ) g f() has an overt pole of ( + iπ) 6 order 6 at iπ. [ ] d 6 res{f(); iπ} d (e ) 5a f() eiπ 7 7 h() k() iπ with h(), k(), k (). There are covert simple poles at ±, ±i. 5b res{f(); } h() k () res{f(); } h( ) k ( ) res{f(); i} i i i i res{f(); i} ( i) ( i) i f() + + ( + )( + ) There are simple poles at,. res{f(); } ( ) ( ) res{f(); } ( ) ( )

17 5c simple poles at the cube roots of. f() 5 + ( + ) (overt) double pole at ; (covert) res{f(); } [ ] d d + [ ( + )( ) ( )( ] ) ( + ) 6 9 5d f() ( ) ( ) ( + ) ( i) ( + i) Double poles at ±, ±i. [ ] d res{f(); } d ( + ) ( + ) [ ( ( + )( + ) + ( + )()( + ) ) ] ( + ) ( + ) ( + ) (6 + ) e + f() ( + )( + 6) g() + where g() +. Thus f() has an + 6 overt simple pole at, with res{f(); } g( ) 7 (There are also (covert) simple poles at ±i.) 6a Set f() + and I f() d. (;) Then f() has (covert) simple poles when + ± i 7

18 and both of these points lie inside the circle. We have res{f(); i } 8(i/) i There are (overt) simple poles at, 6i ; the only pole inside the given circle is. res{f(); } 6i i 6 res{f(); i/} i i 8( i/) So (;) f() d πi i 6 π Hence I πi( i + i ). 6b f() poles at ± i. has (covert) simple + 9 6d ( + ) f() + ( + ) ( + ) Hence 6c res{f(); i/} i/ 8(i/) 8 res{f(); i/} i/ 8( i/) 8 (;) ( f() d πi 8 + ) 8 πi 8 πi 9 f() + 6i ( + 6i) Simple poles at (overt) and ± i (covert). res{f(); } 5 ( () + ) () + ( ) (i) + res{f(); i/} (i) 8(i) (i + ) 8 + i 8 i 8 ( ) ( i) + res{f(); i/} ( i) 8( i) i 8 8

19 (;) ( πi 5 + i + i ) 8 8 ( πi i ) 8 ( ) 9 πi i 9πi + π 6e f() sin at ±. (;) has covert simple poles res{f(); } sin 8( ) 6 sin res{f(); } sin 8( ) 6 sin ( f() d πi 6 sin + 6 sin ) πi sin 6f f() e has covert simple poles at sin πk ( k Z ). The only pole in the unit circle is. res{f(); } e cos (;) f() d πi πi 6g f() e has covert simple poles cos π at k + π ( k Z ). 7 res{f(); } e π sin π/ res{f(); /} (;) e/ π e / π sin( π/) e/ π ( ) e / f() d πi + e/ π π ( ) e / e / i i sinh f() + + ( ) ( + )( ) Simple poles at, ±. res{f(); } ( ) () + () res{f(); } ()() 8 9

20 () + ( ) res{f(); } ( )( ) 8 res{f(); } ()( ) 7a 7b (i;) ( f() d πi + + ) πi() 6πi 8a res{f(); } ()() (; ) f() d πi( ) πi 7c (+i;) (+i;) f() d πi πi ( ) f() d πi + 5πi + 8 f() ( )( ) poles at,,. res{f(); } has overt simple ( )( ) 8b 8c (; ) f() d πi πi (; ) f() d πi πi ( ) ( ) 8d f() d by Cauchy s ( ; ) Theorem. Sheet For these questions, I shall use the notation R to denote the (positively oriented) semicircular contour Γ R [ R, R]. (This is the contour I used in the two examples from the last lecture.) I shall also use I to denote the integral in the question and I R to denote the integral f() d of f() along the semicircular Γ R arc. I trust this won t be too confusing... a We integrate f() about the + semicircular contour R. f() has (covert) simple poles at ±i. res{f(); i} (i) i

21 So R f() d πi π ( ) i π f() d Rie it Γ R R e it + dt R R π R x + dx R as R R x + dx f() d f() d R Γ R π f() d Γ R π as R Hence I π. b f() ( + ) ( + i) ( i) There are (now overt) double poles at ±i. [ ] d res{f(); i} d ( + i) i [ ( + i) ] i (i) 8i i So R f() d πi π ( ) i π f() d Rie it Γ R ( + R e it ) dt O(R ) R as R f() d ( ) f() d f() d R Γ R π as R c Put f() +. There are covert + simple poles at cis ( π + kπ ) for k,,,, i.e., ± + i, ± i. res{f(); cis (π/)} [ ] + + cis (π/) cis (π/) cis (π/) ( + cis (π/))(cis ( π/) (cis ( π/) + cis ( π/)) ( i + i ) ( i) i

