Mathematics 312 (Fall 2012) October 26, 2012 Prof. Michael Kozdron Lecture #22: The Cauchy Integral Formula
Recall that the Cauchy Integral Theorem, Basic Version states that if D is a domain and f(z)isanalyticinD with f (z)continuous,then
f(z)dz =0 C for any closed contour C lying entirely in D having the property that C is continuously deformable to a point. We also showed that if C is any closed contour oriented counterclockwise in C and a is inside C,then 1 dz =2πi. ( ) z a ∗ C − Our goal now is to derive the celebrated Cauchy Integral Formula which can be viewed as a generalization of ( ). ∗ Theorem 22.1 (Cauchy Integral Formula). Suppose that D is a domain and that f(z) is analytic in D with f (z) continuous. If C is a closed contour oriented counterclockwise lying entirely in D having the property that the region surrounded by C is a simply connected subdomain of D (i.e., if C is continuously deformable to a point) and a is inside C, then 1 f(z) f(a)= dz. 2πi z a C − Proof. Observe that we can write f(z) f(a) f(z) f(a) f(z) f(a) dz = dz + − dz =2πf(a)i + − dz z a z a z a z a C − C − C − Ca − where C = z a = r oriented counterclockwise since ( )implies a {| − | } ∗ f(a) 1 dz = f(a) dz =2πf(a)i z a z a C − C − and f(z) f(a) f(z) f(a) − dz = − dz z a z a C − Ca − since the integrand f(z) f(a) − z a − is analytic everywhere except at z = a and its derivative is continuous everywhere except at z = a so that integration over C can be continuously deformed to integration over Ca. However, if we write f(z) f(z) f(a) dz 2πf(a)i = − dz, z a − z a C − Ca − 22–1 and note that the left side of the previous expression does not depend on r,thenweconclude f(z) f(z) f(a) dz 2πf(a)i =lim − dz. z a − r 0 z a C − ↓ Ca − Hence, the proof will be complete if we can show that f(z) f(a) lim − dz =0. r 0 z a ↓ Ca − To this end, suppose that Mr =max f(z) f(a) ,zon Ca .Therefore,ifz is on Ca = z a = r ,then {| − | } {| − | } f(z) f(a) f(z) f(a) f(z) f(a) M − = | − | = | − | r z a z a r ≤ r − | − | so that f(z) f(a) f(z) f(a) M M M − dz − dz r dz = r 1dz = r (C ) z a ≤ z a ≤ r r r a Ca − Ca − Ca Ca M = r 2πr r · =2πMr
since the arclength of Ca is (Ca)=2πr. However, since f(z)isanalyticinD,weknowthat f(z) is necessarily continuous in D so that
lim f(z) f(a) =0 or,equivalently, limMr =0. z a r 0 → | − | ↓ Therefore, f(z) f(a) lim − dz lim(2πMr)=0 r 0 z a ≤ r 0 ↓ Ca ↓ − as required. Example 22.2. Compute 1 zez dz 2πi z i C − where C = z =2 is the circle of radius 2 centred at 0 oriented counterclockwise. {| | } z z z Solution. Observe that f(z)=ze is entire, f (z)=ze + e is continuous, and i is inside C.Therefore,bytheCauchyIntegralFormula, 1 zez 1 f(z) dz = dz = f(i)=iei. 2πi z i 2πi z i C − C − Example 22.3. Compute zez dz z + i C where C = z =2 is the circle of radius 2 centred at 0 oriented counterclockwise. {| | } 22–2 z z z Solution. Observe that f(z)=ze is entire, f (z)=ze + e is continuous, and i is inside C.Therefore,bytheCauchyIntegralFormula, −
z ze f(z) i i dz = dz =2πif( i)=2πi ie− =2πe− . z + i z + i − ·− C C Example 22.4. Compute zez dz z2 +1 C where C = z =2 is the circle of radius 2 centred at 0 oriented counterclockwise. {| | } Solution. Observe that partial fractions implies
1 1 1 i/2 i/2 = = = z2 +1 z2 i2 (z + i)(z i) z + i − z i − − − and so zez i zez i zez dz = dz dz. z2 +1 2 z + i − 2 z i C C C − z z z Let f(z)=ze . Note that f(z)isentireandf (z)=ze + e is continuous. Since both i and i are inside C,theCauchyIntegralFormulaimplies − z z ze i i ze i i dz =2πif( i)=2πi ie− =2πe− and dz =2πif(i)=2πi ie = 2πe z + i − ·− z i · − C C − so that
z i i ze i i i i i i e + e− dz = 2πe− 2πe = πie− + πie =2πi =2πi cos 1. z2 +1 2 · − 2 ·− 2 C
22–3 Mathematics 312 (Fall 2012) October 29, 2012 Prof. Michael Kozdron
Lecture #23: Consequences of the Cauchy Integral Formula
The main result that we will establish today is that an analytic function has derivatives of all orders. The key to establishing this is to first prove a slightly more general result.
Theorem 23.1. Let g be continuous on the contour C and for each z0 not on C, set g(ζ) G(z )= dζ. 0 ζ z C − 0 Then G is analytic at z0 with g(ζ) G(z )= dζ. ( ) 0 (ζ z )2 ∗ C − 0 Remark. Observe that in the statement of the theorem, we do not need to assume that g is analytic or that C is a closed contour.
Proof. Let z0 not on C be fixed. In order to prove the differentiability of G and the desired formula for G(z0), we must show that G(z +∆z) G(z ) g(ζ) lim 0 − 0 = dζ. ∆z 0 ∆z (ζ z )2 → C − 0 Observe that G(z +∆z) G(z ) 1 g(ζ) g(ζ) 0 − 0 = dζ ∆z ∆z ζ (z +∆z) − ζ z C − 0 − 0 1 1 1 = g(ζ) dζ ∆z ζ (z +∆z) − ζ z C − 0 − 0 g(ζ) = dζ (ζ z ∆z)(ζ z ) C − 0 − − 0 and so (with a bit of algebra) G(z +∆z) G(z ) g(ζ) g(ζ) g(ζ) 0 − 0 dζ = dζ dζ ∆z − (ζ z )2 (ζ z ∆z)(ζ z ) − (ζ z )2 C − 0 C − 0 − − 0 C − 0 g(ζ) =∆z dζ ( ) (ζ z ∆z)(ζ z )2 † C − 0 − − 0 The next step is to show that g(ζ) dζ (ζ z ∆z)(ζ z )2 C 0 0 − − − is bounded.
