Solutions to Tutorial 12 (Week 13) Material Covered Outcomes Summary of Essential Material

The University of Sydney School of Mathematics and Statistics Solutions to Tutorial 12 (Week 13)

MATH2962: Real and Complex Analysis (Advanced) Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2962/ Lecturer: Florica C. Cˆırstea

Questions with * are more diﬃcult questions.

Material covered (1) Cauchy Integral theorem and integral formula; (2) Laurent expansions; (3) Isolated singularities, their classiﬁcation and residues; (4) The residue theorem and application to real integrals.

Outcomes This tutorial helps you to

(1) know how to apply the basic theorems of complex analysis such as the Cauchy integral theorem and formula, Laurent expansions, the residue theorem with application to real integrals. (2) have a working knowledge of (real and complex) power series and their properties; (3) know the basic facts on analytic functions including the uniqueness theorem;

Summary of essential material

Cauchy integral formula and analyticity. Let D ⊆ C be open and let z0 ∈ D. Suppose 1 N that r > 0 is such that B(z0, r) ⊆ D. Such r > 0 exists since D is open. Let f ∈ C (D, C ) it and γr := z0 + e , t ∈ [0, 2π]. If |z − z0| < r, then the Cauchy integral formula applies: 1 Z f(w) f(z) = dw. 2πi γr w − z 1 Expanding about z by means of the geometric series, and applying the Weierstrass w − z 0 M-Test we derive the power (Taylor) series expansion

∞ Z X k 1 (k) 1 f(w) f(z) = ak(z − z0) with ak = f (z0) = k+1 dw. (1) k! 2πi (w − z0) k=0 γr

The radius r can be chosen arbitrarily so that B(z0, r) ⊆ D. The radius of convergence, ρ, of the series satisﬁes the estimate

ρ ≥ sup{r > 0: B(z0, r) ⊆ D} > 0.

Copyright c 2017 The University of Sydney 1 Cauchy integral formula on annulus and Laurent expansions. Let D ⊆ C be open and let z0 ∈ C (not necessarily in D). Suppose that ∞ > r2 > r1 > 0 is such that the annulus 1 N it {z ∈ C: r1 ≤ |z − z0| ≤ r2} is contained in D. Let f ∈ C (D, C ) and γr := z0 + e , t ∈ [0, 2π]. If r1 < |z − z0| < r2, then the Cauchy integral formula for an annulus applies: 1 Z f(w) 1 Z f(w) f(z) = dw − dw. 2πi w − z 2πi w − z γr2 γr1 1 Using an inner and an outer expansion of about z by means of the geometric series, w − z 0 and applying the Weierstrass M-Test we derive the Laurent series expansion ∞ ∞ X k X 1 f(z) = ak(z − z0) + a−k k (2) (z − z0) k=0 k=1 | {z } | {z } regular part singular part with 1 Z f(w) ak = k+1 dw (k ∈ Z) 2πi γr (w − z0) The radius r can be chosen arbitrarily so that r1 ≤ r ≤ r2. The the annulus on which the series converges is at least the largest annulus contained in D centred at z0. Note that if f is analytic on B(z0, r2), then the Laurent expansion (2) reduces to the power (Taylor) series expansion (1) of f about z0.

Classiﬁcation of singularities. We say z0 is an isolated singularity of an analytic function f if f is analytic on B(z0, r) \{z0} for some r > 0. We can represent f by its Laurent series in the annulus 0 < |z − z0| < r. We classify the isolated singularities depending on the nature of the Laurent series (2).

Removable singularity. a−k = 0 for all k ≥ 1, that is, the singular part of (2) vanishes. Then f(z0) := lim f(z) z→z0

exists and f can be extended to an analytic function on B(z0, r). Pole of order m. The singular part of (2) only has ﬁnitely many non-zero terms. m Let m ≥ 1 be the largest index such that am 6= 0. Setting g(z) := (z −z0) f(z),

m g(z0) := lim (z − z0) f(z) = a−m 6= 0 z→z0

exists and g can be extended to an analytic function on B(z0, r). It is essential that the limit is non-zero! In that case we can write g(z) f(z) = m , (z − z0)

where g is an analytic function on B(z0, r) with g(z0) 6= 0. If m = 1 we say z0 is a simple pole.