22 res{f(); cis (π/)} + cis (π/) cis (9π/) ( + cis (π/))(cis ( π/)) (cis ( π/) + cis (5π/)) i ( f() d πi i ) R π π π π f() d + R e it Γ R + R Rieit eit dt ( + R )(R) π R dt Hence R d + x + x dx R as R f(x) dx R ( ) f() d f() d R Γ R π f() d Γ R π f() as R ( + ) ( + i) ( i) Overt triple poles at ±i. res{f(); i} [ ] d! d ( + i) [ ] ( + i) 5 i [ ] (i) 5 [ ] i 6i i 6 So i R f() d Rie it Γ R ( + R e it ) dt O(R 5 ) R e f(x) hence f(x) dx R as R f(x) dx R ( ) f() d f() d R Γ R ( ( ) ) i πi I R 6 π 6 I R π 6 as R f(x) dx x is an odd function, + x8 f(x) dx f(x) dx + f(x) dx and f(x) dx I m not quite sure what the intention of this question was...

23 You can also show that this integral is ero by the much more convoluted route of finding the residue at each of the (simple) poles inside the standard semicircular contour, and discovering that they all add up to ero. f + ( ) +, so f() ( ) has covert double poles + when ( ), that is, ± i. To find the residues, we need to express f() in such a way as to make the poles overt: f() ( + ) ( ( ) + ) ( ( ) + i ) ( ( ) i ) ( ( i) ) ( ( + i) ) using a standard Estimation Theorem argument on I R. g f() ( + )( + 9) Covert simple poles at ±i, ±i. res{f(); i} (i) + 9 (i) i 8 i 6 res{f(); + i} [ ] d ( ) d ( i) +i [( ) ( )] ( i) ( i) ( ) ( i) (i) ( + i)(i) (i) i + 6 i R x R x x + ) dx f() d f() d R Γ R πi i I R π as R +i R res{f(); i} (i) + (i) 6i 8 i 8 R f(x) dx f(x) dx R f() d f() d R Γ R ( πi i ) I R πi ( i 8 π I R 6 + i 8 ) I R π as R

24 Solutions to Sheet a ± + i g + i (x + iy) + i b ± i c i cis (π/). cis (π/6), cis (5π/6), cis ( π/) + i, + i, i. d i cis (π/). cis (π/8), cis (5π/8), cis ( π/8) or cis ( 7π/8). x y xy y x x 6x 6x 8x 6x 8x (x )(x + ) x (because x ) x ± y (± /) ± + i ± h + i cis (π/). cis (π/), cis (π/), cis (7π/). e ±5i i cis π. cis (±π/), cis (±π/) ± + i, ± i. j cis. cis (πk/7) for k,..., 6 or for k,...,. f 8 cis. cis (kπ/) for k,,...,. ±, ±i, ± + i, ± i k 6 cis π. cis ( π 6 + πk ) for k,..., 5 or for k,...,. cis (±π/6), cis (±π/), cis (±5π/6) + i ±, ±i, ± + i

25 Solutions to Sheet Solutions to Sheet a (t) ( i)t : t b (t) t( + 5i) + ( t)( + i) t + 5it + i + t it t + i(t + ) (t ) c (t) ( + i)t : t 5, or (t) (5 + i)t : t d e (t) ( + 5i)t + ( t)( + i) t + 5it + + i t it t + + i(t + ) ( t ) (t) (9 5i)t + ( t)( i) 9t 5it + i t + it 8t + i(t + ) ( t ) or (s) s + i(s + ) i + ( i)s for s. f (t) ( 7 + 8i)t + ( t)( i) 7t + 8it i + it 7t + i(t ) ( t ) a Straight line segment from + i to i b Straight line segment from to + 6i c Anticlockwise circle, centre + i, radius, traced from the positive real direction. d Anticlockwise circle, centre i, radius, traced from the negative real direction. e Put x cos t, y sin t. Then cos t + sin t ( y x + ) a i + e it ( t π) b i + e it ( t π) c t + it ( t ) d t + t e ( t ) x + y ( x ) + y Put x cos t, y sin t. Then cos t + i sin t ( t π). 6c f() Im (t) e it (for t π ). f() d π π cos t + i sin t (t) ie it i cos t sin t sin t(i cos t sin t) dt (i sin t cos t sin t) dt π [ ( )] cos t i sin t cos t dt [ i sin t t ] π sin t + π x + y This is the equation of the (anticlockwise) ellipse traced out. 5