23–1 To this end, let M =max g(ζ) : ζ C be the maximum value of g(ζ) on C,andletd = {| | ∈ } | | min dist(z0,w):w C be the minimal distance from z0 to C. Note that ζ z0 d>0 for all{ ζ on C. Without∈ } loss of generality, assume that ∆z so that ( ) holds. Note that we have proved G(z )isdifferentiableatz C for z fixed. ∗ 0 0 ∈ 0 Since z0 was arbitrary, we conclude that G(z0) is differentiable at any z0 C implying that G is analytic at z C as required. ∈ 0 ∈ It is important to note that exactly the same method of proof yields the following result. Corollary 23.2. Let g be continuous on the contour C and for each z0 not on C, set g(ζ) H(z )= dζ 0 (ζ z )n C − 0 where n is a positive integer. Then H is analytic at z0 with g(ζ) H(z )=n dζ. ( ) 0 (ζ z )n+1 ∗∗ C − 0 Now we make a very important observation that follows immediately from Theorem 23.1 and Corollary 23.2. Theorem 23.3. If f(z) is analytic in a domain D, then all of its derivatives f (z), f (z), f (z), . . . exist and are themselves analytic. Remark. This theorem is remarkable because it is unique to complex analysis. The analogue for real-valued functions is not true. For example, f(x)=9x5/3 for x R is differentiable 2/3 ∈ 1/3 for all x, but its derivative f (x)=15x is not differentiable at x =0(i.e.,f (x)=10x− does not exist when x =0). 23–2 Moreover, if the function in the statement of Theorem 23.1 happens to be analytic and C happens to be a closed contour oriented counterclockwise, then we arrive at the follow- ing important theorem which might be called the General Version of the Cauchy Integral Formula. Theorem 23.4 (Cauchy Integral Formula, General Version). Suppose that f(z) is analytic inside and on a simply closed contour C oriented counterclockwise. If z is any point inside C, then n! f(ζ) f (n)(z)= dζ, 2πi (ζ z)n+1 C − n =1, 2, 3,.... For the purposes of computations, it is usually more convenient to write the General Version of the Cauchy Integral Formula as follows. Corollary 23.5. Suppose that f(z) is analytic inside and on a simply closed contour C oriented counterclockwise. If a is any point inside C, then f(z) 2πif (m 1)(a) dz = − . (z a)m (m 1)! C − − Example 23.6. Compute e5z dz (z i)3 C − where C = z =2 is the circle of radius 2 centred at 0 oriented counterclockwise. {| | } Solution. Let f(z)=e5z so that f(z)isentire,andleta = i which is inside C.Therefore, e5z 2πif (i) dz = =25πie5i (z i)3 2! C − 5z since f (z)=25e . Applications to Harmonic Functions Suppose that f(z)=u(z)+iv(z)=u(x, y)+iv(x, y)isanalyticinadomainD.From Theorem 23.3 we know that all of the derivatives of f are also analytic in D.Inparticular, this implies that all the partials of u and v of all orders are continuous. This means that we can replace Example 13.9 and Proposition 16.2 with the following. Theorem 23.7. If f(z)=u(z)+iv(z)=u(x, y)+iv(x, y) is analytic in a domain D, then u = u(x, y) is harmonic in D so that uxx(x0,y0)+uyy(x0,y0)=0 for all (x ,y ) D, and v = v(x, y) is harmonic in D so that 0 0 ∈ vxx(x0,y0)+vyy(x0,y0)=0 for all (x ,y ) D. 0 0 ∈ 23–3 Proof. Since f(z)=u(z)+iv(z)isanalyticatz D,weknow 0 ∈ f (z )=u (z )+iv (z )=v (z ) iu (z ). 0 x 0 x 0 y 0 − y 0 By Theorem 23.3 we know that all of the derivatives of f are also analytic in D and so the Cauchy-Riemann equations for f imply f (z )=u (z )+iv (z )=v (z ) iu (z ) 0 xx 0 xx 0 yx 0 − yx 0 and f (z )=v (z ) iu (z )= u (z ) iy (z ). 0 xy 0 − xy 0 − yy 0 − yy 0 In particular, we obtain uxy = uyx, vxy = vyx,and uxx(z0)+uyy(z0)=0 and vxx(z0)+vyy(z0)=0 so that u and v are harmonic at z D as required. 0 ∈ Suppose that C = C = z = R is the circle of radius R>0centredat0oriented R {| | } counterclockwise. We know that if f(z)isanalyticinsideCR,then 1 f(ζ) f(z)= dζ 2πi ζ z CR − for z is harmonic inside CR. Thus, we want to determine an expression for 1 f(ζ) Re(f(z)) = Re dζ . 2πi ζ z CR − The trick to doing so is to consider the function f(ζ)¯z g(ζ)= R2 ζz¯ − which is an analytic function of ζ inside and on CR. (Note that the denominator, as a function of ζ, is never 0. Why?) Hence, by the Cauchy Integral Theorem, 1 f(ζ)¯z g(ζ)dζ =0 or,equivalently, dζ =0. 2πi R2 ζz¯ CR CR − Therefore, if we add this 0 to f(z)weobtain, 1 f(ζ) 1 f(ζ)¯z 1 f(ζ) f(ζ)¯z f(z)= dζ + dζ = + dζ 2πi ζ z 2πi R2 ζz¯ 2πi ζ z R2 ζz¯ CR − CR − CR − − 1 1 z¯ = + f(ζ)dζ 2πi ζ z R2 ζz¯ CR − − 1 R2 z 2 = −| | f(ζ)dζ. 2πi (ζ z)(R2 ζz¯) CR − − 23–4 If we now parametrize C by ζ = Reit,0 t 2π,thenweobtain R ≤ ≤ 1 2π R2 z 2 f(z)= −| | f(Reit) iReit dt 2πi (Reit z)(R2 Reitz¯) · 0 − − R2 z 2 2π f(Reit) = −| | dt 2π (Reit z)(Re it z¯) 0 − − − R2 z 2 2π f(Reit) = −| | dt. 2π Reit z 2 0 | − | If we now write the analytic function f(z)asf(z)=u(z)+iv(z), and then write z = reiθ as the polar form of z,weobtain f(Reit) u(Reit)+iv(Reit) u(Reit) Re =Re = . Reit z 2 Reit reiθ 2 R2 + r2 2rR cos(t θ) | − | | − | − − Thus, we conclude R2 r2 2π u(Reit) u(z)=u(reiθ)= − dt 2π R2 + r2 2rR cos(t θ) 0 − − is harmonic for z = reiθ with z = r 1 2π u(0) = u(eit)dt 2π 0 which is sometimes called the circumferential mean value theorem.Thisresultalsohasa probabilistic interpretation. The value of the harmonic function at the centre of the unit disk is the uniform average of the values of that function around the unit circle. 23–5 Mathematics 312 (Fall 2012) October 31, 2012 Prof. Michael Kozdron Lecture #24: Applications to Harmonic Functions We know from Theorem 23.7 that if a function f(z)isanalyticinadomainD,thenitsreal part is harmonic in D.Wewillnowproveapartialconverse,namelythatifafunctionu is harmonic in a simply connected domain D,thenthereisananalyticfunctioninD whose real part is u. Theorem 24.1. Suppose that D is a simply connected domain. If u = u(z)=u(x, y) is harmonic in D, then there is a function f(z) which is analytic in D with Re(f(z)) = u(z). Proof. Suppose that u is harmonic in D so that, by assumption, the second partials uxx, uyy, and uxy = uyx are continuous in D,and uxx(z0)+uyy(z0)=0 for every z D.Supposethatwenowset 0 ∈ g(z)=u (z) iu (z). x − y Observe that g satisfies the Cauchy-Riemann equations in D;thatis, ∂ ∂ ∂ ∂ Re(g(z)) = u (z)=u (z)= u (z)= ( u (z)) = Im(g(z)) ∂x ∂x x xx − yy ∂y − y ∂y using the fact that uxx + uyy =0and ∂ ∂ ∂ ∂ Re(g(z)) = u (z)=u (z)=u (z)= (u (z)) = Im(g(z)) ∂y ∂y x xy yx ∂x y −∂x using the fact that uxy = uyx.Sincethepartialsofu are continuous, we conclude from Theorem 13.8 that g(z) is analytic in D.Therefore,weconcludethereexistsananalytic function G(z)suchthatG(z)=g(z). If we write G(z)=ϕ(z)+iψ(z), then G(z)satisfies the Cauchy-Riemann equations (since G(z)isanalytic)sothatϕx(z)=ψy(z)andϕy(z)= ψ (z). Moreover, − x G(z)=ϕ (z)+iψ (z)=ϕ (z) iϕ (z) x x x − y so that G(z)=g(z)implies ϕ (z) iϕ (z)=u (z) iu (z). x − y x − y That is, ux(z)=ϕx(z)anduy(z)=ϕy(z). This means that u ϕ = c for some real constant c. Hence, the required analytic function f(z)is − f(z)=G(z)+c and the proof is complete. 