Essential singularity. The singluar part of (2) has inﬁnitely many non-zero terms. This is equivalent to

m lim (z − z0) f(z) does not exist for all m ∈ N. z→z0

2 Residues and the residue theorem. If z0 is an isolated singularity of the analytic function f, then the coeﬃcient a−1 in the Laurent expansion (2) is called the residue of f at z0. We have 1 Z Res[f, z0] := a−1 = f(z) dz, 2πi γr where γr is as before.

We can compute the residues as follows, depending on the nature of the singularity:

(i) Simple pole at z0: Res[f, z0] = lim (z − z0)f(z); z→z0

d 2 (ii) Pole of order 2 at z0: Res[f, z0] = lim (z − z0) f(z) ; z→z0 dz

m−1 1 d m (iii) Pole of order m at z0: Res[f, z0] = lim m−1 (z − z0) f(z) ; z→z0 (m − 1)! dz g(z) Alternatively, if f(z) = m with g analytic, and g(z0) 6= 0, then (z − z0)

1 dm−1 Res[f, z ] = g(z ). 0 (m − 1)! dzm−1 0

1 Z (iv) Essential singularity: Compute Laurent series expansion, or f(z) dz. 2πi γr This can be diﬃcult in practice.

The Residue Theorem lets us compute path integrals over a path γ enclosing ﬁnitely many isolated singularities z1, . . . , zn of an analytic function f:

n Z X f(z) dz = 2πi Res[f, zk]. γ k=1

Questions to complete during the tutorial f 1. Let f, g be analytic functions deﬁned on suitable open subsets of . Explain why fg, , C g f ◦ g and f −1 if they exist are analytic on their natural domains. Solution: The product, quotient and chain rules, as well as the formula for the derivative of an inverse function, show that the derivatives of the given functions are continuous function. By a result from lectures they are analytic.

1 2. Consider f(z) := . Determine the Laurent expansion about z = 0 for z(z − 1)(z − 2i) 0 the following annuli. To get the Laurent expansions we use the partial fraction decomposition 1 1 1 1 = − (3) (z − 1)(z − 2i) 1 − 2i z − 1 z − 2i

and then we expand the fractions using the geometric series in zk, k ∈ Z.

3 (a)0 < |z| < 1; Solution: If 0 < |z| < 1 the geometric series expansions

∞ ∞ 1 1 X 1 1 1 1 X 1 = − = − zk and = − = − zk (4) z − 1 1 − z z − 2i 2i 1 − z 2i (2i)k k=0 2i k=0 both converge. (The second one in fact converges for |z| < 2.) Using (3) we get the Laurent expansion

∞ ∞ ∞ 1 1 X 1 X 1 1 1 1 X 1 f(z) = − zk + zk = + − 1 zk. z 1 − 2i 2i (2i)k 2i z 1 − 2i (2i)k+2 k=0 k=0 k=0

(b)1 < |z| < 2; Solution: If |z| > 1, then |1/z| < 1 and so

∞ ∞ 1 1 1 1 X1k X = = = z−k (5) z − 1 z 1 − 1 z z z k=0 k=1 converges. Using (4) and (5), jointly with (3), we ﬁnd the Laurent expansion

∞ ∞ ∞ ∞ 1 1 X 1 X 1 1 X 1 1 X 1 f(z) = z−k+ zk = z−k+ + zk z 1 − 2i 2i (2i)k 1 − 2i 2i z (2i)k+2 k=1 k=0 k=2 k=0 valid for 1 < |z| < 2. (c)2 < |z| < ∞; Solution: For |z| > 2 we have |2i/z| < 1 and thus we use the outer expansion

∞ ∞ 1 1 1 1 X2ik X = = = (2i)k−1z−k. (6) z − 2i z (1 − 2i ) z z z k=0 k=1 Hence, using (3), (5) and (6), we obtain the Laurent expansion

∞ ∞ ∞ 1 1 X X 1 X f(z) = z−k − (2i)k−1z−k = 1 − (2i)k−2z−k. z 1 − 2i 1 − 2i k=1 k=1 k=2 valid for |z| > 2.

