8. Solutions to Part 1

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1 MATH Complex Analysis 8. Solutions to Part 8. Solutions to Part Solution. (i) (3+4i) 9+4i 6 7+4i. (ii) (iii) (iv) (v) +3i 3 4i +3i 3+4i 3 4i3+4i (+3i)(3+4i) 5 5i +3i 8 5 +i 5. i i+ i i+ i. +i i i i7 5. Solution. First note that i and arg( i) π/4 (draw a picture!). Similarly, 3 i and arg( 3 i) π/6. Hence i e iπ/4, 3 i e iπ/6. Hence ( i) 3 ( 3 i) 3/ e 3iπ/4 3 3 e 3iπ/6 3/ 3 e 3iπ/4+3iπ/6 3/ e 43iπ/ 3/ e 5iπ/. Note that 5e 3iπ/4 +e iπ/6 5cos(3π/4)+5isin(3π/4)+cos( π/6)+isin( π/6) i + i i 5. Solution.3 (i) Write z x+iy. Then x +ixy y 5+i. Comparing real and imaginary parts gives the simultaneous equations x y 5, xy. The second equation gives y 6/x and substituting this into the first gives x 4 +5x 36, a quadratic in x. Solving this quadratic equation gives x 4, hence x ±. When x we have y 3; when x we have y 3. Hence z +3i, 3i are the solutions. c University of Manchester 8 99

2 MATH Complex Analysis 8. Solutions to Part (ii) A bare-hands computation as in (i) will work, but is very lengthy. The trick is to instead first complete the square. Write z +4z + 6i (z +) +8 6i. Write z+ x+iy. Then (x+iy) 8+6i. Solving this as in (i) gives x,y 3 or x,y 3. Hence z +3i or z 3 3i. Solution.4 (i) Letz a+ib, w c+id. ThenRe(z+w) Re(a+ib+c+id) Re((a+c)+i(b+d)) a+c Re(z)+Re(w). Similarly Re(z w) Re(z) Re(w). (ii) Note that Im(z+w) Im(a+ib+c+id) Im((a+c)+i(b+d)) b+d Im(z)+Im(w). Similarly Im(z w) Im(z) Im(w). Almost any two complex numbers picked at random will give an example for which Re(zw) Re(z)Re(w). For example, choose z i, w i. Then zw. However, Re(zw) Re(z)Re(w). Similarly, Im(zw) Re(z)Re(w) ( ). Solution.5 Throughout write z a+ib, w c+id. (i) z +w (a+ib)+(c+id) (a+c)+i(b+d) (a+c) i(b+d) (a ib)+(c id) z + w. Similarly for z w. (ii) zw (a+ib)(c+id) (ac bd)+i(ad+bc) (ac bd) i(ad+bc) (a ib)(c id) zw. (iii) First note that ( ) ( ) ( ) z a+ib a ib a+ib. We also have z a +b a +b a ib a+ib, so a +b the result follows. (iv) z + z (x+iy)+(x iy) x Re(z). (v) z z (x+iy) (x iy) iy iim(z). Solution.6 Let z,w C. Then by the reverse triangle inequality. Hence Similarly, w z z w. Hence z z w +w z w + w z w z w. z w z w. Solution.7 (i) Writing z x+iy we obtain Re(z) {(x,y) x > }, i.e. a half-plane. (ii) Here we have the open strip {(x,y) < y < }. c University of Manchester 8

3 MATH Complex Analysis 8. Solutions to Part (iii) The condition z < 3 is equivalent to x +y < 9; hence the set is the open disc of radius 3 centred at the origin. (iv) Write z x+iy. We have x+iy < x+iy+, i.e. (x ) +y < (x+) +y. Multiplying this out (and noting that the ys cancel) gives x >, i.e. an open halfplane. Solution.8 (i) We have zw rs((cosθcosφ sinθsinφ)+i(cosθsinφ+sinθcosφ)) rs(cos(θ+φ)+isin(θ +φ)). Hence argzw θ +φ argz +argw. (ii) From (i) we have that argz argz. By induction argz n n(argz). Put z cosθ +isinθ so that argz θ. Note that z cos θ +sin θ. Hence z n. Hence z n cosnθ+isinnθ. (iii) Applying De Moivre s theorem in the case n 3 gives (cosθ+isinθ) 3 cos 3 θ +3icos θsinθ 3cosθsin θ isin 3 θ cos3θ +isin3θ. Hence, comparing real and imaginary parts and using the fact that cos θ+sin θ, we obtain cos3θ cos 3 θ 3cosθsin θ 4cos 3 θ 3cosθ, sin3θ 3cos θsinθ sin 3 θ 3sinθ 4sin 3 θ. Similarly, cos4θ sin 4 θ 6cos θsin θ+cos 4 θ, sin4θ 4cosθsin 3 θ +4cos 3 θsinθ. Solution.9 Let w re iθ and suppose that z n w. Write z ρe iφ. Then z n ρ n e inφ. Hence ρ n r and nφ θ +kπ, k Z. Thus we have that ρ r /n and we get distinct values of the argument φ of z when k,,...,n. Hence z r /n e i(θ+kπ n ), k,,...,n. Solution. Take z z + i. Then Arg(z ) Arg(z ) 3π/4. However, z z i and Arg(z z ) π/. (Draw a picture!) In this case, Arg(z z ) Arg(z )+Arg(z ). (More generally, any two points z,z for which Arg(z )+Arg(z ) ( π,π] will work.) Solution. As far as I know it isn t possible to evaluate this integral using some combination of integration by substitution, integration by parts, etc. However, there is one technique that may work (I haven t tried it), and it s one that was a favourite of Richard Feynman (Nobel laureate in physics, safe-cracker, and bongo-player, amongst many other talents). Feynman c University of Manchester 8

4 MATH Complex Analysis 8. Solutions to Part claimed to have never learned complex analysis but could perform many real integrals using a trick called differentiation under the integral sign. See ~kconrad/blurbs/analysis/diffunderint.pdf for an account of this, if you re interested. c University of Manchester 8

5 MATH Complex Analysis 9. Solutions to Part 9. Solutions to Part Solution. (i) This set is open. Let D {z C Im(z) > }. Let z D. We have to find ε > such that B ε (z ) D. To do this, write z x +iy and let ε y / >. Suppose that z x+iy B ε (z ). Then y y x x + y y z z y. Hence y < y y < y so that y > y / i.e. Im(z) >. Hence z D. See Figure 9.(i). (ii) This set is open. Let D {z C Re(z) >, z < }. Let z D. We have to find ε > such that B ε (z ) D. That one can do this is clear from Figure 9.(ii). In order to produce ε we argue as follows. Let { x ε min, z } > (note that z < as z D). Let z x+iy B ε (z ) so that z z < x / and z z ( z )/. Then arguing as in (i) we see that x x x x + y y z z x so that x < x x < x from which it follows that x > x / >, i.e. Re(z) >. We also have that z z z +z z z + z z as z <. Hence z <. It follows that z D. + z + z < (iii) Let D {z C z 6}. This set is not open. If we take the point z 6 on the real axis, then no matter how small ε > is, there are always points in B ε (z ) that are not in D. See Figure 9.(iii). Solution. (i) For any z C we have so that f (z) z +. f z +z (z (z ) lim +z ) z z z z (z z )(z +z +) lim z z z z lim z z z +z + z + c University of Manchester 8 3