24–1 We can now prove a very interesting property of harmonic functions known as the Maximum Principle. Suppose that u = u(z)=u(x, y)isharmonicinasimplyconnecteddomainD. Let f(z)=u(z)+iv(z)beananalyticfunctioninD whose existence is guaranteed by Theorem 24.1. If we now consider the function ef(z),thenweobservethat ef(z) = eu(z)+iv(z) = eu(z) eiv(z) = eu(z) | | | | | || | since eiv(z) =1andeu(z) > 0. The fact that the exponential is a monotonically increasing function| of| a real variable implies that the maximum points of u(z)mustcoincidewiththe maximum points of the modulus of the analytic function ef(z). Theorem 24.2 (Maximum Principle for Harmonic Functions). If u = u(x, y) is harmonic in a simply connected domain D and u(z) achieves its maximum value at some point z0 in D, then u(z) is constant in D. To state the theorem in slightly different language, a harmonic function u(z)cannotachieve its maximum at an interior point z D unless u(z)isconstant. 0 ∈ Of course, the minimum points of u(z)arejustthemaximumpointsof u(z). This means that we also have a minimum principle for harmonic functions. − Theorem 24.3 (Minimum Principle for Harmonic Functions). If u = u(x, y) is harmonic in a simply connected domain D and u(z) achieves its minimum value at some point z0 in D, then u(z) is constant in D. Combining these results, we arrive at the following theorem. Theorem 24.4. A function u(x, y) that is harmonic in a bounded simply connected domain and continuous up to and including the boundary attains its maximum and minimum values on the boundary. 24–2 Supplement: The Dirichlet Problem Let D be a domain and suppose that g(z)forz C = ∂D is a given continuous function. The Dirichlet Problem for D is to find a function∈ u(z)=u(x, y)thatiscontinuouson D = D C,harmonicinD,andsatisfiesu(z)=g(z)forz C. ∪ ∈ In the case when D is the simply connected domain D = z 2 2 2π it R r g(Re ) iθ − 2 2 dt, for z = re with z = r lim u(z0)=g(z), z0 z, z0 < z =R → | | | | or, equivalently, that R2 r2 2π g(Reit) lim − dt = g(Reiθ). r R 2π R2 + r2 2rR cos(t θ) ↑ 0 − − Without loss of generality, assume that R =1andθ =0sothatwemustshow 1 r2 2π g(eit) lim − dt = g(1). r 1 2π 1+r2 2r cos t ↑ 0 − 24–3 Unfortunately, a completely rigorous proof of this fact is beyond the scope of Math 312. We can, however, give the correct intuition for why it is true. Observe that 1 1 r2 lim − 2π r 1 1+r2 2r cos t ↑ − depends on the value of t [0, 2π]. There are two possibilities: (i) if t =0ort =2π,then cos t =1sothat ∈ 1 1 r2 lim − = , 2π r 1 1+r2 2r ∞ ↑ − and (ii) if 0 ∞ δ0(t)h(t)dt = h(0). −∞ Thus, assuming that we can interchange limits and integrals, we arrive at 1 r2 2π g(eit) 2π 1 1 r2 lim − dt =lim − g(eit)dt r 1 2π 1+r2 2r cos t r 1 2π 1+r2 2r cos t ↑ 0 − ↑ 0 − 2π 1 1 r2 = lim − g(eit) dt r 1 2π 1+r2 2r cos t 0 ↑ − 2π 1 1 r2 = g(eit) lim − dt r 1 2 0 ↑ 2π 1+r 2r cos t 2π − it = g(e )δ0(t)dt 0 = g(ei0)=g(1). as required. 24–4 Finally, we should note that the integrating kernel 1 R2 r2 − 2π R2 + r2 2rR cos(t θ) − − has a special name. It is the Poisson kernel;thatis,ifz = reiθ D and w = Reit C = ∂D, then ∈ ∈ 1 w 2 z 2 1 R2 r2 P (w; z)= | | −| | or, equivalently, P (R, t; r, θ)= − . 2π w z 2 2π R2 + r2 2rR cos(t θ) | − | − − Thus, the solution to the Dirichlet problem for z 2π u(reiθ)= P (R, t; r, θ)g(Reit)dt. 0 This representation of u(z)isoftencalledPoisson’s integral formula. Remark. Note that we have introduced the Poisson kernel from a purely analytic point-of- view. However, as a result of Exercise 24.5 we can view the Poisson kernel as a probability density function on the boundary of the disk of radius R.Thatis,P (R, t; r, θ) 0and satisfies ≥ 2π P (R, t; r, θ)dt =1. 0 Moreover, it turns out that Brownian motion and the Poisson kernel are intimately con- nected. The density of the first exit from the disk of radius R by two dimensional Brownian motion started at the interior point z = reiθ is exactly P (R, t; r, θ). Although the Poisson kernel has been studied for over 150 years, and the relationship between Brownian motion and the Poisson kernel has been understood for over 60 years, the Poisson kernel is still vital for modern mathematics. In fact, the Poisson kernel plays a significant role in the Fields Medal winning work of Wendelin Werner (2006) and Stas Smirnov (2010) on the Schramm-Loewner evolution. 24–5 Mathematics312(Fall2012) November2,2012 Prof. Michael Kozdron Lecture #25: Taylor Series Our primary goal for today is to prove that if f(z) is an analytic function in a domain D,thenf(z) can be expanded in a Taylor series about any point a D.Moreover,the Taylor series for f(z) converges uniformly to f(z)foranyz in a closed disk∈ centred at a and contained entirely in D. Theorem 25.1. Suppose that f(z) is analytic in the disk z a (n) n (j) f (a) 2 f (a) n f (a) j T (z; f,a)=f(a)+f (a)(z a)+ (z a) + + (z a) = (z a) , n − 2! − ··· n! − j! − j=0 converges to f(z) for all z in this disk. Furthermore, the convergence is uniform in any closed subdisk z a R Proof. It is sufficient to prove uniform convergence in every subdisk z a R 25–1 and 1 1 z a z a n ζ a z a n+1 = 1+ − + + − + − − . ( ) ζ z ζ a ζ a ··· ζ a ζ z ζ a ∗∗ − − − − − − Substituting ( )into( )weconclude ∗∗ ∗ 1 f(ζ) z a z a n ζ a z a n+1 f(z)= 1+ − + + − + − − dζ 2πi ζ a ζ a ··· ζ a ζ z ζ a C − − − − − 1 f(ζ) (z a) f(ζ) (z a)n f(ζ) = dζ + − dζ + + − dζ 2πi ζ a 2πi (ζ a)2 ··· 2πi (ζ a)n+1 C − C − C − (z a)n+1 f(ζ) + − dζ. 2πi (ζ z)(ζ a)n+1 C − − However, from the Cauchy Integral Formula, we know 1 f(ζ) 1 f(ζ) 1 f(ζ) f (a) dζ = f(a), dζ = f (a), dζ = , 2πi ζ a 2πi (ζ a)2 2πi (ζ a)3 2! C − C − C − and in general 1 f(ζ) f (j)(a) dζ = 2πi (ζ a)j+1 j! C − so that (n) n+1 f (a) n (z a) f(ζ) f(z)=f(a)+f (a)(z a)+ + (z a) + − dζ − ··· n! − 2πi (ζ z)(ζ a)n+1 C − − = Tn(z; f,a)+Rn(z; f,a). Thus, we see that in order to show that T (z; f,a)convergestof(z)uniformlyfor z a R, n | − |≤ it suffices to show that R (z; f,a)convergesto0uniformlyfor z a R. Suppose, n | − |≤ therefore, that z a R and ζ a = R where R =(R + R)/2asbefore.Bythe triangle inequality,| − |≤ | − | R + R R R ζ z R R = R = − , | − |≥ − 2 − 2 and so (z a)n+1 f(ζ) 1 f(ζ)(z a)n+1 R (z; f,a) = − dζ − dζ | n | 2πi (ζ z)(ζ a)n+1 ≤ 2π (ζ z)(ζ a)n+1 C C −n+1 − − − 1 f(ζ) (R) | | dζ ≤ 2π (R )n+1(R R )/2 C − 1 R n+1 1 max f(ζ) (C) ≤ π ζ C | | R R R ∈ − where (C)=2πR is the arclength of C.Thatis,aftersomesimplification,weobtain n 2R 2R Rn(z; f,a) max f(ζ) . | |≤ R + R R R ζ C | | − ∈ 25–2 Notice that the right side of the previous inequality is independent of z.Since2R z2 z3 ∞ zj ez =1+z + + + = 2! 