3. Find the isolated singularities of the following functions and determine their nature.

sin z (a) z2 sin z Solution: The only singular point is z = 0. Since → 1 as z → 0, we z conclude that z = 0 is a pole of order one. cos z − 1 (b) z3 cos z − 1 Solution: The only singular point is z = 0. As → −1/2 as z → 0 we z2 conclude that z = 0 is a pole of order one.

4 exp(z) − 1 (c) z2(z − 1)2 exp(z) − 1 Solution: The singular points are z = 0 and z = 1. As → 1 as z → 0, z(z − 1)2 exp(z) − 1 we conclude that z = 0 is a pole of order one. Since → e − 1 6= 0, we z2 infer that z = 1 is a pole of order two. 1 (d) z4 sin z ∞ X z2k+1 Solution: Since sin z = (−1)k we have (2k + 1)! k=0 ∞ 1 X z−(2k+1) z4 sin = z4 (−1)k . z (2k + 1)! k=0 Hence, z = 0 is an essential singularity. 1 (e) Log(1 + z) z 1 Solution: The only isolated singularity is z = 0. Moreover, Log(1 + z) → 1 z as z → 0, so zero is a removable singularity. Note that z = −1 is not an isolated singularity, since Log is not continuous along the negative real axis. 1 + z (f) 1 − z4 Solution: As 1 − z4 = (1 − z2)(1 + z2) = −(z − 1)(z + 1)(z − i)(z + i), we ﬁnd that the points z = 1, −1, i, −i are isolated singularities. The point z = −1 is removable, the others are poles of order 1.

4. Determine the singularities and the corresponding residues for the following functions

1 (a) exp ; z Solution: The Laurent expansion about the singularity z = 0 is

∞ X z−k f(z) = . k! k=0 Since the coefficient of z−1 is 1, it follows that Res[f, 0] = 1. 1 (b) , where a > 0. a2 − z2 Solution: Since 1 1 f(z) := = − a2 − z2 (z − a)(z + a) we find that a is a pole of order one. Hence the residue is 1 1 Res[f, a] = lim(z − a)f(z) = − lim = − z→a z→a z + a 2a Similarly, −a is a pole of order one and hence the residue is 1 1 Res[f, −a] = lim (z + a)f(z) = − lim = . z→−a z→−a z − a 2a 5 1 1 Remark. For a = 0, the function f(z) = = − has z = 0 as an isolated a2 − z2 z2 2 singularity, which is a pole of order 2. Indeed, limz→0 z f(z) = −1 6= 0. In this case, we have Res[f, 0] = 0. We can either use the formula for a pole of order 2, d that is Res[f, 0] = lim (z2f(z)) = 0 or observe that f(z) = −1/z2 represents z→0 dz the Laurent series of f around 0 for 0 < |z| < ∞ and the coefficient of 1/z is 0. z + 1 (c) . z2 − π √ Solution: There are two simple poles, namely z = ± π. Hence the residues are √ √ √ z + 1 π + 1 Res[f, π] = lim√ (z − π)f(z) = lim√ √ = √ z→ π z→ π z + π 2 π and √ √ √ √ z + 1 − π + 1 π − 1 Res[f, − π] = lim√ (z + π)f(z) = lim√ √ = √ = √ . z→− π z→− π z − π −2 π 2 π