6 MATH Complex Analysis 9. Solutions to Part z ε z ε z ε (i) (ii) (iii) Figure 9.: See Solution.. (ii) For z we have so that f (z) /z. (iii) For each z C we have f /z /z (z ) lim z z z z lim z z lim z z z z z z z z z(z z ) f (z 3 z ) (z 3 (z ) lim z ) z z z z (z z )(z +z z +z lim z z ) z z z z lim z +z z +z z z z z 3z z so that f (z) 3z z. Notice that the complex derivatives are identical to and can be computed in the same way as their real analogues ( bring down the power and knock one off the power, etc). Solution.3 (i) Throughout write z x+iy. (a) Note that f(z) (x+iy) x +ixy y. Hence u(x,y) x y, v(x,y) xy. (b) Note that for z f(x+iy) x+iy x iy x +y so that u(x,y) x/(x +y ), v(x,y) y/(x +y ). x x +y +i y x +y c University of Manchester 8 4

7 MATH Complex Analysis 9. Solutions to Part (ii) (a) Here u v x x y, u y v y x so that the Cauchy-Riemann equations are satisfied. (b) Here u x x +y u (x +y ), y xy (x +y ), v x xy (x +y ), v y x +y (x +y ). Hence u/ x v/ y and u/ y v/ x so that the Cauchy-Riemann equations hold. (iii) When f(z) z we have f(x+iy) x +y so that u(x,y) x +y, v(x,y). Then for (x,y) (,) we have u x x u (x +y ) /, y y v (x +y ) /, x, v y. If the Cauchy-Riemann equations hold then x/(x +y ) /, y/(x +y ) /, which imply that x y, which is impossible as we are assuming that (x,y) (,). At (x,y) (,) we have u x lim h h h which does not exist. (To see this, note that if h, h >, then h /h ; however, if h, h <, then h /h h/h.) Hence f is not differentiable anywhere. Solution.4 (i) Here u x 3x 3y, v y 3x 3y, and u y 6xy, v x 6xy so that the Cauchy-Riemann equations hold. (ii) Here and u x 4 (x +y ) 5( x5 +x 3 y 5xy 4 ), v 4 y (x +y ) 5( x5 +x 3 y 5xy 4 ), u 4 y (x +y ) 5( 5x4 +x y 3 y 5 ), v x 4 (x +y ) 5(5x4 x y 3 +y 5 ) c University of Manchester 8 5

8 MATH Complex Analysis 9. Solutions to Part so that the Cauchy-Riemann equations hold. Let z C. In both cases, the partial derivatives of u and v exist at z. The partial derivatives of u and v are continuous at z. The Cauchy-Riemann equations holds at z. Thus u and v satisfy the hypotheses of Proposition.5.. Hence f u+iv is differentiable at z. As z C is arbitrary, we see that f is holomorphic on C. Solution.5 (i) Recall that f f(z) f() () lim. z z As f(z) for the function in the question we need to investigate the limit f(z) lim z z. Put z x+ix with x >. Then f(x) x and However, if z x ix, x > then f(z) x lim lim z z x x+ix +i. f(z) x lim lim z z x x ix i. Hence there is no limit of (f(z) f())/z as z. (ii) For f(x+iy) xy we have u(x,y) xy and v(x,y). Then clearly Now and v v (,) x y (,). u u(h,) u(,) (,) lim lim x h h h h u u(,k) u(,) (,) lim y k k Hence the Cauchy-Riemann equations are satisfied. lim k k. This does not contradict Proposition.5. because the partial derivative u/ x is not continuous at (,). To see this, note that for x >,y >, u x (x,y) y x so that u lim (x,y) (,) x (x,y) does not exist. c University of Manchester 8 6

9 MATH Complex Analysis 9. Solutions to Part Solution.6 By the Cauchy-Riemann equations we have that so that Similarly u x u x x x v y v x y v y x y u x + u y. v x u x y u y so that v x x y v x x u x y u y u x y u y x v y v y v x + v y. Solution.7 Let f(z) z 3 and write z x+iy so that f(x+iy) (x+iy) 3 x 3 3xy +i(3x y y 3 ). Hence u(x,y) x 3 3xy and v(x,y) 3x y y 3. Now u x 6x, u y 6x and so that Hence both u and v are harmonic. v x 6y, v y 6y u x + u y, v x + v y. Solution.8 Suppose we know that u(x,y) x 5 x 3 y +5xy 4. Then Integrating with respect to y gives u x 5x4 3x y +5y 4 v y. v(x,y) 5x 4 y x y 3 +y 5 +α(x) (9..) for some function α(x) that depends only on x and not on y. (Recall that we are looking for an anti-partial derivative and α(x)/ y.) Similarly, u y x3 y +xy 3 v x c University of Manchester 8 7

10 MATH Complex Analysis 9. Solutions to Part and integrating with respect to x gives v(x,y) 5x 4 y x y 3 +β(y) (9..) for some arbitrary function β(y). Comparing (9..) and (9..) we see that y 5 +α(x) β(y), i.e. α(x) β(y) y 5. The right-hand side depends only on y and the left-hand side depends only on x. This is only possible if both α(x) and β(y) y 5 is a constant. Hence for some constant c R. v(x,y) 5x 4 y x y 3 +y 5 +c Solution.9 From Exercise.6, we know that if u is the real part of a holomorphic function then u is harmonic, i.e. u satisfies Laplace s equation. Note that so that u x 6x, u y kx u x + u y (6 k)x. Hence k 3. It remains to show that in the case k 3, u is the real part of a holomorphic function. We argue as in Exercise.8. First note that if u(x,y) x 3 3xy +xy x then u x 3x 3y +y v y. Hence v(x,y) 3x y y 3 +6y y +α(x) (9..3) for some arbitrary function α(x) depending only on x. Similarly u y 6xy +x v y so that v(x,y) 3x y 6x +β(y) (9..4) for some arbitrary function β(y) depending only on y. Comparing (9..3) and (9..4) we see that α(x)+6x β(y)+y 3 6y +y; as the left-hand side depends only on x and the right-hand side depends only on y, the above two expressions must be equal to a constant c R. Hence v(x,y) 3x y 6x y 3 +6y y +c. Note that the partial derivatives for both u and v exist and are continuous at every point in C and the Cauchy-Riemann equations hold at every point in C, it follows from the converse of the Cauchy-Riemann Theorem that f(x + iy) u(x,y) + iv(x,y) is a holomorphic function on C. c University of Manchester 8 8

11 MATH Complex Analysis 9. Solutions to Part Solution. Suppose that f(x+iy) u(x,y)+iv(x,y) and u(x,y) c, a constant. Then u/ x. Hence by the Cauchy-Riemann equations v/ y. Integrating with respect to y gives that v(x,y) α(x) for some function α(x) that depends only on x. Similarly, u/ y. Hence by the Cauchy-Riemann equations v/ x. Integrating with respect to x gives that v(x,y) β(y) for some function β(y) that depends only on y. Hence v(x,y) α(x) β(y). As α(x) depends only on x and β(y) depends only on y, this is only possible if both α(x) and β(y) are constant. Hence v(x,y) is constant and it follows that f is constant. Solution. Suppose that f(x + iy) u(x) + iv(y) where the real part depends only on x and the imaginary part depends only on y. Then u x u (x), v y v (y). By the Cauchy-Riemann equations, u (x) v (y). As the left-hand side of this equation depends only on x and the right-hand side depends only on y, we must have that u (x) v (y) λ for some real constant λ. From u (x) λ we have that u(x) λx+c, for some constant c R. From v (y) λ we have that v(y) λy + c for some constant c R. Let a c +ic. Then f(z) λz +a. Solution. Suppose that f(z) u(x,y)+iv(x,y) and u(x,y) +v(x,y) 5. Partially differentiating the latter expression with respect to x gives u x + v x and using the Cauchy-Riemann equations gives u x u y. Similarly, partially differentiating u(x,y) + v(x,y) 5 with respect to y and using the Cauchy-Riemann equations gives u x + u y. This gives us two simultaneous equations in u/ x and u/ y. Solving these equations gives u x, u y. From u/ x it follows that u(x,y) α(y), an arbitrary function of y. From u/ y it follows that u(x,y) β(x), an arbitrary function of x. This is only possible if u is constant. c University of Manchester 8 9