3! ··· j! j=0 for every z C. ∈ Example 25.3. Find the Taylor series for both f1(z)=sinz and f2(z)=cosz about a =0, and then show that the Taylor series for eiz equals the sum of the Taylor series for cos z and i sin z. Solution. Observe that f1 (z)=cosz = f2(z)andf2 (z)= sin z = f1(z). Since cos 0 = 1 and sin 0 = 0, we obtain − − z3 z5 z7 ∞ z2j+1 sin z = z + + = ( 1)j − 3! 5! − 7! ··· − (2j +1)! j=0 and z2 z4 z6 ∞ z2j cos z =1 + + = ( 1)j − 2! 4! − 6! ··· − (2j)! j=0 for every z C.Observethat ∈ (iz)2 (iz)3 (iz)4 (iz)5 (iz)6 eiz =1+(iz)+ + + + + + 2! 3! 4! 5! 6! ··· (iz)2 (iz)4 (iz)6 (iz)3 (iz)5 (iz)7 = 1+ + + + + iz + + + + 2! 4! 6! ··· 3! 5! 7! ··· z2 z4 z6 i2z3 i4z5 i6z7 = 1 + + + i z + + + + − 2! 4! − 6! ··· 3! 5! 7! ··· z2 z4 z6 z3 z5 z7 = 1 + + + i z + + − 2! 4! − 6! ··· − 3! 5! − 7! ··· =cosz + i sin z as expected. It is worth noting that term-by-term manipulations of the sum of Taylor series are justified by Theorem 25.1 since the Taylor series involved converge uniformly in closed disks about the point a =0. Remark. Sometimes the phrase Maclaurin series is used in place of Taylor series when a =0. Theorem 25.4. If f(z) is analytic at z0, then the Taylor series for f (z) at z0 can be obtained by termwise differentiation of the Taylor series for f(z) about z0 and converges in the same disk as the Taylor series for f(z). 25–3 Proof. Since f(z)isanalyticatz0,theTaylorseriesforf(z)aboutz0 is given by ∞ f (j)(z ) f(z)= 0 (z z )j. j! − 0 j=0 By termwise differentiation, we obtain ∞ (j) ∞ (j) f (z0) j 1 f (z0) j 1 f (z)= j (z z ) − = (z z ) − . ( ) j! − 0 (j 1)! − 0 ∗ j=0 j=1 − Suppose now that g(z)=f (z). By Theorem 23.3, we know that g(z)isanalyticatz0 so that its Taylor series is ∞ g(j)(z ) g(z)= 0 (z z )j. j! − 0 j=0 (j) (j+1) However, g (z0)=f (z0)sothat ∞ (j) ∞ (j+1) g (z0) j f (z0) j f (z)=g(z)= (z z ) = (z z ) . ( ) j! − 0 j! − 0 ∗∗ j=0 j=0 But by a change of index, it is clear that ( )and( )areequalasrequired. ∗ ∗∗ 25–4 Mathematics312(Fall2012) November5,2012 Prof. Michael Kozdron Lecture #26: Taylor Series and Isolated Singularities Recall from last class that if f(z)isanalyticata,then ∞ f j(a) f(z)= (z a)j j! − j=0 for all z in some neighbourhood of a.Thisneighbourhood,denotedby z a Example 26.1. We know from Lecture #5 that 1 =1+z + z2 + z3 + 1 z ··· − 1 for z < 1. Since (1 z)− is analytic for z < 1, we conclude this must be its Taylor series expansion| | about a =0.Moreover,since− | |z2 < 1whenever z < 1, we find |− | | | 1 =1+( z2)+( z2)2 +( z2)3 + =1 z2 + z4 z6 + z8 z10 + 1+z2 − − − ··· − − − ··· for z < 1. Observe now that | | d 1 arctan z = . dz 1+z2 Consequently we can use Theorem 25.4, which tells us that the derivative of an analytic function f(z)canbeobtainedbytermwisedifferentiationoftheTaylorseriesoff(z), to conclude that the Taylor series expansion about 0 for arctan z must be z3 z5 z7 arctan z = z + + − 3 5 − 7 ··· for z < 1since | | d d z3 z5 z7 1 arctan z = z + + =1 z2 + z4 z6 + z8 z10 + = . dz dz − 3 5 − 7 ··· − − − ··· 1+z2 We can now recover the famous Leibniz formula for π from 1682; that is, since arctan(1) = π/4, we find π 1 1 1 =arctan(1)=1 + + , 4 − 3 5 − 7 ··· or equivalently, ∞ ( 1)j π =4 − . 2j +1 j=0 26–1 Observe that in the last example we obtained the Taylor series for 1 1+z2 by formally plugging z2 into the Taylor series for − 1 . 1 z − Suppose that we try do the same thing with a point at which the Taylor series is not analytic. For instance, we know that z2 z3 ez =1+z + + + 2! 3! ··· for all z C. However, is it true that ∈ 1 1 1 e1/z =1+ + + + , z 2!z2! 3!z3 ··· at least for z =0?OurgoalistonowdevelopthetheoryofLaurentserieswhichwillprovide us with the means to understand the series expansion for e1/z given above. Consider again the function e1/z.Observethatthepointz =0isspecialbecauseitisthe only point at which e1/z fails to be analytic. We call such a point an isolated singularity. Definition. Apointz0 is called an isolated singular point of f(z)iff(z)isnotanalyticat z0 but is analytic at all points in some small neighbourhood of z0. Example 26.2. If f(z)=e1/z,thenz =0isanisolatedsingularpointoff(z). Example 26.3. If 1 f(z)= , z then z0 =0isanisolatedsingularpointoff(z). Example 26.4. If 1 f(z)=cscz = , sin z then f(z)hasisolatedsingularpointsatz = nπ for n Z. ∈ Example 26.5. If 1 f(z)= , sin(1/z) then f(z)hasisolatedsingularpointsatthosepointsforwhichsin(1/z)=0,namely 1 1 = nπ or equivalently z = z nπ for n = 1, 2, 3,.... Note that z =0isnotanisolatedsingularpointsincethereare points of± the± form± 1/(nπ) arbitrarily close to 0 for n sufficiently large. In other words, 0 is a cluster point or an accumulation point of the sequence of isolated singular points 1/(nπ), n =1, 1, 2, 2, 3, 3,... − − − 26–2 The basic idea is as follows. For a function f(z)analyticfor z z This expansion is the so-called Laurent series expansion of f(z)aboutz0 =0.Thenext several lectures will be devoted to the development of the theory of Laurent series. Here is one important use of the series expansion just determined. Example 26.7. Suppose that C = z =1/2 oriented counterclockwise. Compute {| | } 1+2z dz. z2 + z3 C Solution. Assuming that 1+2z 1 1 = + 1+z z2 + , z2 + z3 z2 z − − ··· we obtain 1+2z 1 1 dz = + 1+z z2 + dz z2 + z3 z2 z − − ··· C C 1 1 = dz + dz 1dz + z dz z2 dz + z2 z − − ··· C C C C C =0+2πi +0+0+0+ ··· =2πi. 26–3 Mathematics312(Fall2012) November16,2012 Prof. Michael Kozdron Lecture #27: Laurent Series Recall from Lecture #26 that we considered the function 1+2z f(z)= z2 + z3 and we formally manipulated f(z)toobtaintheinfiniteexpansion 1 1 f(z)= + 1+z z2 + . z2 z − − ··· Observe that f(z)isanalyticintheannulus0< z < 1. Does | | 1+2z 1 1 = + 1+z z2 + z2 + z3 z2 z − − ··· for all 0 < z < 1? The answer turns out to be yes. Thus, our goal for today is to prove that if a function| |f(z) is analytic in an annulus, then it has an infinite series expansion which converges for all z in the annulus. This expansion is known as the Laurent series for f(z). Theorem 27.1. Suppose that f(z) is analytic in the annulus r< z z0 ∞ ∞ j j f(z)= aj(z z0) + a j(z z0)− ( ) − − − ∗ j=0 j=1 where 1 f(ζ) a = dζ, j =0, 1, 2,..., j 2πi (ζ z )j+1 ± ± C − 0 and C is any closed contour oriented counterclockwise that surrounds z0 and lies entirely in the annulus. The proof is very similar to the proof of Theorem 25.1 for the Taylor series representation of an analytic function in a disk z z ∞ 1 j ak k j f(z)(z z )− = (z z ) − 2πi − 0 2πi − 0 k= −∞ 27–1 and so 1 f(z) ∞ ak 1 j dz = j k dz 2πi (z z0) 2πi (z z0) C k= C − − −∞ − where C is any closed contour oriented counterclockwise that surrounds z0 and lies entirely in the annulus. We now observe from Theorem 21.1 that 1 dz =2πi if k = j 1 (z z )j k − C − 0 − and 1 dz =0 if k = j 1 (z z )j k − C − 0 − so that 1 f(z) ∞ ak 1 j dz = j k dz = aj 1. 