COMPLEX ANALYSIS: SOLUTIONS 5
1. Find the poles and residues of the following functions 1 1 π cot(πz) 1 , , , (m, n ∈ ) z4 + 5z2 + 6 (z2 − 1)2 z2 zm(1 − z)n Z>0
Solution: Throughout we use the following formula for calculating residues: If f(z) has a pole of order k at z = z0 then 1 dk−1 k res(f, z0) = k−1 (z − z0) f(z) . (k − 1)! dz z=z0
In particular, if f(z) has a simple pole at z0 then the residue is given by simply evaluating the non-polar part: (z −z0)f(z), at z = z0 (or by taking a limit if we have an indeterminate form).
Let 1 1 1 f(z) := = = √ √ √ √ . z4 + 5z2 + 6 (z2 + 2)(z2 + 3) (z + i 2)(z − i 2)(z + i 3)(z − i 3) √ √ This has simple poles at z = ±i 2, ±i 3 with residues
√ √ 1 1 1 res(f, i 2) = (z − i 2) √ = √ √ = √ , z4 + 5z2 + 6 z=i 2 (z + i 2)(z2 + 3) z=i 2 2i 2
√ √ 1 1 res(f, −i 2) = (z + i 2) √ = − √ , z4 + 5z2 + 6 z=−i 2 2i 2
√ 1 res(f, i 3) = − √ , 2i 3
√ 1 res(f, −i 3) = √ . 2i 3 1 COMPLEX ANALYSIS: SOLUTIONS 5 2
For the second one let 1 1 f(z) = = . (z2 − 1)2 (z + 1)2(z − 1)2 This has double poles at ±1. From the formula we get d 1 res(f, 1) = = −1/4, dz (z + 1)2 z=1 d 1 res(f, −1) = = 1/4. dz (z − 1)2 z=−1
For the third let π cot(πz) f(z) = . z2 Now, cot(πz) has poles wherever sin(πz) = 0, so at z = n ∈ Z. About these points we have (z − n)2 (z − n)3 sin(πz) = sin(πn) + π cos(πn)(z − n) − π2 sin(πn) − π3 cos(πn) + ··· 2! 3! (z − n)2 =(−1)nπ(z − n)1 − π2 + O((z − n)4) 3! and (z − n)2 (z − n)3 cos(πz) = cos(πn) − π sin(πn)(z − n) − π2 cos(πn) + π3 sin(πn) + ··· 2! 3! (z − n)2 =(−1)n1 − π2 + O((z − n)4). 2! Hence, for z close to n ∈ Z, we have 1 − π2(z − n)2/2 + O((z − n)4) cot(πz) = π(z − n)1 − π2(z − n)2/6 + O((z − n)4) 1 − π2(z − n)2/2 + O((z − n)4) = 1 + π2(z − n)2/6 + O((z − n)4) π(z − n) 1 π = − (z − n) + O((z − n)3) π(z − n) 3 Therefore, f(z) = π cot(πz)/z2 has simple poles at z = n 6= 0 and a triple pole at z = 0. For the simple poles we have
π cot(πz) 1 res(f, n) = (z − n) 2 = 2 . z z=n n COMPLEX ANALYSIS: SOLUTIONS 5 3
For the triple pole at at z = 0 we have 1 π2 1 f(z) = − + O(z) z3 3 z so the residue is −π2/3.
Finally, the function 1 f(z) = zm(1 − z)n has a pole of order m at z = 0 and a pole of order n at z = 1. From the formula for residues we have m−1 1 d 1 res(f, 0) = m−1 n (m − 1)! dz (1 − z) z=0 n(n + 1) ··· (n + m − 2) = (m − 1)! (n + m − 2)! = (n − 1)!(m − 1)! and n−1 n 1 d (−1) res(f, 1) = n−1 m (n − 1)! dz z z=1 −m(m + 1) ··· (m + n − 2) = (n − 1)! (m + n − 2)! = − (m − 1)!(n − 1)! = − res(f, 0).