12 MATH Complex Analysis 9. Solutions to Part If u is constant then (so that v depends only on x) and u x v y u y v x (so that v depends only on y). Hence v must also be constant. c University of Manchester 8

13 MATH Complex Analysis. Solutions to Part 3. Solutions to Part 3 Solution 3. Let z n C. Let s n n z k, x n k n Re(z k ), y n k n Im(z k ) denote the partial sums of z n, Re(z n ), Im(z n ), respectively. Let s k z n, x k Re(z k), y k Im(z k), if these exist. Supposethat n z n is convergent. Let ε >. Then there exists N such that if n N we have s s n < ε. As x x n s s n < ε, and y y n s s n < ε (usingthefacts that Re(w) w and Im(w) w forany complex numberw), provided n N, it follows that k Re(z k) and k Im(z k) exist. Conversely, suppose that k Re(z k) and k Im(z k) exist. Let ε >. Choose N such that if n N then x x n < ε/. Choose N such that if n N then y y n < ε/. Then if n max{n,n } we have that Hence k z k converges. k z z n x x n + y y n < ε. Solution 3. Recall that a formula for the radius of convergence R of a n z n is given by /R lim n a n+ / a n (if this limit exists). (i) Here a n n /n so that a n+ a n n n+ n n n+ R n+ as n. Hence the radius of convergence is R /. (ii) Here a n /n! so that a n+ a n n! (n+)! n+ R as n. Hence the radius of convergence is R and the series converges for all z C. c University of Manchester 8

14 MATH Complex Analysis. Solutions to Part 3 (iii) Here a n n! so that a n+ a n (n+)! n! n R as n. Hence the radius of convergence is R and the series converges for z only. (iv) Here a n n p so that a n+ a n (n+)p n p ( n+ as n. Hence the radius of convergence is R. n ) p p R Solution 3.3 To see that the expression in Proposition 3..(i) does not converge, note that a n+ n a n if n is even, 3n+ 3 n if n is odd. n+ Hence lim n a n+ /a n if we let n through the subsequence of even values of n but lim n a n+ /a n if we let n through the subsequence of odd values of n. Hence lim n a n+ /a n does not exist. To see that the expression in Proposition 3..(ii) does not converge, note that { a n /n / if n is even, /3 if n is odd. Hence lim n a n+ /a n does not exist. Note, however, that a n / n for all n. Hence a n z n z n n z n, n n which converges provided that z/ <, i.e. if z <. Hence, by the comparison test, n a nz n converges for z <. n Solution 3.4 (i) We know that for z < n z n z (this is the sum of a geometric progression). Hence ( ) ( )( ) ( )( ) z n z n. z z z Using Proposition 3.. we can easily see that the coefficient of z n in the above product is equal to n. Hence ( ) nz n. z n n n c University of Manchester 8

15 MATH Complex Analysis. Solutions to Part 3 (ii) Using Proposition 3.. we see that n z n n! n w n n! n c n where c n n r The claimed result follows by noting that c n n! by the Binomial Theorem. n r ( n r r!(n r)! zr w n r. ) z r w n r (z +w)n n! Solution 3.5 Let f(z) n zn. Then f(z) defines a power series with radius of convergence. Moreover, f(z) /( z) by summing the geometric progression. By Theorem 3.3. we can differentiate n zn /( z) k-times and obtain a power series that converges for z <. We obtain, for each k, nk n(n ) (n (k ))z n (k ) (k )!( z) k for z <. Dividing both sides by (k )! gives the result. Solution 3.6 (i) We have that e iz cosz+isinz and e iz cosz isinz. Adding these expressions gives cosz e iz +e iz so that cosz (e iz +e iz )/. (ii) Subtracting the above expressions for e iz and e iz gives isinz e iz e iz so that sinz (e iz e iz )/i. (iii) It is easier to start with the right-hand side: (iv) Similarly, sinzcosw+coszsinw 4i (eiz e iz )(e iw +e iw )+ 4i (eiz +e iz )(e iw e iw ) (e i(z+w) e i(z+w)) i sin(z +w). coszcosw sinzsinw 4 (eiz +e iz )(e iw +e iw ) 4i (eiz e iz )(e iw e iw ) (e i(z+w) +e i(z+w)) cos(z +w). c University of Manchester 8 3

16 MATH Complex Analysis. Solutions to Part 3 Solution 3.7 (i) We have sinz sin(x+iy) (e i(x+iy) e i(x+iy)) i i (eix e y e ix e y ) i ((e y cosx e y cosx)+i(e y sinx+e y sinx)) ( e y sinx+e y ) ( sinx e y cosx e y ) cosx +i sinxcoshy +icosxsinhy. Hence the real and imaginary parts of sinz are u(x,y) sinxcoshy and v(x,y) cos x sinh y, respectively. Now and u x cosxcoshy, v y cosxcoshy, u y sinxsinhy, v x sinxsinhy, so that the Cauchy-Riemann equations are satisfied. (ii) Here we have that cosz cos(x+iy) (e i(x+iy) +e i(x+iy)) (eix e y +e ix e y ) ((e y cosx+e y cosx)+i(e y sinx e y sinx)) cosxcoshy isinxsinhy. Hence the real and imaginary parts of cosz are u(x,y) cosxcoshy and v(x,y) sinxsinhy, respectively. Now and u x sinxcoshy, v y sinxcoshy, u y cosxsinhy, v x cosxsinhy, so that the Cauchy-Riemann equations are satisfied. Alternatively, one could note that ( cosz sin z + π ) ( sin x+ π ) ( coshy +icos x+ π ) sinhy cosxcoshy isinxsinhy. c University of Manchester 8 4

17 MATH Complex Analysis. Solutions to Part 3 (iii) Here we have that sinhz sinh(x+iy) (e x+iy e (x+iy)) (ex e iy e x e iy ) ((ex cosy e x cosy)+i(e x siny +e x siny)) sinhxcosy +icoshxsiny. Hence the real and imaginary parts of sinhz are u(x,y) sinhxcosy and v(x,y) cosh x sin y, respectively. Now and u x coshxcosy, v y coshxcosy, u y sinhxsiny, v x sinhxsiny, so that the Cauchy-Riemann equations are satisfied. (One can also argue, assuming the results of (i), by using the fact that sinh z isiniz.) (iv) Here we have that coshz cosh(x+iy) (e x+iy +e (x+iy)) (ex e iy +e x e iy ) ((ex cosy +e x cosy)+i(e x siny e x siny)) coshxcosy +isinhxsiny. Hence the real and imaginary parts of coshz are u(x,y) coshxcosy and v(x,y) sinh x sin y, respectively. Now and u x sinhxcosy, v y sinhxcosy, u y coshxsiny, v x coshxsiny, so that the Cauchy-Riemann equations are satisfied. (Alternatively, using the results of (ii), one can use the fact that coshz cosiz to derive this.) Solution 3.8 (i) A complex-valued function takes real values if and only its imaginary part equals. c University of Manchester 8 5