2πi (z z0) 2πi (z z0) − C k= C − − −∞ − In other words, 1 f(z) 1 f(z) aj 1 = dz or, equivalently, aj = dz. − 2πi (z z )j 2πi (z z )j+1 C − 0 C − 0 Remark. Observe that 1 f(ζ) a = dζ j 2πi (ζ z )j+1 C − 0 so, at least for j =0, 1, 2,...,itmightbetemptingtousetheCauchyIntegralFormula (Theorem 23.4) to try and conclude that aj is equal to (j) f (z0) j! as was the case in the Taylor series derivation. This is not true, however, since the assumption on f(z)isthatitisanalyticintheannulusr< z z Example 27.2. Suppose that 1+2z f(z)= z2 + z3 which is analytic for 0 < z < 1. Show that the Laurent series expansion of f(z)for 0 < z < 1is | | | | 1 1 + 1+z z2 + . z2 z − − ··· 27–2 Solution. Suppose that C is any closed contour oriented counterclockwise lying entirely in 0 < z < 1 and surrounding z =0.Consider { | | } 0 1 f(ζ) 1 1+2ζ 1 1 (1 + 2ζ)/(1 + ζ) a = dζ = dζ = dζ. j 2πi (ζ z )j+1 2πi ζ2 + ζ3 · ζj+1 2πi ζj+3 C − 0 C C The reason for writing it in this form is that now we can apply the Cauchy Integral Formula to compute 1 (1 + 2ζ)/(1 + ζ) dζ. 2πi ζj+3 C Observe that the function 1+2z g(z)= 1+z is analytic inside and on C.Thus,theCauchyIntegralTheoremimpliesthatifj 3, then ≤− 1 (1 + 2ζ)/(1 + ζ) dζ =0 2πi ζj+3 C so that a 3 = a 4 = = 0. The Cauchy Integral Formula implies that if j 2, then − − ··· ≥− 1 (1 + 2ζ)/(1 + ζ) 1 g(ζ) g(j+2)(0) dζ = dζ = . 2πi ζj+3 2πi ζj+3 (j +2)! C C Note that if j = 2, then a 2 = g(0) = 1. In order to compute successive derivatives of − g(z), observe that− 1+2z 1 2z g(z)= = + 1+z 1+z 1+z Now, if k =1, 2, 3,...,then dk 1 k! =( 1)k dzk 1+z − (1 + z)k+1 and dk z k! k!z =( 1)k+1 +( 1)k dzk 1+z − (1 + z)k − (1 + z)k+1 so that g(k)(0) = ( 1)kk!+2( 1)k+1k!=( 1)k+1k!fork =1, 2, 3,.... − − − This implies g(j+2)(0) ( 1)j+3(j +2)! a = = − =( 1)j+3 =( 1)j+1 for j = 1, 0, 1, 2 ... j (j +2)! (j +2)! − − − so that the Laurent series expansion of f(z)for0< z < 1is | | 1 1 1 ∞ + 1+z z2 + = + ( 1)j+1zj. z2 z − − ··· z2 − j= 1 − Remark. We will soon learn other methods for determining Laurent series expansions that do not require the computation of contour integrals. 27–3 A first look at residue theory as an application of Laurent series One important application of the theory of Laurent series is in the computation of contour integrals. Suppose that f(z)isanalyticintheannulus0< z z0 ∞ ∞ j j f(z)= aj(z z0) + a j(z z0)− . − − − j=0 j=1 Let C be any closed contour oriented counterclockwise lying entirely in the annulus and surrounding z0 so that ∞ ∞ j j f(z)dz = aj (z z0) dz + a j (z z0)− dz. − − − C j=0 C j=1 C We know from Theorem 21.1 that j 2πi, if j = 1, (z z0) dz = − − 0, if j = 1, C − so that f(z)dz =2πia 1. − C Thus, we see that the coefficient a 1 in the Laurent series expansion of f(z)inanannulus − of the form 0 < z z 27–4 Mathematics312(Fall2012) November19,2012 Prof. Michael Kozdron Lecture #28: Calculating Laurent Series Example 28.1. Determine the Laurent series for 1 f(z)= (z 1)(2 z) − − for (i) 1 < z < 2, and (ii) z > 2. | | | | Solution. Note that the only singular points of f(z)occurat1and2.Thismeansthat(i) f(z) is analytic in the annulus 1 < z < 2, and (ii) f(z)isanalyticfor z > 2. Hence, in either case, the Laurent series for f(|z)willnecessarilybeoftheform| | | ∞ ∞ j j f(z)= ajz + a jz− , − j=1 j=1 and so the idea is that we will find the coefficients aj directly rather than by contour inte- gration. (It is worth stressing that the coefficients in the Laurent series expansions for (i) and (ii) will not necessarily be the same.) (i) Using partial fractions, we find 1 1 1 f(z)= = . (z 1)(2 z) z 1 − z 2 − − − − Now consider 1 1 ∞ = = zj. z 1 −1 z − j=0 − − We know this series converges for z < 1. However, we are interested in determining a series which converges for z > 1. Thus,| we| write | | 1 1/z = z 1 1 1/z − − and observe that the series 1 ∞ = (1/z)j 1 1/z j=0 − converges for 1/z < 1, or equivalently, z > 1. This implies | | | | ∞ ∞ ∞ 1 1/z 1 j j 1 j = = z− = z− − = z− for z > 1. z 1 1 1/z z | | j=0 j=0 j=1 − − Now consider 1 1/2 1 ∞ ∞ zj = = (z/2)j = −z 2 1 z/2 2 2j+1 j=0 j=0 − − 28–1 which converges for z/2 < 1, or equivalently, z < 2. Thus, when we add these two series, we obtain | | | | ∞ ∞ j 1 1 j z = z− + . z 1 − z 2 2j+1 j=1 j=0 − − Note that the first series converges for z > 1whilethesecondseriesconvergesfor z < 2. This means that they BOTH converge when| | 1 < z < 2, and so we have found the Laurent| | series for f(z)for1< z < 2, namely | | | | ∞ j ∞ 1 z j f(z)= = + z− . (z 1)(2 z) 2j+1 j=0 j=1 − − (ii) We now want the Laurent series for f(z)toconvergefor z > 2. We already know that | | ∞ 1 j = z− z 1 j=1 − converges for z > 1. However, the series that we found for | | 1 z 2 − converges for z < 2. This means that we need to manipulate it differently, say | | ∞ ∞ ∞ 1 1/z 1 j j j 1 j 1 j = = (2/z) = 2 z− − = 2 − z− z 2 1 2/z z j=0 j=0 j=1 − − which converges for 2/z < 1, or equivalently, z > 2. Thus, | | | | ∞ ∞ ∞ 1 1 1 j j 1 j j 1 j f(z)= = = z− 2 − z− = (1 2 − )z− (z 1)(2 z) z 1 − z 2 − − j=1 j=1 j=1 − − − − for z > 2. | | Example 28.2. Determine the Laurent series for e2z f(z)= (z 1)3 − for all z 1 > 0. | − | Solution. Observe that f(z)isanalyticfor z 1 > 0. Now | − | e2z e2(z 1+1) e2(z 1) f(z)= = − = e2 − . (z 1)3 (z 1)3 (z 1)3 − − − Recall that the Taylor series for ew is w2 w3 ∞ wj ew =1+w + + + = 2! 3! ··· j! j=0 28–2 which converges for all w C. This implies ∈ 2 2 3 3 ∞ j j 2(z 1) 2 (z 1) 2 (z 1) 2 (z 1) e − =1+2(z 1) + − + − + = − . − 2! 3! ··· j! j=0 Hence, 2(z 1) ∞ j j ∞ j j 3 e − 3 2 (z 1) 2 (z 1) − =(z 1)− − = − (z 1)3 − j! j! j=0 j=0 − so that the Laurent series of f(z)for z 1 > 0is | − | 2z j j 3 e ∞ 2 (z 1) − 1 2 4 8 f(z)= = e2 − = e2 + + + + . (z 1)3 j! (z 1)3 (z 1)2 2(z 1) 6 ··· j=0 − − − − Example 28.3. Let z z e e− sinh z = − = i sin(iz) 2 − and let 1 f(z)= . z2 sinh z Determine the first few terms of the Laurent series for f(z)in0< z <π, and then calculate | | 1 dz z2 sinh z C where C = z =1 is the unit circle centred at 0 oriented counterclockwise. {| | } Solution. On Assignment #9 you will show that the Taylor series for the entire function sinh z is z3 z5 ∞ z2j+1 sinh z = z + + + = 3! 