26 Solutions to Sheet 5 b f is holomorphic everywhere, so Cauchy s Theorem applies and the integral is ero. d f is holomorphic inside and on the given circle (the only singularity is outside the curve), so Cauchy s Theorem does apply, and the integral is ero. f (;) Re d π π iπ cos t(i cos t sin t) dt (i cos t sin t cos t) dt h i (;) π i i d π π (cos t i sin t)(i cos t sin t) dt ( cos( t) + i sin( t) )( cos t + i sin t ) dt dt πi (;) d (;) (;) (;) Re d i( π) iπ π ( i Im ) d (i cos t sin t) dt d i Im d (;) +i [ d ] +i ( + i) + i i i (i ) (Note: This only works because the integrand is antidifferentiable; if you calculate the different path integrals you should get the same answer.) 6

27 5a (;) π d i πi π e it (ieit ) dt dt 5b (;) π d 8π π e it ie it dt dt 5 The principle of deformation of path does not apply because the integrand is not holomorphic. Solutions to Sheet 6 Note: The integrands in these questions are all holomorphic (and also integrable) everywhere, and consequently the given integrals are independent of the path used to get from the initial point to the final point. We thus use the Fundamental Theorem of Calculus for complex functions. b a +i [ d ] +i ( + i) + i i + i i i [ ( + ) (x + ) d ] i i (i + ) (i + ) 7i.9. +.i i.. +.i ( 7i 8 + i i ) ( ) i(5 96 7) (6 + 7i) + 9i 7

28 c i [ ( + ) d + i + i i d Warning: There is a mistake somewhere in this answer. It should come out to i Kudos to the first person who can provide a correct solution. i i ( ) d i i ( 6 + ) d [ ] 7 i i [ 8i.i ] [ i 8i i i ] 5 i i 8i 96i 7 5 i + i 7 + i 5 i ( i ) 5 i 76 5 ] i e g πi e d [e ] πi e πi e πi πi [ e /] πi πi [ e πi] [ e πi] ( )( ) ( )( ) f πi π [ e e d ] πi π e6πi e π e π h i e d [ ] i e e e (e e ) sinh 8

29 i j i i i [ cosh π sinh π d π cosh πi π ] i eπi + e πi π π π π sin d [ cos ] i i cosh π π cos i + cos( i) cosh + cosh( ) (because cosh is an even function) k l πi π [ ] cos πi sin d π cos πi cos π + cosh π eπ + e π eπ + + e π + ( e π + e π ) (cosh π) πi πi cos d [sin ] πi πi sin πi sin( πi) sin πi i sinh π i(e π e π ) m n o i [ cosh π sinh π d π cosh πi π +i πi e d cos π π π [ π ] e +i ] i cosh π e e+i 6 i+ e e e e ] πi [ sin cos d sin( π ) sin π f() + ( + i)( i). Simple poles at ±i. Unless otherwise stated, these answers involve Cauchy s Integral Formula. a (; ). f() is holomorphic inside and on, so f() d by Cauchy s Theorem. b ( i; ). [ ] f() d πi i ( ) πi i π i 9

30 c (i; ). [ ] f() d πi + i i πi π d (i; ) [ ] f() d πi + i i π ( ) i (You could also use the Principle of Deformation of Path to obtain this result from the previous one.) f() ( + )( )( + i)( i) Again we use Cauchy s Integral Formula unless otherwise indicated. a ( ; ) f() d [ ] πi ( )( + i)( i) πi ( )() πi b ( ; ). f() d by Cauchy s Theorem. c (i; ) [ ] f() d πi ( + )( )( + i) πi ( )(i) π i d (; ) [ ] f() d πi ( + )( + i)( i) πi ()() πi Again we use Cauchy s Integral Formula, with (; ). a b cos (Cauchy s Theorem) c d e d πi cos πi + d e e d πie πi d πi(e ) i d i d πi [ ] i/ ( ) i πi 8 π 8

31 f (Cauchy s Theorem) g h cos π d sin d sin d πi sin e i d e i d πi e(i/) πiei 5 (; ) ; these questions use Cauchy s Integral Formula for Derivatives. 5a [ ] πi d d ( i)! d i πi.i i 5d 5e 5f 5g 5h cos [ ] d d πi d cos πi( sin ) [ ] πi d d ( + )! d πi.6( ) sin π cos πi! 6πi [ ] d sin π d πi( π ) sin d πi! πi [ ] d d cos [ ] e d d πi ( ) d e πie 5b 5c e π [ ] πi d d! d eπ πi.πe (Cauchy s Theorem) π i ( i) d 5i 5j e sin [ ] e πi d d! d e [ ] d d πi d e sin πi [e cos + e sin ] πi

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