18 MATH Complex Analysis. Solutions to Part 3 For expz: note that if z x+iy then e z e x cosy +ie x siny and this is real if and only e x siny. As e x > for all x R, this is zero if and only if siny, i.e. y kπ, k Z. Using the results of the previous exercise, sinz is real if and only if cosxsinhy, i.e. either x π +kπ, k Z or y. Similarly, the imaginary part of cosz is sinxsinhy and this equals zero if and only if x kπ, k Z, or y. The imaginary part of coshz is sinhxsiny and this equals zero if and only if x or y kπ, k Z. The imaginary part of sinhz is coshxsiny and this equals zero if and only if y kπ, k Z (as coshx > for all x R). (ii) A complex-valued function takes purely imaginary values if and only if its real part is zero. Now e z e x cosy + ie x siny has zero real part if and only if e x cosy, i.e. if cosy. Hence e z takes purely imaginary values when y π +kπ, k Z. The real part of sinz is sinxcoshy and this equals zero if and only if sinx, i.e. x kπ, k Z. The real part of cosz is cosxcoshy and this equals zero if and only if cosx, i.e. x π +kπ, k Z. The real part of sinhz is sinhxcosy and this equals zero if and only if either x or y π +kπ, k Z. The real part of coshz is coshxcosy and this equals zero if and only if y π +kπ, k Z. Solution 3.9 Let z x+iy. Then sinz if and only if both the real parts and imaginary parts of sinz are equal to. This happens if and only if sinxcoshy and cosxsinhy. We know coshy > for all y R so the first equation gives x kπ, k Z. Now coskπ ( ) k so the second equation gives sinhy, i.e. y. Thus the solutions to sinz are z kπ, k Z. For cosz, we note that the real part of cosz is cosxcoshy and the imaginary part is sinxsinhy. Now cosxcoshy implies cosx so that x (k + /)π, k Z. As sin(k + /)π ( ) k, it follows that the second equation gives sinhy, i.e. y. Hence the solutions to cosz are z (k +/)π, k Z. Solution 3. (i) Let z x+iy and suppose that +e z. Then e z e x cosy +ie x siny. Comparing real and imaginary parts we have that e x cosy and e x siny. As e x > for all x R the second equation gives that siny, i.e. y kπ, k Z. Substituting this into the first equation gives ( ) k e x. When k is even this c University of Manchester 8 6

19 MATH Complex Analysis. Solutions to Part 3 gives e x which has no real solutions. When k is odd this gives e x, i.e. x. Hence the solutions are z (k +)πi, k Z. (ii) Let z x+iy and suppose that +i e z. Then e z e x cosy +ie x siny +i and comparing real and imaginary parts gives e x cosy, e x siny. As e x > for x R, it follows that cosy siny, i.e. either y π/4 + kπ or y 5π/4+kπ, k Z. In the first case, cos(π/4+kπ) sin(π/4 +kπ) / and so we have e x ; hence x log. In the second case, cos(5π/4 +kπ) sin(5π/4 + kπ) / so that e x, which has no real solutions. Hence z log +i(π/4+kπ), k Z. Solution 3. (i) Write z x + iy. Suppose that sin(z + p) sinz for all z C, for some p C. Putting z we get sinp sin, so that p kπ, k Z. Now sin(z +nπ) sin(z +(n )π +π) sin(z +(n )π)cosπ +cos(z +(n )π)sinπ sin(z +(n )π). Continuing inductively, we see that sin(z +nπ) ( ) n sinz. Hence sin(z +nπ) sinz if and only if n is even. Hence the periods of sin are p πn, n Z. (ii) Suppose that exp(z+p) expz for all z C. Putting z gives expp exp. Put p x+iy. Then expp e x cosy +ie x siny and comparing real and imaginary parts gives e x cosy, e x siny. As e x > for all x R, the second equation gives siny, i.e. y nπ, n Z. The first equation then gives ( ) n e x. When n is odd this has no real solutions. When n is even this gives e x, i.e. x. Hence the periods of exp are nπi, n Z. Solution 3. Let z r e iθ, z r e iθ C\{}. We choose θ,θ ( π,π] to be the principal value of the arguments of z,z. Hence Logz lnr +iθ, Logz lnr +iθ. Then z z r e iθ r e iθ r r e i(θ +θ ) so that z z has argument θ +θ. However, θ +θ ( π,π] and is not necessarily a principle value of the argument for z z. However, by adding or subtracting π to θ +θ we can obtain the principal value of the argument of z z. Thus Logz z lnr r +i(θ +θ +πn) Logz +Logz +πn c University of Manchester 8 7

20 MATH Complex Analysis. Solutions to Part 3 for some integer n {,,}. For example, take z z +i. Then z z and Argz Argz 3π/4. Moreover z z i and the principal value of the argument of z z is π/. Hence so that Hence Logz Logz ln + 3πi 4, Logz z ln πi Logz +Logz ln+ 3πi. Logz z Logz +Logz πi. (Any two complex numbers z,z where the principal values of the arguments of z,z add to either more than π or less than π will also work.) Solution 3.3 Take b z i. Then i, argi π/+nπ and the principal value of the argument is π/. Hence Hence and the subsiduary values are Log(i) ln()+i π iπ ( π ) log(i) ln()+i +nπ i i exp(ilogi) exp exp(ilogi) exp ( π ) i +nπ. ( ) π ( ) π +nπ. Solution 3.4 (i) Using Proposition 3..(i), the radius of convergence of this power series is given by R lim n α(α ) (α n+)(α n) (n+)! n! α(α ) (α n+), if this limit exists. Note that lim α(α ) (α n+)(α n) n! n (n+)! α(α ) (α n+) lim α n n n+ Hence the power series has radius of convergence R. (ii) Recall from Theorem 3.3. that a power series is holomorphic on its disc of convergence and can be differentiated term by term. Hence for z < we have f (z) α+α(α )z+ α(α )(α )! z + + α(α )(α ) (α n) z n +. n! c University of Manchester 8 8

21 MATH Complex Analysis. Solutions to Part 3 Multiply this by +z. Using Proposition 3.. we see that the coefficient of z n for n in (+z)f (z) is given by α(α )(α ) (α (n ))(α n) n! which can be rearranged to α(α )(α ) (α (n )) (n )! ( α n n ) + + α(α )(α ) (α (n )) (n )! α (α )(α ) (α (n )), n! which is α times the coefficient of z n in the power series f(z). Clearly the constant term of (+z)f (z) is α. Hence (+z)f (z) αf(z) for z <. (iii) Let g(z) f(z)/(+z) α. Then for z < we have that g (z) (+z)α f (z) α(+z) α f(z) (+z) α, using (ii). Hence g (z) on {z C z < }. By Lemma 3.4. we have that g(z) is equal to a constant on {z C z < }. Noting that g() we have that g(z) for all z with z <. Hence f (z) (+z) α for z <. c University of Manchester 8 9

22 MATH Complex Analysis. Solutions to Part 4. Solutions to Part 4 Solution 4. (i) We have γ(t) e it cost isint, t π, so the path is the semicircle of radius centre that starts at and travels clockwise to. See Figure.(i). (ii) Here γ(t) x(t) + iy(t), x(t) + cost, y(t) + sint, t π. Hence (x ) +(y ), i.e. γ is the circle centred at +i with radius, where we start at + i and travel anticlockwise. See Figure.(ii). (iii) Here γ(t) x(t) + iy(t) where x(t) t, y(t) cosht, t, i.e. y coshx. Hence γ describes the piece of the graph of cosh from to. See Figure.(iii). (iv) Here γ(t) x(t)+iy(t) where x(t) cosht, y(t) sinht, t. Hence x(t) y(t) cosh t sinh t, i.e. γ describes a hyperbola from (cosh( ),sinh( )) to (cosh(), sinh()). See Figure.(iv). +i (i) (ii) (iii) (iv) Figure.: See Solution 4.. Solution 4. Let f(z) x y +ix where z x+iy. c University of Manchester 8