2. Use the substitution eiθ = z along with the residue theorem to show that Z 2π dθ 2π = √ . 0 2 + cos θ 3
Solution: As suggested we let eiθ = z so that dθ = dz/(iz) and the integral becomes 1 Z dz 2 Z dz −1 = 2 . i |z|=1 z(2 + (z + z )/2) i |z|=1 z + 4z + 1 √ 2 Now z + 4z + 1 has zeros of order 1 at z = z± = −2 ± 3 and so the integrand has simple poles at z+ and z−. Only z+ lies in the unit disk and therefore by the residue COMPLEX ANALYSIS: SOLUTIONS 5 4 theorem 2 Z dz 2 1 2 = × 2πi × res 2 , z+ i |z|=1 z + 4z + 1 i z + 4z + 1
1 =4π(z − z+) z2 + 4z + 1 z=z+
1 =4π(z − z+) (z − z )(z − z ) + − z=z+ 1 =4π z+ − z− 1 =4π √ 2 3 2π =√ . 3 3. Evaluate the following integrals via residues. Show all estimates. (i) Z ∞ x2 4 2 dx 0 x + 5x + 6 (ii) Z ∞ x sin x 2 2 dx, a real 0 x + a (iii) Z ∞ log x 2 dx 0 1 + x
Solution: (i) Since the integrand is an even function the integral in question is equal to I/2 where Z ∞ x2 I = 4 2 dx. −∞ x + 5x + 6 √ √ As a function of a complex variable, the integrand has simple poles at ±i 2, ±i 3. We will be considering a semicircular√ √ contour in the upper half plane so we only need calculate the residues at z = i 2, i 3. A slight modification of the first calculation in question 1 gives √ √ z2 √ (i 2)2 i 2 res , i 2 = √ = z4 + 5z2 + 6 2i 2 2 COMPLEX ANALYSIS: SOLUTIONS 5 5 and √ √ z2 √ (i 3)2 i 3 res , i 3 = − √ = − . z4 + 5z2 + 6 2i 3 2
Now, Consider the semicircular contour ΓR, which starts at R, traces a semicircle in the upper half plane to −R and then travels back to R along the real axis. Then, on taking R large enough, by the residue theorem √ √ Z z2 X i 2 i 3 √ √ 4 2 dz = 2πi× residues inside ΓR = 2πi( − ) = π( 3− 2). ΓR z + 5z + 6 2 2 On the other hand Z z2 Z z2 lim dz = I + lim dz R→∞ 4 2 R→∞ 4 2 ΓR z + 5z + 6 γR z + 5z + 6 where γR is the semicircle in the upper half plane. But by the Estimation Lemma Z z2 z2 1
4 2 dz 6 πR max 4 2 → 0 z∈γR γR z + 5z + 6 z + 5z + 6 R √ √ as R → ∞. Hence, I = π( 3 − 2) and so Z ∞ x2 π √ √ 4 2 dx = ( 3 − 2). 0 x + 5x + 6 2
The last estimate in the inequality for the integral over γR should be clear since the dominant term is the z4 term in the denominator, but for completeness: z2 π 1
πR max 4 2 = max −2 −4 z∈γR z + 5z + 6 R z∈γR 1 + 5z + 6z π 1 6 max −2 −4 R z∈γR 1 − 5|z| − 6|z| π 1 = R 1 − 5R−2 − 6R−4 where we have used |z + w| > |z| − |w| in the second line. On taking R > 5, say, this is 6 2π/R and the constant 2π is absorbed into the symbol.