5! ··· (2j +1)! j=0 which converges for all z C.Moreover,itisnottoodifficulttoshowthatsinhz =0ifand only if z 0, iπ, 2iπ,...∈ . This implies that f(z)isanalyticfor0< z <π. Now ∈{ ± ± } | | 1 1 1 1 f(z)= = = . z2 sinh z z2 z + z3 + z5 + z3 1+ z2 + z4 + 3! 5! ··· 3! 5! ··· Using the identity 1 =1 w + w2 w3 + , 1+w − − ··· we obtain 1 z2 z4 z2 z4 2 z2 7z4 =1 + + + + + + =1 + + 1+ z2 + z4 + − 3! 5! ··· 3! 5! ··· ··· − 6 360 ··· 3! 5! ··· so that 1 1 z2 7z4 1 1 7z = 1 + + = + + z2 sinh z z3 − 6 360 ··· z3 − 6z 360 ··· for 0 < z <π. Hence, | | 1 1 1 7z 2πi πi dz = + + dz = = . z2 sinh z z3 − 6z 360 ··· − 6 − 3 C C 28–3 Mathematics312(Fall2012) November21,2012 Prof. Michael Kozdron Lecture #29: Laurent Series and Residue Theory Example 29.1. Determine the Laurent series of z2 2z +3 f(z)= − z 2 − for z 1 > 1. | − | Solution. Note that f(z) is analytic for z 1 > 1sincetheonlysingularityforf(z)occurs | − | at z =2.Sinceweareexpandingaboutthepointz0 =1,theLaurentserieswillnecessarily have the form ∞ f(z)= a (z z )j. j − 0 j= −∞ Therefore, if we want to expand in powers of z 1, we need to turn both our numerator and denominator into functions of z 1. Observe that− − j ∞ ∞ ∞ 1 1 1 1 1 1 j 1 j = = = = (z 1)− − = (z 1)− z 2 (z 1) 1 z 1 1 1 z 1 z 1 − − z 1 j=0 j=0 j=1 − − − − − − − − for 1/(z 1) < 1, or equivalently, z 1 > 1, and | − | | − | z2 2z +3=(z 1)2 +2. − − This yields j 2 j ∞ ∞ ∞ 2 j 1 − 1 f(z)= (z 1) +2 (z 1)− = +2 − − z 1 z 1 j=1 j=1 − j=1 − j 2 j ∞ 1 − ∞ 1 =(z 1) + 1 + +2 − z 1 z 1 j=3 − j=1 − j ∞ 1 =(z 1) + 1 + 3 − z 1 j=1 − for z 1 > 1astherequiredLaurentseries. | − | Classifying isolated singularities We will now focus on functions that have an isolated singularity at z0.Therefore,suppose that f(z)isanalyticintheannulus0< z z0 ∞ ∞ j j f(z)= aj(z z0) + a j(z z0)− . − − − j=0 j=1 29–1 We call the part with the negative powers of (z z ), namely − 0 ∞ j a j(z z0)− , − − j=1 the principal part of the Laurent series. There are three mutually exclusive possibilities for the principal part. (i) If aj =0forallj<0, then we say that z0 is a removable singularity of f(z). (ii) If a m =0forsomem N,butaj =0forallj< m,thenwesaythatz0 is a pole − of order m for f(z). ∈ − (iii) If aj =0forinfinitelymanyj<0, then we say that z0 is an essential singularity of f(z). Example 29.2. Suppose that sin z f(z)= z for z > 0. Since the Laurent series expansion of f(z)is | | 1 1 z3 z5 z7 z2 z4 z6 f(z)= sin z = z + + =1 + + , z z − 3! 5! − 7! ··· − 3! 5! − 7! ··· we conclude that z0 =0isaremovablesingularity. Example 29.3. Suppose that ez f(z)= zm for z > 0wherem is a positive integer. Since | | 1 z2 z3 f(z)= 1+z + + + zm 2! 3! ··· 1 1 1 1 1 z = + + + + + + + , zm zm 1 2!zm 2 ··· (m 1)!z m! (m +1)! ··· − − − we conclude that z0 =0isapoleoforderm. Example 29.4. Suppose that f(z)=e1/z for z > 0. Since | | (1/z)2 (1/z)3 1 1 1 f(z)=e1/z =1+(1/z)+ + + =1+ + + + , 2! 3! ··· z 2!z2 3!z3 ··· we conclude that z0 =0isanessentialsingularity. 29–2 Recall from Lecture #27 that we took a first look at residue theory as an application of Laurent series. The basic idea is that if we have the Laurent series expansion of a function f(z)for0< z z f(z)dz =2πia 1 − C 1 where a 1 is the coefficient of the (z z0)− term in the Laurent series expansion of f(z). − − Definition. Suppose that the function f(z)hasanisolatedsingularityatz0.Thecoefficient 1 a 1 of (z z0)− in the Laurent series expansion of f(z)aroundz0 is called the residue of − − f(z) at z0 and is denoted by a 1 =Res(f; z0). − Example 29.5. Suppose that C = z =1 denotes the unit circle oriented counterclock- wise. Compute the following three integrals:{| | } sin z (a) dz, z C ez (b) dz,wherem is a positive integer, and zm C (c) e1/z dz. C Solution. In order to compute all three integrals, we use the fact that z0 =0isanisolated singularity so that f(z)dz =2πi Res(f;0). C (a) From Example 29.2, we know that a 1 =0sothat − sin z dz =0. z C (b) From Example 29.3, we know that a 1 =1/(m 1)! so that − − ez 2πi dz = . zm (m 1)! C − (c) From Example 29.4, we know that a 1 =1sothat − e1/z dz =2πi. C Note that although we could have used the Cauchy Integral Formula to solve (a) and (b), we could not have used it to solve (c). 29–3 Question. Given the obvious importance of the coefficient a 1 in a Laurent series, it is − natural to ask if there is any way to determine a 1 without computing the entire Laurent − series. Theorem 29.6. A function f(z) has a pole of order m at z0 if and only if g(z) f(z)= (z z )m − 0 for some function g(z) that is analytic in a neighbourhood of z and has g(z ) =0. 0 0 Proof. Suppose that f(z)hasapoleoforderm at z0.Bydefinition,theLaurentseriesfor f(z)hastheform ∞ a m j − f(z)= m + aj(z z0) (z z0) − j= (m 1) − −− and so ∞ ∞ m j+m j (z z0) f(z)=a m + aj(z z0) = a m + aj m(z z0) . − − − − − − j= (m 1) j=1 −− Therefore, if we let ∞ j g(z)=a m + aj m(z z0) , − − − j=1 then g(z)isanalyticinaneighbourhoodofz0.Byassumption,a m =0sincef(z)hasa − pole of order m at z0,andsog(z0)=a m =0. − Conversely, suppose that g(z) f(z)= (z z )m − 0 for some function g(z)thatisanalyticinaneighbourhoodofz and has g(z ) =0.Since 0 0 g(z)isanalytic,itcanbeexpandedinaTaylorseriesaboutz0,say ∞ g(z)=b + b (z z )+b (z z )2 + = b (z z )j. 0 1 − 0 2 − 0 ··· j − 0 j=0 Since g(z )=b =0byassumption,weobtain 0 0 1 ∞ b b f(z)= b (z z )j = 0 + 1 + . (z z )m j − 0 (z z )m (z z )m 1 ··· 0 j=0 0 0 − − − − Therefore, by definition, f(z)hasapoleoforderm at z0. 29–4 Mathematics312(Fall2012) November23,2012 Prof. Michael Kozdron Lecture #30: The Cauchy Residue Theorem Recall that last class we showed that a function f(z)hasapoleoforderm at z0 if and only if g(z) f(z)= (z z )m − 0 for some function g(z)thatisanalyticinaneighbourhoodofz and has g(z ) =0. 0 0 Example 30.1. Suppose that sin z f(z)= . (z2 1)2 − Determine the order of the pole at z0 =1. Solution. Observe that z2 1=(z 1)(z +1)andso − − sin z sin z sin z/(z +1)2 f(z)= = = . (z2 1)2 (z 1)2(z +1)2 (z 1)2 − − − Since sin z g(z)= (z +1)2 2 is analytic at 1 and g(1) = 2− sin(1) =0,weconcludethatz =1isapoleoforder2. 