23 MATH Complex Analysis. Solutions to Part 4 (i) The straight line from to +i has parametrisation γ(t) t+it, t. Here γ (t) +i and f(γ(t)) t t+it it. Hence γ x y +ix it (+i)dt t +it dt [ 3 t3 + i 3 t3 3 + i 3. (ii) The imaginary axis from to i has parametrisation γ(t) it, t. Here γ (t) i and f(γ(t)) t. Hence γ x y +ix [ it i. itdt (iii) The line parallel to the real axis from i to + i has parametrisation γ(t) t + i, t. Here γ (t) and f(γ(t)) t +it. Hence γ x y +ix ] ] (t +it )dt [ ] t it3 t+ 3 + i 3. Solution 4.3 The path γ is the circle of radius, centre, described anticlockwise. The path γ is the arc of the circle of radius, centre i, from i+ to, described clockwise. (i) Let f(z) /(z ). Note that and Hence γ f(γ (t)) e it γ (t) ie it. dz π z e itieit dt πi. c University of Manchester 8

24 MATH Complex Analysis. Solutions to Part 4 (ii) Let f(z) /(z i) 3. Note that and f(γ (t)) e 3it γ (t) ie it. Hence γ dz π/ (z i) 3 e 3it( ie it )dt i π/ e it dt i i [ e it ] π/. Solution 4.4 The circle z described anticlockwise has parametrisation γ(t) + e it + cost+isint, t π. Here γ (t) ie it sint+icost. Note that γ(t) (+cost) +sin t +cost+cos t+sin t (+cost). Hence γ z dz π π π (+cost)( sint+icost)dt ( sint sintcost)+i(cost+cos t)dt ( sint sint)+i(cost++cost)dt [cost+ cost+i(sint+t+ sint) ] π (+ +πi ) ( + ) πi. Solution 4.5 (i) We want to find a function F such that F (z) f(z). We know that differentiation for complex functions obeys the same rules (chain rule, product rule, etc) as for real functions, so we firstfind an anti-derivative for the real function f(x) x sinx. Note that by integration by parts we have x sinxdx x cosx+ xcosxdx x cosx+xsinx sinxdx x cosx+xsinx+cosx. Let F(z) z cosz+zsinz+cosz. Then clearly F is defined on C and one can check that F (z) z sinz. Hence, by the Fundamental Theorem of Contour Integration (Theorem 4.3.3), if γ is any smooth path from to i then f F(i) F() i cosi+isini+cosi cos γ 3cosh sinh. c University of Manchester 8

25 MATH Complex Analysis. Solutions to Part 4 (ii) Again, let us first find an anti-derivative for f(x) xe ix. Integrating by parts gives xe ix dx ixe ix + ie ix dx ixe ix +e ix (noting that /i i). Hence F(z) ize iz +e iz is an anti-derivative for f. Hence if γ is any smooth path from to i then f F(i) F() ( i e +e ) (+) e. γ Solution 4.6 (i) The contour γ that goes vertically from to i and then horizontally from i to +i is the sum of the two paths γ (t) it, t, γ (t) t+i, t. Note that γ (t) t, γ (t) i, γ (t) t +, γ (t). Hence z dz z dz + z dz γ γ γ [ it 3 3 it dt+ ] i 3. [ t t t +dt ] (ii) Similarly, the contour γ that goes horizontally from to and then vertically from to +i is the sum of the paths γ 3 (t) t, t γ 4 (t) +it, t. Here γ 3 (t) t, γ 3 (t), γ 4(t) +t, γ 4 (t) i. Hence z dz z dz + z dz γ 3 γ 4 γ [ 3 t3 t dt+ ] 3 + 4i 3. + [it+ it3 3 (+t )idt As the integral from to + i depends on the choice of path, this tells us (by the Fundamental Theorem of Contour Integration) that z does not have an anti-derivative. ] c University of Manchester 8 3

26 MATH Complex Analysis. Solutions to Part 4 Figure.: See Solution 4.7. Solution 4.7 See Figure.. Solution 4.8 Let γ : [a,b] D be a parametrisation of the contour γ. Then γ has γ(t) γ(a+b t) : [a,b] D as a parametrisation. Note that ( γ) (t) γ (a+b t) (γ (a+b t)), using the chain rule for differentiation. Hence b f f( γ(t))( γ (t))dt γ a b a a b b f(γ(a+b t))γ (a+b t)dt f(γ(u))γ (u)du using the substitution u a+b t a γ f(γ(t))γ (t)dt f. Solution 4.9 Let γ : [a,b] C be a parametrisation of γ. Using the formula for integration by parts from real analysis, we can write b fg f(γ(t))g (γ(t))γ (t)dt γ a b a f(γ(t)) d dt (g(γ(t)))dt c University of Manchester 8 4

27 MATH Complex Analysis. Solutions to Part 4 b f(γ(t))g(γ(t)) b ta d dt (f(γ(t)))g(γ(t))dt f(z )g(z ) f(z )g(z ) f(z )g(z ) f(z )g(z ) a b a γ f (γ(t))g(γ(t))γ (t)dt Solution 4. Note that the function f(z) /(z ) is not differentiable at z ± (because it is not defined at z ±). To use the Generalised Cauchy Theorem, we need a domain that excludes these points. Let D be the domain f g. D {z C z < 3, z > /3, z + > /3}. (There are lots of choices of D that will work. The 3 may be replaced by anything larger than ; the /3 by anything less than / the point is that D should contain γ,γ and γ but not ±. Alternatively, one could take D C \ {±}.) Obviously in this case D contains γ,γ and γ. Let γ 3 (t) e it, t π, i.e. γ 3 is γ but described in the opposite direction. Suppose z D. If z 3 then w(γ,z) w(γ,z) w(γ 3,z). If z /3 then z is inside the contours γ and γ 3 so that w(γ,z), w(γ,z) +, w(γ 3,z). Similarly if z + /3 then w(γ,z), w(γ,z), w(γ 3,z) +. Hence for each z D we have w(γ,z)+w(γ,z)+w(γ 3,z). Furthermore, since ± D, the function f is holomorphic on D. Applying the Generalised Cauchy Theorem we have that f + f + f γ γ γ 3 and the claim follows by noting that γ 3 f γ f. Solution 4. Let γ (t) e it, t π. Then γ f π f(γ(t))γ (t)dt π e itieit dt πi. Let D C\{}. We apply the Generalised Cauchy Theorem to the contours γ, γ. The only point not in D is z. Note that w(γ,) (so that w( γ,) ) and w(γ,). Hence w(γ,z)+w( γ,z) for all z D. By the Generalised Cauchy Theorem we have that f + f. γ γ Hence f f f πi. γ γ γ c University of Manchester 8 5

28 MATH Complex Analysis. Solutions to Part 4 Solution 4. WeapplytheGeneralisedCauchyTheorem(Theorem4.5.7)totheclosedcontoursγ, γ,γ. There are only two points not in D, namely z and z. Note that and w(γ,z ),w( γ,z ),w(γ,z ) w(γ,z ),w( γ,z ),w(γ,z ). Hence w(γ,z)+w( γ,z)+w(γ,z) whenever z D. By the Generalised Cauchy Theorem, we have f + f + f. γ γ Re-arranging this and noting that γ f γ f we have that γ γ f f f (3+4i) (5+6i) i. γ γ c University of Manchester 8 6