(ii). We have Z ∞ x sin x 1 Z ∞ x sin x 1 Z ∞ xeix 2 2 dx = 2 2 dx = = 2 2 dx . 0 x + a 2 −∞ x + a 2 −∞ x + a Denote this last integral by J. Again, we will consider J as the horizontal section of the contour ΓR from part (i). COMPLEX ANALYSIS: SOLUTIONS 5 6
In the upper half plane the integrand has a simple pole at z = ia with residue
iz iz −a −a ze ze iae e res 2 2 , ia = (z − ia) 2 2 = = . z + a z + a z=ia 2ia 2 Hence, by the residue theorem Z zeiz Z zeiz πie−a = lim dz = J + lim dz. R→∞ 2 2 R→∞ 2 2 ΓR z + a γR z + a Thus it remains to show that this last integral vanishes in the limit. This is similar to question 7 (ii) of Problems 3; a trivial estimate of the integrand is 1/R which is not enough for the Estimation Lemma. Instead we apply integration by parts which is probably the quickest way (see Problems 3). Integrating eiz and differentiating the rest gives
Z iz iz −R Z 2 ze ze 1 1 2z iz 2 2 dz = 2 2 − 2 2 − 2 2 2 e dz. γR z + a i(z + a ) R i γR z + a (z + a ) The first term on the right is −2R cos R/i(R2 + a2) 1/R. For the integrals we use the Estimation Lemma to give Z iz iz e e 1 1 2 2 dz 6 πR max 2 2 6 πR 2 2 , z∈γR γR z + a z + a R − a R
Z 2 iz 2 iz z e z e 3 1 1 2 2 2 dz 6 πR max 2 2 2 6 πR 2 2 2 , z∈γR γR (z + a ) (z + a ) (R − a ) R as R → ∞. Hence, J = πie−a and so Z ∞ x sin x 1 π −a 2 2 dx = =(J) = e . 0 x + a 2 2
(iii) This is quite hard and, as I discovered recently, the solution is in Conway’s book anyway (pg. 117–118). My bad. It’s still instructive to attempt this before reading Conway though. We see that, as a function of a complex variable, the integrand has a branch cut and simple poles at z = i, −i. Taking the branch of the log with −π < arg(z) < π, we would like to choose a contour which lies just above and below the cut and that also picks up the residues at i, −i. A natural choice is the contour we used in the lectures. This consisted of a small circle about z = 0, horizontal lines just above and below the negative real axis, and a large circle completing the contour. COMPLEX ANALYSIS: SOLUTIONS 5 7
Unfortunately, with this choice the integrals over the horizontal lines L1, L2 are given by Z log z Z log z 2 dz + 2 dz L1 1 + z L2 1 + z Z 0 log(x + iδ) Z −∞ log(x − iδ) = 2 dx + 2 dx −∞ 1 + (x + iδ) 0 1 + (x − iδ) Z 0 log |x| + πi Z −∞ log |x| − πi → 2 dx + 2 dx −∞ 1 + x 0 1 + x Z 0 1 =2πi 2 dx −∞ 1 + x and the integral we’re after has disappeared. This motivates the choice of a new contour: we want something with a horizontal line over the whole real axis, since then the integral over this line is given by Z ∞ log |x| Z ∞ log x 2 dx = 2 2 dx −∞ 1 + x 0 1 + x and we avoid the problem of cancellation. Consequently, we choose a branch of log z with a branch cut along the negative imaginary axis. To avoid z = 0 and the branch cut we indent our contour with a small circle in the upper half plane. We complete the contour with a large circle. Thus, let Γr,R be the contour consisting of a line from r to R, a semicircle in the upper half plane traced from R to −R, a line from −R to −r, and a semicircle in the upper half plane traced from −r to r. Follwing the details in Conway we find Z ∞ log x 2 dx = 0. 0 1 + x
4. Let ΓN be the square which crosses the real axis at ±(N + 1/2) with N ∈ N. (i) Show that cot(πz) is bounded on ΓN and hence show that Z π cot(πz) lim dz = 0. N→∞ 2 ΓN z (ii) For a given N compute the above integral via residues. Conclude something interesting.