0 Suppose that f(z)hasapoleoforderm at z0 so that we can write ∞ a m a (m 1) a 1 j f(z)= − + − − + + − + a (z z ) . (z z )m (z z )m 1 ··· z z j − 0 0 0 − 0 j=0 − − − Therefore, ∞ m m 1 j+m (z z0) f(z)=a m + a (m 1)(z z0)+ + a 1(z z0) − + aj(z z0) − − − − − ··· − − − j=0 which is analytic at z0 by Theorem 29.6. This means we can pick out a 1 by taking successive − derivatives. That is, differentiating m 1timesgives − m 1 ∞ d − m j+1 (z z0) f(z)=(m 1)!a 1 + bj(z z0) dzm 1 − − − − − j=0 for some coefficients bj.Theexactvalueofthesecoefficientsisunimportantsinceweare going to make the non-constant terms disappear now. If we evaluate the previous expression at z = z (which is justified since (z z )mf(z)=g(z)isanalyticatz ), then we find 0 − 0 0 m 1 d − m m 1 (z z0) f(z) =(m 1)!a 1, dz − − − − z=z0 30–1 and so we’ve found our formula for a 1 in the case that f(z)hasapoleoforderm at z0. − Note that if m =1,inwhichcasef(z)hasasimplepoleatz0,thentakingderivativesis unnecessary. We just need to evaluate (z z )f(z)atz . − 0 0 Theorem 30.2. If f(z) is analytic for 0 < z z0 m 1 m 1 1 d − m 1 d − m Res(f; z0)= m 1 (z z0) f(z) = lim m 1 (z z0) f(z). (m 1)! dz − − (m 1)! z z0 dz − − z=z0 → − − In particular, if z0 is a simple pole, then Res(f; z0)=(z z0)f(z) =lim(z z0)f(z). − z z0 − z=z0 → Example 30.3. Determine the residue at z0 =1of sin z f(z)= (z2 1)2 − and compute f(z)dz C where C = z 1 =1/2 is the circle of radius 1/2centredat1orientedcounterclockwise. {| − | } Solution. Since we can write (z 1)2f(z)=g(z)where − sin z g(z)= (z +1)2 is analytic at z =1withg(1) =0,theresidueoff(z)atz =1is 0 0 1 d2 1 d Res(f;1)= − (z 1)2f(z) = (z 1)2f(z) (2 1)! dz2 1 − dz − − − z=1 z=1 d sin z = dz (z +1)2 z=1 (z +1)2 cosz 2(z +1)sinz = − (z +1)4 z=1 4cos1 4sin1 = − 16 cos 1 sin 1 = − . 4 Observe that if C = z 1 =1/2 oriented counterclockwise, then the only singularity of {| − | } f(z)insideC is at z0 =1.Therefore, sin z (cos 1 sin 1)πi dz =2πi Res(f;1)= − . (z2 1)2 2 C − 30–2 It is worth pointing out that we could have also obtained this solution using the Cauchy Integral Formula; that is, sin z g(z) cos 1 sin 1 (cos 1 sin 1)πi dz = dz =2πig(1) = 2πi − = − (z2 1)2 (z 1)2 · 4 2 C − C − as above. Remark. Suppose that C is a closed contour oriented counterclockwise. If f(z)isanalytic inside and on C except for a single point z0 where f(z)hasapoleoforderm,thenboth the Cauchy Integral Formula and the residue formula will require exactly the same work, namely the calculation of the m 1derivativeof(z z )mf(z). − − 0 Recall that there are two other types of isolated singular points to consider, namely removable singularities and essential singularities. If the singularity is removable, then the residue is obviously 0. Unfortunately, there is no direct way to determine the residue associated with an essential singularity. The coefficient a 1 of the Laurent series must be determined explicitly. − In summary, suppose that f(z)isanalyticfor0< z z n f(z)dz =2πi Res(f; zj). C j=1 Proof. Observe that if C is a closed contour oriented counterclockwise, then integration over C can be continuously deformed to a union of integrations over C1,C2,...,Cn where Cj is a circle oriented counterclockwise encircling exactly one isolated singularity, namely zj,and not passing through any of the other isolated singular points. This yields f(z)dz = f(z)dz + + f(z)dz. ··· C C1 Cn Since f(z)dz =2πi Res(f; zj), Cj the proof is complete. Remark. Note that if the isolated singularities of f(z)insideC are all either removable or poles, then the Cauchy Integral Formula can also be used to compute f(z)dz. C 30–3 If any of the isolated singularities are essential, then the Cauchy Integral Formula does not apply. Moreover, even when f(z)hasonlyremovablesingularitiesorpoles,theResidue Theorem is often much easier to use than the Cauchy Integral Formula. Example 30.5. Compute 3z3 +4z2 5z +1 − dz (z 2i)(z3 z) C − − where C = z =3 is the circle of radius 3 centred at 0 oriented counterclockwise. {| | } Solution. Observe that 3z3 +4z2 5z +1 3z3 +4z2 5z +1 f(z)= − = − (z 2i)(z3 z) z(z 1)(z +1)(z 2i) − − − − has isolated singular points at z1 =0,z2 =1,z3 = 1, and z4 =2i.Moreover,eachisolated singularity is a simple pole, and so − 3z3 +4z2 5z +1 1 i Res(f;0)= − = = , (z 1)(z +1)(z 2i) 1 2i −2 z=0 − − − ·− 3 2 3z +4z 5z +1 3+4 5+1 3 3(1 + 2i) Res(f;1)= − = − = = , z(z +1)(z 2i) 1 2 (1 2i) 2(1 2i) 10 z=1 − · · − − 3z3 +4z2 5z +1 3+4+5+1 7 7(2i 1) Res(f; 1) = − = − = = − , − z(z 1)(z 2i) 1 2 ( 1 2i) −2(1 + 2i) 10 z= 1 − − − − ·− · − − 3z3 +4z2 5z +1 3(2i)3 +4(2i)2 5(2i)+1 34 15i Res(f;2i)= − = − = − . z(z 1)(z +1) 2i(2i 1)(2i +1) 10 z=2i − − By the Cauchy Residue Theorem, 3z3 +4z2 5z +1 i 3(1 + 2i) 7(2i 1) 34 15i − dz =2πi + + − + − =6πi. (z 2i)(z3 z) −2 10 10 10 C − − 30–4 Mathematics312(Fall2012) November26,2012 Prof. Michael Kozdron Lecture #31: Computing Real Trigonometric Integrals Suppose that C is a closed contour oriented counterclockwise. Last class we proved the Residue Theorem which states that if f(z) is analytic inside and on C,exceptforafinite number of isolated singular points z1,...,zn,then n f(z)dz =2πi Res(f; zj). C j=1 In order to compute Res(f; zj), we need to determine whether or not the isolated singular point zj is removable, essential, or a pole of order m.Ifzj is a pole of order m,thenwe know m 1 m 1 1 d − m 1 d − m Res(f; zj)= m 1 (z zj) f(z) = lim m 1 (z zj) f(z). (m 1)! dz − − (m 1)! z zj dz − − z=zj → − − In particular, if zj is a simple pole, then Res(f; zj)=(z zj)f(z) =lim(z zj)f(z). − z zj − z=zj → Example 31.1. Compute 2 dz z2 +4z +1 C where C = z =1 is the circle of radius 1 centred at 0 oriented counterclockwise. {| | } Solution. Clearly 2 f(z)= z2 +4z +1 has two simple poles. Notice that z2 +4z +1=(z2 +4z +4) 3=(z +2)2 3 = 0 implies − − z = √3 2andz = √3 2 1 − 2 − − are simple poles. However, z1 < 1and z2 > 1whichmeansthatf(z)onlyhasoneisolated singularity inside C.Since| | | | z2 +4z +1=(z z )(z z )=(z √3+2)(z + √3+2), − 1 − 2 − we find 2 2 1 Res(f; z )=Res(f; √3 2) = = = , 1 − √ √ √ z + 3+2z=√3 2 2 3 3 − so by the Cauchy Residue Theorem, 2z 1 2 dz =2πi = πi. z2 +4z +1 · √3 √3 C 31–1 Suppose that we now parametrize C by z(t)=eit,0 t 2π, and attempt to compute this same contour integral using ≤ ≤ 2π f(z)dz = f(eit) ieit dt. · C 0 That is, 2 2π 2 2π eit dz = ieit dt =2i dt z2 +4z +1 e2it +4eit +1· e2it +4eit +1 C 0 0 2π 1 =2i dt eit +4+e it 0 − 2π 1 =2i dt 4+2cost 0 2π 1 = i dt 2+cost 0 and so 2 2π 1 2π 1 2 πi = i dt or, equivalently, dt = π. √3 2+cost 2+cost √3 0 0 Notice that we were able to compute a definite integral by relating it to a contour integral that could be evaluated using the Residue Theorem. If we have a definite integral, the limits of integration are 0 to 2π,andtheintegrandisafunctionofcosθ and sin θ,thenwecan systematically convert it to a contour integral as follows. Let C = z =1 denote the circle of radius 1 centred at 0 oriented counterclockwise and parametrize{| | C by} z(θ)=eiθ, iθ 0 θ 2π,sothatz(θ)=ie = iz(θ). Since ≤ ≤ iθ 1 iθ z(θ)=e =cosθ + i sin θ and = e− =cosθ i sin θ, z(θ) − we find iθ iθ iθ iθ e + e− 1 1 e e− 1 1 cos θ = = z(θ)+ and sin θ = − = z(θ) . 