29 MATH Complex Analysis. Solutions to Part 5. Solutions to Part 5 Solution 5. (i) Since sin z ( cosz)/ and cosz n ( ) n zn (n)! (..) we have that sin z cosz n ( ) n+ 4 n z n. (..) (n)! As the radius of convergence for (..) is R, it follows that the radius of convergence for (..) is also R. (Alternatively, one could check this using the fact that a n+ /R lim lim n a n n where a n denote the coefficients in (..).) (ii) Here 4 n+ (n)! ((n+))! 4 n lim n +z z +4z 8z 3 + ( z) n n 4 n(n+) (by recognising this as a sum of a geometric progression). The radius of convergence is given, using Proposition 3..(ii), by so R /. (iii) We have /R lim n (n ) /n e z and the radius of convergence is R. Solution 5. Let f(z) Log(+z). This is defined and holomorphic on the domain C\{z C Im(z),Re(z) < }. By Taylor s Theorem, we can expand f as a Taylor series at valid on some disc centred at as f(z) a n z n n n z n n! c University of Manchester 8 7

30 MATH Complex Analysis. Solutions to Part 5 where a n n! f(n) (). Here and in general f (z) +z, f (z) (+z),... f (n) (z) ( )n+ (n )! (+z) n. Hence a n ( ) n+ (n )!/n! ( ) n+ /n for n and a. By the ratio test, this power series has radius of convergence. Solution 5.3 Let p(z) be a polynomial of degree at least. Let a C. We want to find z C such that p(z ) a. Let q(z) p(z) a. Then q is a polynomial of degree at least. By the Fundamental Theorem of Algebra, there exists z C such that q(z ). Hence p(z ) a. Solution 5.4 Since f is differentiable everywhere, for any r > and any m we have that f (k+m) () M r(k +m)! r k+m where M r sup{ f(z) z r}. Applying the bound on f(z) we have that M r Kr k. Hence f (k+m) () K(k+m)!rk r k+m K(k+m)! r m. Since this holds for r arbitrarily large, by letting r we see that f (k+m) (). Substituting this into the Taylor expansion of f shows that f is a polynomial of degree at most k. Solution 5.5 (i) Let f(z) z + and let z x+iy. Then f(z) (x+)+iy (x+) +y. Writing f(z) u(x,y)+iv(x,y) we have that u(x,y) (x+) +y and v(x,y). Hence u x (x+), v y and u y y, v x. Suppose that z is a point on the unit circle γ and that z ; then at least one of u/ x v/ y, u/ y v/ x holds. (Note that the Cauchy-Riemann equations do hold at the point z and that the partial derivatives are continuous at x,y, hence by Proposition.5. f is differentiable at z.) Hence f is not holomorphic on any domain that contains the unit circle γ. c University of Manchester 8 8

31 MATH Complex Analysis. Solutions to Part 5 (ii) Let z be a point on the unit circle γ. Then z /z. Hence ( ) z + (z +)(z +) (z +) z + (z +) z (note that this only holds on the unit circle γ). Let g(z) (z + ) /z. Then g is holomorphic on C\{}. (iii) Let h(z) (z +). Then γ z + dz γ γ (z +) dz z h(z) z dz πih() by Cauchy s Integral Formula πi. c University of Manchester 8 9

32 MATH Complex Analysis 3. Solutions to Part 6 3. Solutions to Part 6 Solution 6. Let f(z) m a mz m denote the Laurent series of f around z. (i) Since we are looking for an expansion valid for z > 3, we should look at powers of /z: z 3 /z (3/z) ) 3z 3 (+ + z z + z + 3 z + 3 3n z3 zn+ (ii) Here and this is valid for z <. Hence is valid for < z <. (iii) For z we have z +z +z + z( z) z ++z +z + Hence e /z m z 3 e /z + m!z m + n!z n + +!z + z +. (n+3)!z n + + 4!z + 3! + z! +z +z 3. (iv) Recall that For z we have cosz z! + z4 4! + ( )n z n +. (n)! cos/z + ( )n + + (n)!zn 4!z 4!z +. Solution 6. Note that z + ( z) ( z) n n ( ) n z n, (3..) n c University of Manchester 8 3

33 MATH Complex Analysis 3. Solutions to Part 6 summing a geometric progression with common ratio z. This expression converges for z <, i.e. for z <. We also have that z + z ( z ) z ( ) n z n ( ) n+ zn, (3..) summing a geometric progression with common ratio /z. This expression converges for /z <, i.e. for z >. Similarly, we have that z 3 ( ) 3 z 3 3 and this is valid when z < 3. We also have that n ( z ) n (3..3) 3 n z 3 z 3 z z n ( ) 3 n z n 3 n z n, (3..4) valid for z > 3. Hence when z < we have the Laurent expansion f(z) n For < z < 3 we have the Laurent expansion ( ( ) n ) 3 n+ z n. f(z) +( ) n+ z n + z + z 3 3z 3 n + zn. For z > 3 we have the Laurent expansion Solution 6.3 (i) First note that f(z) z (z ) n ( ( ) n+ +3 n ) z n. z z z (+z +z + +z n + ) and that this expansion is valid for < z <. Hence f(z) has Laurent series valid for < z <. z z z z z n c University of Manchester 8 3

34 MATH Complex Analysis 3. Solutions to Part 6 (ii) Let w z. Then z w + so that Note that, using the hint, w(+w) w z (z ) (w +) w w(+w) n( w) n n w ( w +3w 4w 3 + +( ) n nw n + ) w +3w 4w + +( ) n nw n + and that this is valid provided that < w <. Substituting in for z we then have that f(z) z +3(z ) 4(z ) + +( ) n n(z ) n + and that this is valid for < z <. Solution 6.4 Let f(z) /(z ). (i) Note that /(z ) is already a Laurent series centred at. Hence f has Laurent series f(z) (z ) valid on the annulus {z C < z < }. (ii) Note that f(z) /(z ) is holomorphic on the disc {z C z < }. Therefore we can apply Taylor s theorem and expand f as a power series f(z) +z +3z + +(n+)z n + valid on the disc {z C z < }. (To calculate the coefficients, recall that if f has Taylor series n a nz n then a n f (n) ()/n!. Here we can easily compute that f (n) (z) ( ) n (n + )!(z ) n so that f (n) () (n + )!. Hence a n n +. Alternatively, use the method given in Exercise 3.4.) As a Taylor series is a particular case of a Laurent series, we see that f has Laurent series valid on the disc {z C z < }. (iii) Note that f(z) +z +3z + +(n+)z n + (z ) z ( ). Replacing z by /z in the first part of the computation in (ii) above, we see that ( ) + z + 3 n+ + + z z n + z c University of Manchester 8 3 z

35 MATH Complex Analysis 3. Solutions to Part 6 provided /z <, i.e. provided z >. Multiplying by /z we see that f(z) z + z n + + z4 z n + valid on the annulus {z C < z < }. Solution 6.5 Recall that a function f(z) has a pole at z if f is not differentiable at z (indeed, it may not even be defined at z ). (i) Thepolesof/(z +)occurwhenthedenominatorvanishes. Nowz + (z i)(z+i) so the denominator has zeros at z ±i and both zeros are simple. Hence the poles of /(z +) occur at z ±i and both poles are simple. (ii) The poles occur at the roots of the polynomial z Let z re iθ. Then we have z 4 r 4 e 4iθ 6 6e iπ. Hence r, 4θ π+kπ, k Z. We get distinct values of z for k,,,3. Hence the poles are at e iπ 4 +ikπ, k,,,3, or in algebraic form (+i), ( i), ( +i), ( i). All the poles are simple. (iii) The poles occur at the roots of z 4 +z + (z +) (z+i) (z i). The roots of this polynomial are at z ±i, each with multiplicity. Hence the poles occur at z ±i and each pole is a pole of order. (iv) The poles occur at the roots of z +z, i.e. at z ( ± 5)/, and both poles are simple. Solution 6.6 (i) Since sin z ( ) m+ (m+)!z m+ m our function has infinitely many non-zero term in the principal part of its Laurent series. Hence we have an isolated essential singularity at z. (ii) By Exercise 5., the function sin z has Taylor series Hence ( ) n+ n n z n (n)!. z 3 sin z z 4 4! z + 6 6! z3 so that z 3 sin z has a simple pole at z. c University of Manchester 8 33