Solution: (i) We have eπiz + e−πiz e2πiz + 1 e2πixe−2πy + 1 cot(πz) = i = i = i . eπiz − e−πiz e2πiz − 1 e2πixe−2πy − 1 COMPLEX ANALYSIS: SOLUTIONS 5 8
Thus, if y > 1 1 + e−2πy 1 + e−2π | cot(πz)| . 6 1 − e−2πy 6 1 − e−2π Since cot(πz) = cot(πz) we obtain the same bound for y 6 −1. It remains to show that cot(πz) is bounded for z = ±(N + 1/2) + iy with |y| < 1. But in this case
±2πi(N+1/2) −2πy −2πy 2π e e + 1 | − e + 1| e − 1 | cot(πz)| = = < =: C. e±2πi(N+1/2)e−2πy − 1 | − e−2πy − 1| e−2π + 1
Since C is bigger than the bound for |y| > 1, we may say that | cot(πz)| < C on all of ΓN .
(ii) By the estimation lemma Z π cot(πz) cot(πz) C 8πC 2 dz 6 8π(N+1/2) max 2 < 8π(N+1/2) 2 = z∈ΓN ΓN z z (N + 1/2) N + 1/2 and this tends to zero as N → ∞. On the other hand, for a fixed N this integral is given by 2πi × the sum of residues inside ΓN . Using the results from question 1 then gives Z π cot(πz) X 1 π2 2 dz = 2πi 2 − . Γ z n 3 N −N6n6N n6=0 Therefore, on letting N → ∞ we have
∞ X 1 π2 X 1 π2 0 = − = 2 − n2 3 n2 3 −∞6n6∞ n=1 n6=0 and so ∞ X 1 π2 ζ(2) = = , n2 6 n=1 which is interesting. 5. (i) Let f(z) = z6 + cos z. Find the change in argument of f(z) as z travels once around the circle of radius 2, center zero, in the positive direction.
z (ii) How many solutions does 3e − z = 0 have in the disk |z| 6 1? (iii) Use Rouche’s Theorem to prove that a polynomial of degree n has n roots in C. COMPLEX ANALYSIS: SOLUTIONS 5 9
Solution: (i) By the argument principle, the change in argument of f(z) as z travels around the circle is equal to (2π ×) the number of zeros minus poles of f(z) inside the circle, so just the number of zeros then. To find the number of zeros of f(z) inside the circle we compare f(z) with its dominant term z6 and apply Rouche’s Theorem. So let g(z) = z6. On the circle we have |g(z)| = |z|6 = 26 = 64 and |eiz| + |e−iz| |f(z) − g(z)| = | cos(z)| e|z| = e2 < 64. 6 2 6 Thus, Rouche’s Theorem applies and f(z) has the same number of zeros inside the circle as z6, which is 6. Hence, the change in argument is 2π × 6 = 12π.
(ii). Rephrasing the question we ask for the number of zeros of f(z) = 3ez − z in the closed unit disk. We apply Rouche’s Theorem again. This time the dominant term is g(z) = 3ez. On the unit circle we have
<(z) −1 |g(z)| = 3e > 3e > 1 and |f(z) − g(z)| = |z| = 1.
By Rouche’s Theorem 3ez − z has the same number of zeros in the unit disk as 3ez, which is none. So the answer is no solutions.
(iii) In the lectures we showed that the polynomial z6 + z + 2 has 6 zeros in the 6 disk |z| 6 2 by comparing it with z and applying Rouche’s Theorem. We generalise this by comparing a general polynomial
n n−1 f(z) = anz + an−1z + ··· + a0, an 6= 0 with its leading term
n g(z) = anz on arbitrarily large circles. On the circle |z| = R > 1 we have