5. Using the residue theorem, compute the following integrals. Z ∞ 1 (a) 2 2 dx −∞ a + x Solution: As shown in lectures, the integral equals the sum of the residues of the integrand in the upper half plane since the integrand decays like x2. The only singularity in the upper half plane is ia (a > 0). Since 1 1 f(z) := = a2 + z2 (z − ia)(z + ia) ia is a pole of order one. Hence the residue is 1 1 Res[f, ia] = lim (z − ia)f(z) = lim = z→ia z→ia z + ia 2ia Hence, Z ∞ 1 1 π 2 2 dx = 2πi Res[f, ia] = 2πi = . −∞ a + x 2ai a Note that this is the same results we get using arctan. Z ∞ x2 (b) 4 2 dx. 0 x + x + 1 Solution: As shown in lectures, the integral equals the sum of the residues of the integrand in the upper half plane since the integrand decays like x2. We need to determine the poles of the integrand. The denominator x4 + x2 + 1 is reducible to a quadratic in x2. It is zero if √ 1 1√ 1 3 x2 = − ± −3 = − ± i = exp(±2πi/3). 2 2 2 2 4 2 Hence the only roots of x + x + 1 = 0 in the upper half plane are z1 = exp(πi/3) and z2 = exp(2πi/3), those in the lower half plane are the corresponding complex conjugates. Hence, We need to compute their residues. Since the polynomial is of z2 order four, all roots are simple poles of f(z) := . The roots lie on the z4 + z2 + 1 unit circle as follows:

6 iR z2 z1

z¯2 z¯1 √ From this we see that z1 − z1 = i 3, z1 − z2 = 1 and z1 − z2 = 2z1, so 2 z1 Res[f, z1] = lim (z − z1)f(z) = z→z1 (z − z )(z − z )(z − z ) 1 1 1 2 1 2 √ z2 z 1 3 = 1√ = √1 = + . 2z1i 3 2i 3 4 12i √ Similarly, we have z2 − z1 = −1 z2 − z1 = 2z2, and z2 − z2 = i 3, so 2 z1 Res[f, z2] = lim (z − z2)f(z) = z→z2 (z − z )(z − z )(z − z ) 2 1 2 1 2 2 √ z2 z 1 3 = 2 √ = − √2 = − + . −2z2i 3 2i 3 4 12i Since f is even the integral is Z ∞ x2 1 Z ∞ x2 dx = dx x4 + x2 + 1 2 x4 + x2 + 1 0 −∞ √ 2πi π 3 = Res[f, z ] + Res[f, z ] = . 2 1 2 6 Extra questions for further practice 6. (a) Suppose that F (x, y) is a rational function of the real variables x and y not having a singularity on the unit circle in R2. Let 1 1 1 1 1 g(z) := F z + , z − . iz 2 z 2i z

Let zk = xk + iyk, k = 1, . . . , n be the isolated singularities of g enclosed by the unit circle γ(t) := eit, t ∈ [0, 2π]. Show that n Z 2π Z X F (cos t, sin t) dt = g(z) dz = 2πi Res[g, zk] 0 γ k=1 1 Solution: If we set z = eit = cos t + i sin t and = e−it = cos t − i sin t. Hence, z 1 1 1 1 cos t = z + and sin t = z − . (7) 2 z 2i z Setting z = γ(t) and so γ0(t) = ieit = iγ(t), by deﬁnition of line integrals, we ﬁnd that Z 2π Z 2π 1 1 1 1 1 F (cos t, sin t) dt = F γ(t) + , γ(t) − γ0(t) dt 0 0 iγ(t) 2 γ(t) 2i γ(t) Z 1 1 1 1 1 Z = F z + , z − dz = g(z) dz. γ iz 2 z 2i z γ The required formula now follows from the residue theorem.