36 MATH Complex Analysis 3. Solutions to Part 6 (iii) Since cosz z! + z4 4! we have cosz z + z 4! so that there are no terms in the principal part of the Laurent series. Hence is a removable singularity. Solution 6.7 Expand f as a Laurent series at z and write f(z) b n (z z ) n + a n (z z ) n, n valid in some annulus centred at z. Notice from Laurent s Theorem (Theorem 6..) that b n f(z)(z z ) n dz πi C r where C r is a circular path that lies on the domain D, centred at z of radius r >, and described anticlockwise. By the Estimation Lemma we have that n b n π πr Mrn Mr n as f(z) M at all points on C r. As r is arbitrary and M is independent of r, we can let r and conclude that b n for all n. Hence there are no terms in the principal part of the Laurent series expansion of f at z, and so f has a removable singularity at z. c University of Manchester 8 34

37 MATH Complex Analysis 4. Solutions to Part 7 4. Solutions to Part 7 Solution 7. (i) The function f(z) /z( z ) is differentiable except when the denominator vanishes. Thedenominator vanisheswhenz,±and theseareall simplezeros. Hence there are simple poles at z,±. Then by Lemma 7.4.(i) we have Res(f, ) lim z z z( z ) lim z z ; z(+z) ; Res(f, ) lim(z ) z z( z ) lim z Res(f, ) lim (z +) z z( z ) lim z z( z). (ii) Let f(z) tanz sinz/cosz. Both sinz and cosz are differentiable on C, so f(z) is differentiable except when the denominator is. Hence f has poles at z where cosz, i.e. there are poles at (n + /)π, n Z. These poles are simple (as (n+/)π is a simple zero of cosz). By Lemma 7.4.(ii) we see that Res(f,(n+/)π) sin(n+/)π sin(n+/)π. (iv) Let f(z) z/( + z 4 ). This has poles when the denominator vanishes, i.e. when z 4. To solve this equation, we work in polar coordinates. Let z re iθ. Then z 4 implies that r 4 e 4iθ e iπ. Hence r and 4θ π +kπ. Hence the four quartic roots of are: e iπ/4, e 3iπ/4, e iπ/4,e 3iπ/4. These are all simple zeros of z 4. Hence by Lemma 7.4.(ii) we have that Res(f,z ) z /4z 3 /4z so that Res(f,e iπ/4 ) Res(f,e 3iπ/4 ) Res(f,e iπ/4 ) Res(f,e 3iπ/4 ) 4e iπ/ 4i i 4 4e 3π/ 4i i 4 4e iπ/ 4i i 4 4e 3iπ/ 4i i 4. (v) Let f(z) (z +) /(z +). Then the poles occur when the denominator is zero, i.e. when z ±i. Note that we can write f(z) (z +) (z +i) (z i). c University of Manchester 8 35

38 MATH Complex Analysis 4. Solutions to Part 7 Hence the poles at z ±i are poles of order. By Lemma 7.4. (with m ) we have that d Res(f,i) lim z i dz (z i) f(z) and Res(f, i) lim z i d (z +) lim z i dz (z +i) lim z i (z +i) (z +) (z +) (z i) (z +i) 4 (i) (i+) (i+) (i) (i) 4 i. lim z i d dz (z +i) f(z) d (z +) dz (z i) (z i) (z +) (z +) (z i) lim z i (z i) 4 i. Solution 7. (i) Let f(z) (sinz)/z. As sinz and z are differentiable on C, the poles occur when z. By considering the Taylor expansion of sinz around we have that sinz z z (z z3 3! + z5 5! z z 3! + z 5!. Hence z is a simple pole and Res(f,). (ii) Let f(z) (sin z)/z 4. Recall that sin z ( cosz)/. Hence z 4 sin z z 4( cosx) z 4 ( + (z)! z 3 4! + 5 6! z + (z)4 4! ) + (z)6 6! ) + Hence f has a pole of order at z. The coefficient of the /z term in the above Laurent series for (sin z)/z 4 is. Hence Res(f,). Solution 7.3 (i) We have z( z) z (+z +3z + ) c University of Manchester 8 36

39 MATH Complex Analysis 4. Solutions to Part 7 z ++3z +4z4 +. Hence f has a simple pole at z (as the most negative power appearing in the Laurent series at is /z). The residue Res(f,) is the coefficient of the term /z in the above Laurent series. Hence Res(f,). To calculate the Laurent series at z we change variables and introduce w z. Then z +w. Hence (noting that w ( w) and summing a geometric progression) z( z) w (+w) w ( ( w)) w ( w+w w 3 + ) w w + w +w (z ) (z ) + (z )+(z ) Hence f has a pole of order at z (as the most negative power appearing in the Laurent series at is /(z ) ). The residue Res(f,) is the coefficient of the term (z ) in the above Laurent series. Hence Res(f,). (ii) By Lemma 7.., f(z) /z( z) has a pole of order at z and a pole of order at z. By Lemma 7.4.(i) we have By Lemma 7.4. we have Res(f,) lim z z d Res(f,) lim z dz ( (z ) z( z) lim z ( z). ) d z( z) lim z dz z lim. z Solution 7.4 Suppose that f has a zero of order n at z. Then we can expand f as a Taylor series valid on a disc {z C z z < r} D: f(z) a n (z z ) n +a n+ (z z ) n+ + with a n. Taking out a common factor of (z z ) n we can write f(z) (z z ) n φ(z) where φ is holomorphic on {z C z z < r} and φ(z ) a n. Similarly, we can write g(z) (z z ) m ψ(z) where ψ is holomorphic on a disc centred at z and ψ(z ). Multiplying these together gives f(z)g(z) (z z ) n+m φ(z)ψ(z) where φ(z)ψ(z) is holomorphic on a disc centred at z and φ(z )ψ(z ). Hence f(z)g(z) has a zero of order n+m at z. c University of Manchester 8 37

40 MATH Complex Analysis 4. Solutions to Part 7 Solution 7.5 (i) Let f denote the integrand. Note that z 5z +6 (z )(z 3) so that f has simple poles at z,z 3. Both of these poles are inside C 4. Hence C 4 z dz πires(f,)+πires(f,3). 5z +6 Now by Lemma 7.4.(i) (z ) Res(f, ) lim z (z )(z 3) lim z z 3 (z 3) Res(f, 3) lim z 3 (z )(z 3) lim z 3 z. Hence C 4 z 5z +6 dz πi πi. (ii) Here we have the same integrand as in (i) but integrated over the smaller circle C 5/. This time only the pole z lies inside C 5/. Hence C 5/ z 5z +6 (iii) Let f denote the integrand. Note that Hence f has simple poles at z ±i. Now dz πires(f,) πi. e az +z e az (z +i)(z i). (z i)e az Res(f,i) lim z i (z i)(z +i) lim e az z i z +i eia i Res(f, i) lim z i (z +i)e az (z i)(z +i) lim z i e az e ia z i i. Hence C e az dz πi(res(f,i)+res(f, i)) +z ( e ia ) πi i e ia i πisina. Solution 7.6 (a) (i) Note that x + x. Hence /(x + ) /x. By Lemma 7.5., it follows that the integral is equal to its principal value. c University of Manchester 8 38