n |g(z)| = |an|R COMPLEX ANALYSIS: SOLUTIONS 5 10 and n−1 n−2 |f(z) − g(z)| =|an−1z + an−2z + ··· + a0| n−1 n−2 6|an−1|R + |an−2|R + ··· + |a0| n−1 n−1 Therefore, on a circle of radius R > n maxj=0 |aj|/|an| we have |f(z)−g(z)| < |g(z)|. By Rouche’s Theorem f(z) has the same number of zeros inside the circle as g(z), which is n. 6. Prove that the function X 1 f(z) = (n + z)2 n∈Z is meromorphic on C. Solution: Clearly, f(z) has poles only at the integers so we need to show that f(z) is analytic on C\Z. By the results given in the lectures, this will follow if we can show that the series is uniformly convergent on compact subsets of said region. Since 1 X 1 f(z) = + z2 (n + z)2 n6=0 it suffices to show the uniform convergence of this last sum. So let |z| 6 R, let z be fixed away from integers: |z + n| > ρ > 0 ∀n ∈ Z, and let P −2 −M be the nearest integer to z. Note that the series looks like 2 n6=0 n which is convergent. So we first pull out the convergent part by writing X 1 X 1 = . (n + z)2 n2|1 + z/n|2 6=0 n6=0 Then if |n| 6 |M| |z + n| > |z + m| =⇒ |1 + z/n| > (|M|/|n|)|1 + z/m| > |1 + z/M| > ρ/|M|. If |n| > |M| + 1 |<(z)| |M| + 1 1 |1 + z/n| |1 + <(z)/n| 1 − > 1 − = . > > |n| |M| + 2 |M| + 2 COMPLEX ANALYSIS: SOLUTIONS 5 11 Therefore, 2 X 1 |M| X 1 2 X 1 1 + (|M| + 2) + (n + z)2 6 ρ2 n2 n2 |z ± (M ± 1)|2 n6=0 06=|n|6|M| |n|>|M|+1 (R + 1)2 4 2ζ(2) + 2ζ(2)(R + 3)2 + 6 ρ2 ρ2 <∞. −2 2 2 2 Hence, by the Weierstrass M test with Mn = n max((R + 3) , (R + 1) /ρ ) the series is uniformly convergent on compact subsets of C\Z. (k) 7. Let ϕ : R → R be a smooth function whose derivatives ϕ (t), k > 0, are of rapid decay at ∞ i.e. lim tAϕ(k)(t) = 0 t→∞ for all A ∈ R and all k > 0. The Mellin transform of ϕ is defined as Z ∞ ϕ˜(z) = ϕ(t)tz−1dt. 0 (i) Prove thatϕ ˜(z) is analytic in the region <(z) > 0. (ii) Use integration by parts to show thatϕ ˜(z) can be analytically continued to C\Z60, with possible simple poles at z = −n. (iii) Find the residues at these poles and compute the formal sum ∞ X res(ϕ ˜(z)t−z, −n). n=0 What does this look like? (iv) Given part (iii), suggest an argument that would prove 1 Z ϕ˜(z)t−zdz = ϕ(t) 2πi <(z)=c where the integral is over the vertical line from c − i∞ to c + i∞ with c > 0. Solution: (i). Since ϕ is smooth on R it is bounded in a neighbourhood of t = 0: |ϕ(t)| 6 C for t ∈ [0, t0], say. Then Z t0 Z t0 z−1 <(z)−1 <(z) |ϕ(t)t |dt 6 C t dt = Ct0 /<(z) < ∞. 0 0 COMPLEX ANALYSIS: SOLUTIONS 5 12 Also, since ϕ is of rapid decay, for any z there exists a t1 = t1(<(z)) such that z−1 −2 |ϕ(t)t | 6 Dt for all t > t1. Then Z ∞ z−1 |ϕ(t)t |dt 6 D/t1 < ∞. t1 Since the remaining integral R t1 |ϕ(t)tz−1|dt is finite the integral is absolutely con- t0 vergent in the region <(z) > 0. Let γ be a closed curve in the region <(z) > 0. By absolute convergence and Fubini’s Theorem Z Z Z ∞ Z ∞ Z ϕ˜(z)dz = ϕ(t)tz−1dtdz = ϕ(t)t−1 tzdzdt. γ γ 0 0 γ But since tz is analytic in the region <(z) > 0 for all t ∈ [0, ∞) this last inner integral is zero by Cauchy’s Theorem. Hence, Z ϕ˜(z)dz = 0. γ The continuity ofϕ ˜(z) follows exactly as it does