7 Z 2π 1 (b) Use the method from part (a) to evaluate dt. 0 3 + sin t Solution: The integrand is of the form 1 F (cos t, sin t) = . 3 + sin t We substitute sin t using (7) to get 1 1 1 1 1 1 1 2 g(z) := F z + , z − = = . iz 2 z 2i z iz z2 − 1 z2 + 6iz − 1 3 + 2iz We now need to ﬁnd the isolated singularities of g(z) inside the unit circle. The zeros of z2 + 6iz − 1 = 0 are √ −6i ± −36 + 4 √ z = = (−3 ± 2 2)i 2 √ Both are simple poles of g, and only z1 := ( 2 − 3)i lies inside the unit circle γ. Hence, Z 2π 1 dt = 2 × 2πi Res[g, z1]. 0 3 + sin t

Now we compute Res[g, z1]:

1 1 Res[g, z1] = lim (z − z1)g(z) = lim √ √ = √ , z→z1 z→(−3i+2 2i) z − (−3 − 2 2)i 4 2i and hence by the formula from part (a) we get

Z 2π 1 π dt = 4πi Res[g, z1] = √ . 0 3 + sin t 2

7. Suppose that f : D → CN is analytic on the open set D ⊆ C. (a) Use the Cauchy integral formula to prove that

1 Z 2π f(z) = f(z + reit) dt 2π 0 for all z ∈ D and r > 0 such that B(z, r) ⊂ D. (This means that f(z) is the mean value of f on circles centred at z.) Solution: We use the path γ(t) := z + reit with t ∈ [0, 2π]. As D is open we can choose r > 0 such that B(z, r) ⊆ D.By the Cauchy integral formula 1 Z f(w) f(z) = dw 2πi γ w − z 1 Z 2π f(γ(t)) = γ0(t) dt 2πi 0 γ(t) − z Z 2π it 1 f(z + re ) it = it ire dt 2πi 0 z + re − z 1 Z 2π = f(z + reit) dt. 2π 0

8 (b) Use part (a) to show that z 7→ kf(z)k cannot have an isolated maximum in D. (This is a version of the maximum modulus theorem.)

Solution: Suppose that kfk attains an isolated maximum at z0 ∈ D. Then there exists r > 0 such that kf(w)k < kf(z0)k whenever 0 < |w − z0| ≤ r. Hence Z 2π Z 2π 1 1 it kf(z0)k = kf(z0)k dt > kf(z0 + re )k dt. 2π 0 2π 0 However, by (a) we have Z 2π 1 it kf(z0)k ≤ kf(z0 + re )k dt. 2π 0 Since this is a contradiction, we infer that kfk cannot have an isolated maximum in D.

8. Suppose that f : C → C is a bounded analytic function. We know that we can represent P∞ k f by the series k=0 akz for all z ∈ C with 1 Z f(w) ak = k+1 dw, 2πi γr w it where γr(t) = re , t ∈ [0, 2π] and r > 0. kfk (a) Show that |a | ≤ ∞ for all r > 0 and all k ∈ . k rk N Solution: From the formula for ak, if k ≥ 0, then 1 Z f(w) 1 Z 2π f(reit) |a | = dw = ireit dt k k+1 k+1 i(k+1)t 2πi γr w 2π 0 r e Z 2π 1 kfk∞ 2π kfk∞ kfk∞ ≤ k+1 r dt = k = k . 2π 0 r 2π r r as required. Alternatively we can estimate the integral by using the fact that a path integral can be estimated by the supremum of the integrand on the contour times the length of the path. The path γr is a circle of length 2πr. (b) Hence, show that f is constant. (This is called Liouville’s Theorem: every bounded analytic function on C is constant.) Solution: From the previous part kfk |a | ≤ ∞ k rk

for r > 0. Hence, if k ≥ 1, then by letting r → ∞ we conclude that |ak| = 0. Hence f(z) = a0 is constant.

9. Suppose that f is an analytic function on an open set D with isolated singularity at z0.

(a) Suppose that z0 is a pole of order m. Show that |f(z)| → ∞ as z → z0.