41 MATH Complex Analysis 4. Solutions to Part 7 (ii) Let R >. Let S R denote the semi-circular path Re it, t π and let Γ R [ R,R] + S R denote the D-shaped contour that travels along the real axis from R to R and then travels around the semi-circle of centre and radius R lying in the top half of the complex plane from R to R. Let f(z) /(z +). Then z + (z +i)(z i) so f has simple poles at z ±i. Only the pole at z i lies inside Γ R (assuming that R > ). Note that Res(f,i) lim z i (z i) (z +i)(z i) lim z i z +i i. By Cauchy s Residue Theorem, f + f f πires(f,i) πi S R Γ R i π. [ R,R] Now we show that the integral over S R tends to zero as R. On S R we have that z + z R. Hence by the Estimation Lemma, R length(s R) πr R S R f as R. Hence x dx lim f π. + R [ R,R] (Without using complex analysis, you could have done this by noting that, in R, (x +) has anti-derivative arctanx.) (b) (i) Let f(z) e iz /(z +). Note that when x is real f(x) e ix x + x + x. By Lemma 7.5., the integral is equal to its principal value. Note that f(z) eiz z + e iz (z i)(z +i) so that f has simple poles at z ±i. Let Γ R be the path as described in (a)(ii) above. Only the pole at z i lies inside this contour. See Figure 4.. Note that Res(f,i) lim z i (z i)e iz (z i)(z +i) lim z i e iz z +i e i. c University of Manchester 8 39

42 MATH Complex Analysis 4. Solutions to Part 7 i R i R Figure 4.: The contour Γ R and the poles at ±i. By Cauchy s Residue Theorem, f + f f πires(f,i) πi S R ΓR e i [ R,R] πe. Now we show that the integral over S R tends to zero as R. On S R we have that z + z z R. Also, let z x+iy be a point on S R. Then y R, so R y e iz e i(x+iy) e y+ix e y. (4..) Hence f(z) /(R ). Hence by the Estimation Lemma, R length(s R) πr R as R. Hence S R f e ix x dx lim f πe. + R [ R,R] (ii) Taking real and imaginary parts in the above we see that cosx x + dx πe, sinx x dx. + That the latter integral is equal to zero is obvious and we do not need to use complex integration to see this. Indeed, note that sinx x + dx sinx x + dx+ sinx x + dx+ sinx x + dx sinx x + dx where we have used the substitution x x in the first integral. c University of Manchester 8 4

43 MATH Complex Analysis 4. Solutions to Part 7 (iii) Now consider f(z) e iz /(z + ). Suppose we tried to use the D-shaped contour used in (ii) to calculate f(x)dx. Then, with S R as the semi-circle defined above, we would have to bound f(z) on S R in order to use the Estimation Lemma. However, if z x + iy is a point on S R then, noting that y R, e i(x+iy) e y ix e y e R. We still have the bound / z + /(R ). So, using the Estimation Lemma, er R length(s R) er πr R S R f which does not tend to as R (indeed, it tends to ). Instead, we usea D-shaped contour with the negative semi-circle S R described by Re it, t π. We need to be careful about winding numbers and ensure that we travel around a contour in the correct direction to ensure that the contour is simple. Consider the contour Γ R which starts at R travels along the real axis to R, and then follows the negative semi-circle S R lying in the bottom half of the plane. If z is outside Γ R then w(γ R,z) ; however, if z is inside Γ R then w(γ R,z), so that Γ R is not a simple closed loop. However, Γ R is a simple closed loop and, moreover, f f. Γ R Γ R See Figure 4.. The poles of f occur at z ±i and both of these are simple poles. The only pole inside Γ R occurs at z i. Here Hence (z +i)e iz Res(f, i) lim z i (z +i)(z i) lim e iz z i z i e i. Γ R f πires(f, i) πe. Note that if z x+iy is a point on S R then and, as R y z + z R e iz e y ix e y. Hence f(z) /(R ) for z on Γ R. By the Estimation Lemma f R length(s R) πr R as R. Hence S R R f(x)dx f f πe R S R Γ R c University of Manchester 8 4

44 MATH Complex Analysis 4. Solutions to Part 7 -R R i -i (i) -R R i -i (ii) Figure 4.: The contours (i) Γ R and (ii) Γ R and the poles at ±i. Note that Γ R a simple closed loop but that Γ R is. is not and letting R gives that e ix x + dx πe. Solution 7.7 We will use the same notation as in 7.5.: S R denotes the positive semicircle with centre radius R, Γ R denotes the contour [ R,R]+S R. (i) Let f(z) /(z +)(z +3). Note that (x +)(x +4) x 4 so that f(x) /x 4. Hence by Lemma 7.5. the integral converges and equals its principal value. Now f(z) has simple poles at z ±i,±i 3. Suppose R > 3. Then the poles at z i,i 3 are contained in the D-shaped contour Γ R. Now Res(f,i) lim z i z i (z +)(z +3) lim z i (z +i)(z +3) i( +3) c University of Manchester 8 4

45 MATH Complex Analysis 4. Solutions to Part 7 4i Hence [ R,R] Res(f,i 3) lim z i 3 lim z i 3 z i 3 (z +)(z +3) ( 3+)i 3 4i 3. (z +)(z +i 3) f + f f S R Γ R ( πi Res(f,i)+Res(f,i ) 3) ( πi 4i ) 4i 3 π ( ). 3 Now we show that the integral over S R tends to as R. For z on S R we have that (z +)(z +3) ( z )( z 3) (R )(R 3). Hence by the Estimation Lemma (R )(R 3) length(s R) S R f πr (R )(R 3) as R. Hence (x +)(x +3) lim R [ R,R]f π ( ). 3 (ii) Note that 8+x +x 4 (x +4)(x +7). Let f(z) /(z +4)(z +7). Note that (x +4)(x +7) x 4 so that f(x) /x 4. Hence by Lemma 7.5. the integral converges and equals its principal value. Now f(z) has simple poles at z ±i,±i 7. Suppose R > 7. Then the poles at z i,i 7 are contained in the D-shaped contour Γ R. Now Res(f, i) lim z i z i (z +4)(z +7) lim z i (z +i)(z +7) 4i( 4+7) c University of Manchester 8 43

46 MATH Complex Analysis 4. Solutions to Part 7 Hence [ R,R] i Res(f,i 7) lim z i 7 lim z i 7 z i 7 (z +4)(z +7) ( 7+4)i 7 6i 7. (z +4)(z +i 7) f + f f S R Γ R ( πi Res(f,i)+Res(f,i ) 7) ( πi i ) 6i 7 π ( 3 ). 7 Now we show that the integral over S R tends to as R. For z on S R we have that (z +4)(z +7) ( z 4)( z 7) (R 4)(R 7). Hence by the Estimation Lemma (R 4)(R 7) length(s R) as R. Hence Solution 7.8 Let S R f πr (R 4)(R 7) (x +4)(x +7) lim R [ R,R]f π ( 3 ). 7 f(z) e iz z +4z +5. Then f(x) C x for some constant C >. Hence by Lemma 7.5. the integral f(x)dx exists and is equal to its principal value. Now f(z) has poles when z +4z +5, i.e. at z ±i. Both of these poles are simple. Let Γ R denote the D-shaped contour [ R,R] + S R. Provided R is sufficiently large, only the pole at +i lies inside Γ R. Now Res(f, +i) lim z +i (z ( +i))e iz (z ( +i))(z ( i)) e iz lim z +i z ( i) c University of Manchester 8 44