for the Gamma function, hence it is analytic by Morera’s Theorem. (ii). Integrating by parts once gives z ∞ Z ∞ Z ∞ ϕ(t)t 1 0 z 1 0 z ϕ˜(z) = − ϕ (t)t dt = − ϕ (t)t dt. z t=0 z 0 z 0 R ∞ 0 z By the same reasoning used in part (i), the integral 0 ϕ (t)t dt is analytic in the region <(z) > −1. Hence, the above expression provides a meromorphic continuation ofϕ ˜(z) to said region. At z = 0 we have a residue of Z ∞ Z ∞ 1 0 z 0 z · − ϕ (t)t dt = − ϕ (t)dt = ϕ(0). z 0 z=0 0 In particular, if ϕ(0) = 0 then there is no pole at z = 0. More generally, integrating by parts n times gives (−1)n Z ∞ (1)ϕ ˜(z) = ϕ(n)(t)tz+n−1dt. z(z + 1) ··· (z + n − 1) 0 Again, this expression is seen to be analytic in the region <(z) > −n, with the exception of possible simple poles at z = 0, −1,..., −n + 1. Since n is arbitrary, we’re done. COMPLEX ANALYSIS: SOLUTIONS 5 13 (iii). From equation (1) we see that n+1 Z ∞ (−1) (n+1) z+n res(ϕ ˜(z), z = −n) =(z + n) ϕ (t)t dt z(z + 1) ··· (z + n) 0 z=−n (−1)n+1 Z ∞ = ϕ(n+1)(t)dt (−n)(−n + 1) ··· (−1) 0 1 Z ∞ = − ϕ(n+1)(t)dt n! 0 1 = ϕ(n)(0). n! In particular, if ϕ(t) is of rapid decay at zero as well as at infinity, thenϕ ˜(z) is entire. Since all the poles are simple, an extra factor of t−z in the above residue calculations −z n gives a factor of t |z=−n = t and hence ∞ ∞ X X 1 res(ϕ ˜(z)t−z, −n) = ϕ(n)(0)tn. n! n=0 n=0 This, of course, looks like ϕ(t) in the form of a Taylor expansion about t = 0. (iv). Assuming that the above series converges and equals ϕ(t) (so ϕ is analytic in the real analysis sense) then we would like a contour integral which captures all these residues. As suggested, we should look at 1 Z ϕ˜(z)t−zdz. 2πi <(z)=c We would like to truncate this integral at heights z = c ± iT and then consider the truncated integral as part of a rectangular contour which encloses some of the residues. To estimate these integrals we need bounds onϕ ˜(z). From equation (1) we see that for fixed <(z),ϕ ˜(z) → 0 as |=(z)| → ∞. More precisely, as =(z) → ∞ equation (1) gives Z ∞ 1 (n) <(z)+n−1 C<(z),n |ϕ˜(z)| 6 |ϕ (t)|t dt 6 n , |z(z + 1) ··· (z + n − 1)| 0 |=(z)| for some constant C<(z),n. Consequently, 1 Z 1 Z c+iT 1 h Z ∞ Z −T i ϕ˜(z)t−zdz = ϕ˜(z)t−zdz + + ϕ˜(c + iy)t−c−iydy 2πi <(z)=c 2πi c−iT 2π T −∞ 1 Z c+iT = ϕ˜(z)t−zdz + O(T −n+1) 2πi c−iT COMPLEX ANALYSIS: SOLUTIONS 5 14 We now consider this last integral as part of a rectangular contour Γ whose left edge crosses the real axis at −N − 1/2. Then, 1 Z c+iT 1 Z 1 Z ϕ˜(z)t−zdz = ϕ˜(z)t−zdz − ϕ˜(z)t−zdz 2πi c−iT 2πi Γ 2πi other edges of Γ N X 1 1 Z = ϕ(n)(0)tn − ϕ˜(z)t−zdz n! 2πi n=0 other edges of Γ by the residue theorem. Using our bound on |ϕ˜(z)| we can estimate the remaining integrals. The largest contribution comes from the integral over the left edge of Γ, and this is seen to be O(tM+1/2). The other integrals are small in size. On letting T → ∞ we get N 1 Z X 1 ϕ˜(z)t−zdz = ϕ(n)(0)tn + O(tN+1/2) = ϕ(t). 2πi n! <(z)=c n=0