Solution: If z0 is a pole of order m, then g(z) f(z) = m (z − z0)

for some analytic function g on D∪{z0} with g(z0) 6= 0. Hence, limz→z0 |f(z)| = ∞.

9 (b) Suppose that |f(z)| → ∞ as z → z0. 1 (i) Show that has a removable singularity at z , and that there exists f(z) 0 m ≥ 1 and an analytic function h such that h(z0) 6= 0 and 1 = (z − z )mh(z). f(z) 0

Solution: Assume that limz→z0 |f(z)| = ∞. Then there exists δ > 0 1 such that |f(z)| > 1 whenever 0 < |z − z | < δ. Hence is well de- 0 f(z) ﬁned and analytic on B(z0, δ) \{z0}, and z0 is an isolated sinularity. Since

limz→z0 |f(z)| = ∞, we have that 1 lim = 0, (8) z→z0 f(z)

z0 is a removable singularity. Therefore, it has a power series expansion of the form ∞ 1 X = a (z − z )k f(z) k 0 k=0

valid for z ∈ B(z0, δ) \{z0}. Due to (8) there exists m ≥ 1 such that ak = 0 for k = 0, . . . , m − 1. We can therefore write

∞ ∞ 1 X X = a (z − z )k. = (z − z )m a (z − z )k−m = (z − z )h(z), f(z) k 0 0 k 0 0 k=m k=m ∞ X k−m where h(z) := ak(z − z0) . Note that h is analytic on B(z0, δ) and k=m h(z0) = am 6= 0.

(ii) Hence show that f has a pole of order m at z0. Solution: From the previous part 1 1 f(z) = m h(z) (z − z0)

for some analytic function on a ball B(z0, δ) with h(z0) 6= 0. Hence z0 is a pole of order m.

10. Consider the path γ consisting of intervals and semicircles as shown below:

iR γr

γδ

δ r R

it The path γ consists of the intervals [−r, −δ] and [δ, r] and the semi-circles γr(t) = re −it and γδ(t) = −δe , where t ∈ [0, π].

10 (a) Show that R 1 dz = iπ for all r > 0. γr z Solution: As Log z is a primitive of 1/z on the upper half plane we have Z 1 dz = Log(−r) − Log(r) = log r + iπ − log r − i0 = iπ γr z for all r > 0. Z eiz − 1 (b) Show that lim dz = 0. δ→0 γδ z eiz − 1 Solution: We have → i as z → 0, so there exists M > 1 such that z eiz − 1 < M whenever |z| ≤ 1. The length of the semi-circle γ is πδ, so z δ Z eiz − 1

dz ≤ Mπδ γδ z Z eiz − 1 whenever δ ≤ 1. Hence, lim dz = 0. δ→0 γδ z Z eiz *(c) Show that lim dz = 0. r→∞ γr z Solution: By deﬁnition of path integrals

Z eiz Z π eireit Z π dz = ireit dz ≤ |eir(cos t+i sin t)| dt it γr z 0 re 0 Z π Z π/2 = e−r sin t dt = 2 e−r sin t dt 0 0 for all r > 0. We note that ( 0 if t ∈ (0, π/2], lim e−r sin t = r→∞ 1 if t = 0.

As the limit function as r → ∞ is not continuous on [0, π] but e−r sin t is continuous for all r > 0, the limit cannot be uniform. We therefore have to be careful when we interchange the limit and the integral. For that purpose ﬁx a ∈ (0, π/2) and note that 0 ≤ sup e−r sin t ≤ e−r sin a → 0 x∈[a,π/2] as r → ∞. Hence e−r sin t → 0 uniformly with respect to t ∈ [a, π/2] as r → ∞. Hence, Z π/2 lim e−r sin t dt = 0 (9) r→∞ a for all a ∈ (0, π/2). Fix ε > 0 and choose a := ε/2. Then Z ε/2 Z ε/2 ε e−r sin t dt ≤ dt = . 0 0 2

By (9) there exists r0 > 0 such that Z π/2 Z ε/2 Z π/2 ε ε 0 ≤ e−r sin t dt ≤ e−r sin t dt + e−r sin t dt < + = ε 0 0 ε/2 2 2

11 for all r > r0. As the above argument is valid for every choice of ε > 0 this shows that the given integral converges to zero as r → ∞. Z eiz − 1 (d) The Cauchy integral theorem implies that dz = 0. Use this fact and the γ 2iz previous parts to show that Z ∞ sin t π dt = 0 t 2 eiz − 1 Solution: The function has a removable singularity at z = 0, hence 2iz extends to an analytic function on C. The Cauchy integral theorem implies that the integral over the given curve is zero. The given curve consist of two line segments [−r, −δ] and [δ, r] and two semi-circles By the Cauchy integral theorem Z eiz − 1 Z ez − 1 Z r eit − 1 Z −δ eit − 1 0 = dz + dz + dt + dt γδ 2iz γr 2iz δ 2it −r 2it Z eiz − 1 Z eiz − 1 Z r eit − e−it = dz + dz + dt γδ 2iz γr 2iz δ 2it Z eiz − 1 Z eiz − 1 Z r sin t = dz + dz + dt, γδ 2iz γr 2iz δ t where we made a substitution s = −t in the last integral on the left hand side. Hence the above identity reduces to Z r sin t Z eiz − 1 Z eiz − 1 dt = − dz − dz δ t γδ 2iz γr 2iz Z eiz − 1 1 Z eiz 1 Z 1 = − dz − dz + dz γδ 2iz 2i γr z 2i γr z whenever 0 < δ < r < ∞. The previous parts imply that Z ∞ sin t 1 Z eiz − 1 1 Z eiz 1 π dt = lim dz − lim dz + πi = . δ→0 r→∞ 0 t 2i γδ z 2i γr z 2i 2

Challenge questions (optional)

*11. Given an N × N matrix A ∈ CN×N let %(A) := {λ ∈ C: det(λI − A) 6= 0} be the set for which λI − A is invertible. In Tutorial 11, Question 8 it was shown that (λI − A)−1 is analytic on %(A). (a) Use the Neumann series expansion to derive the Laurent expansion of (λI − A)−1 about λ0 = 0 valid for |λ| large. Solution: Using the Neumann series expansion we get that ∞ 1 1 −1 1 X 1 (λI − A)−1 = I − A = Ak λ λ λ λk k=0 n −n 1/n n 1/n if limn→∞ kA λ k < 1, that is, if |λ| > limn→∞ kA k = r(A). Hence the required Laurent expansion about λ = 0 is ∞ ∞ 1 X 1 X (λI − A)−1 = Ak = Ak−1λ−k. λ λk k=0 k=1 It is valid if |λ| > r(A). The series diverges if |λ| < r(A) (by the root test or the Cauchy–Hadamard Theorem).

12 n 1/n (b) Let r(A) := limn→∞ kA k . We know from Tutorial 6 Question 7(c) that |λ| ≤ r(A) for all eigenvalues λ of A. Show that there exists an eigenvalue λ of A with r(A) = |λ|. Solution: Since the Laurent expansion converges and represents (λI − A)−1 if |λ| > r(A) it follows that λI −A is invertible if |λ| > r(A), so there is no eigenvalue of A with |λ| > r(A). We also know that the Laurent expansion converges on the largest annulus in the domain of the function. On the other hand, we have shown in (a) that the Laurent expansion converges if and only if |λ| > r(A), so there must be a point λ0 with |λ0| = r(A) such that λ0I − A is not invertible. But the only such points are the eigenvalues of A, proving the claim. iR

λ0

) A r(

All eigenvalues (the dots in the ﬁgure above) lie inside the circle with radius r(A). At least one is on the circle, and outside the circle (shaded) is where the Laurent series converges.