2 2 z(θ) 2i 2i − z(θ) Therefore, 2π 2π 1 1 1 1 F (cos θ, sin θ)dθ = F z(θ)+ , z(θ) dθ 2 z(θ) 2i − z(θ) 0 0 2π 1 1 1 1 z (θ) = F z(θ)+ , z(θ) dθ 2 z(θ) 2i − z(θ) z (θ) 0 2π 1 1 1 1 1 = F z(θ)+ , z(θ) z(θ)dθ 2 z(θ) 2i − z(θ) iz(θ) 0 1 1 1 1 1 = F z + , z dz. 2 z 2i − z iz C 31–2 Example 31.2. Compute 2π cos(2θ) dθ. 5 4cosθ 0 − Solution. Let C = z =1 oriented counterclockwise be parametrized by z(θ)=eiθ, 0 θ 2π. Note that{| | } ≤ ≤ e2iθ + e 2iθ 1 1 z(θ)4 +1 cos(2θ)= − = z(θ)2 + = 2 2 z(θ)2 2z(θ)2 and 1 1 2 2z(θ)2 5z(θ)+2 5 4cosθ =5 4 z(θ)+ =5 2z(θ) = − , − − · 2 z(θ) − − z(θ) − z(θ) so that z(θ)4+1 4 4 cos(2θ) 2z(θ)2 z(θ) +1 z(θ) +1 = 2z(θ)2 5z(θ)+2 = 2 = . 5 4cosθ − −2z(θ)(2z(θ) 5z(θ)+2) −2z(θ)(2z(θ) 1)(z(θ) 2) − − z(θ) − − − This implies 2π cos(2θ) 2π z(θ)4 +1 z4 +1 1 dθ = dθ = dz 5 4cosθ − 2z(θ)(2z(θ) 1)(z(θ) 2) − 2z(2z 1)(z 2) · iz 0 − 0 − − C − − i z4 +1 = dz. 2 z2(2z 1)(z 2) C − − Let z4 +1 f(z)= z2(2z 1)(z 2) − − so that f(z)clearlyhasadoublepoleatz1 =0,asimplepoleatz2 =1/2, and a simple pole at z3 =2.Ofthesethreesingularitiesoff(z), only two of them are inside C.Therefore, d z4 +1 4z3(2z 1)(z 2) (z4 +1)(4z 5) 5 Res(f;0)= = − − − − = dz (2z 1)(z 2) (2z 1)2(z 2)2 4 z=0 z=0 − − − − and 1 z4 +1 z4 +1 17 Res(f;1/2) = z = = . − 2 z2(2z 1)(z 2) 2z2(z 2) −12 z=1/2 z=1/2 − − − By the Residue Theorem, 2π cos(2θ) i z4 +1 i 5 17 π dθ = dz = 2πi = . 5 4cosθ 2 z2(2z 1)(z 2) 2 · 4 − 12 6 0 − C − − Remark. It is possible to compute both 2π 1 2π cos(2θ) dθ and dθ 2+cosθ 5 4cosθ 0 0 − by calculating indefinite Riemann integrals. However, such calculations are very challenging. The contour integral approach is significantly easier. 31–3 Mathematics312(Fall2012) November28,2012 Prof. Michael Kozdron Lecture #32: Computing Real Improper Integrals Example 32.1. Compute 2 ∞ x dx. (x2 +1)(x2 +4) 0 Solution. Although it is possible to compute this particular integral using partial fractions easily enough, we will solve it with complex variables in order to illustrate a general method which works in more complicated cases. Observe that by symmetry, 2 2 ∞ x ∞ x 2 2 2 dx = 2 2 dx. 0 (x +1)(x +4) (x +1)(x +4) −∞ Suppose that C = C [ R, R]denotestheclosedcontourorientedcounterclockwiseob- R ⊕ − tained by concatenating CR,thatpartofthecircleofradiusR in the upper half plane parametrized by z(t)=Reit,0 t π,with[ R, R], the line segment along the real axis connecting the point R to the≤ point≤ R.Therefore,if− − z2 f(z)= , (z2 +1)(z2 +4) then f(z)dz = f(z)dz + f(z)dz. ( ) C [ R,R] C ∗ − R We now observe that we can compute z2 f(z)dz = dz (z2 +1)(z2 +4) C C using the Residue Theorem. That is, since z2 z2 f(z)= = , (z2 +1)(z2 +4) (z + i)(z i)(z +2i)(z 2i) − − we find that f(z)hassimplepolesatz = i, z = i, z =2i, z = 2i. However, only z 1 2 − 3 4 − 1 and z3 are inside C (assuming, of course, that R is sufficiently large). Thus, z2 i2 i Res(f; z )= = = 1 (z + i)(z +2i)(z 2i) (2i)(3i)( i) 6 z=z1=i − − and z2 (2i)2 i Res(f; z )= = = , 3 (z + i)(z i)(z +2i) (3i)(i)(4i) −3 z=z3=2i − so by the Residue Theorem, z2 i i π f(z)dz = dz =2πi = . (z2 +1)(z2 +4) 6 − 3 3 C C 32–1 In other words, we have shown that for R sufficiently large ( )becomes ∗ π z2 z2 = 2 2 dz + 2 2 dz. 3 [ R,R] (z +1)(z +4) C (z +1)(z +4) − R The next step is to observe that since [ R, R]denotesthelinesegmentalongtherealaxis connecting the point R to the point −R,ifweparametrizethelinesegmentbyz(t)=t, − R t R,thensincez(t)=1,weobtain − ≤ ≤ z2 R t2 R x2 2 2 dz = 2 2 dt = 2 2 dx [ R,R] (z +1)(z +4) R (t +1)(t +4) R (x +1)(x +4) − − − where the last equality follows by a simple change of dummy variable from t to x.Thus, π R x2 z2 = 2 2 dx + 2 2 dz 3 R (x +1)(x +4) C (z +1)(z +4) − R and so by taking the limit as R of both sides we obtain →∞ π R x2 z2 =lim 2 2 dx +lim 2 2 dz. 3 R R (x +1)(x +4) R C (z +1)(z +4) →∞ − →∞ R We now make two claims. Claim 1. R 2 2 x ∞ x lim 2 2 dx = 2 2 dx R R (x +1)(x +4) (x +1)(x +4) →∞ − −∞ Claim 2. z2 lim dz =0 R (z2 +1)(z2 +4) →∞ CR Assuming that both claims are true, we obtain 2 2 π ∞ x ∞ x π = 2 2 dx and so 2 2 dx = 3 (x +1)(x +4) 0 (x +1)(x +4) 6 −∞ which is in agreement with what one obtains by using partial fractions. Hence, the next task is to address these two claims. We will begin with the second claim which follows immediately from this result. Theorem 32.2. Suppose that CR denotes the upper half of the circle connecting R to R and parametrized by z(t)=Reit, 0 t π.If − ≤ ≤ P (z) f(z)= Q(z) is the ratio of two polynomials satisfying deg Q deg P +2, then ≥ lim f(z)=0. R →∞ CR 32–2 Proof. The fact that deg Q deg P + 2 implies that if z is sufficiently large, then ≥ | | P (z) K f(z) = ( ) | | Q(z) ≤ z 2 † | | for some constant K< .(Seethesupplementforderivationof( ).) Hence, ∞ † K K Kπ f(z)dz f(z) dz = dz = (C )= ≤ | | z 2 R2 R R CR CR CR | | since (C )=πR is the arclength of C .TakingR then yields the result. R R →∞ Thus, we conclude from this theorem that z2 lim dz =0 R (z2 +1)(z2 +4) →∞ CR since P (z)=z2 has degree 2 and Q(z)=(z2 +1)(z2 +4)hasdegree4. Review of Improper Integrals In order to discuss Claim 1, it is necessary to review improper integrals from first year calculus. Suppose that f : R R is continuous. By definition, → b ∞ f(x)dx =lim f(x)dx. b 0 →∞ 0 Thus, assuming the limit exists as a real number, we define ∞ f(x)dx 0 to be this limiting value. By definition, 0 0 f(x)dx =lim f(x)dx. a a −∞ →−∞ Thus, assuming the limit exists as a real number, we define 0 f(x)dx −∞ to be this limiting value. By definition, 0 b 0 ∞ ∞ f(x)dx = f(x)dx + f(x)dx =lim f(x)dx +lim f(x)dx. 0 b 0 a a −∞ −∞ →∞ →−∞ Thus, in order for ∞ f(x)dx ( ) ∗ −∞ 32–3 to exist it must be the case that both b 0 lim f(x)dx and lim f(x)dx b a →∞ 0 →−∞ a exist as real numbers. However, if one of these limits fails to exist as a real number, then the improper integral ( )doesnotexist.Sometimes,wemightwrite ∗ b ∞ f(x)dx =lim f(x)dx ( ) a ,b a † −∞ →−∞ →∞ instead which just writes the two separate limits in a single piece of notation. It is important to stress that this notation still implies that two separate limits are being taken: a and b .Itmightbetemptingtotryandcombinethetwoseparatelimitsintoasingle→−∞ →∞ limit as follows: c ∞ f(x)dx =lim f(x)dx. ( ) c c ‡ −∞ →∞ − However, ( )and() are not the same! As we will now show, it is possible for † ‡ c lim f(x)dx c c →∞ − to exist, but for b lim f(x)dx a ,b →−∞ →∞ a not to exist. Example 32.3. Observe that c x2 c c2 ( c)2 x dx = = − =0 c 2 c 2 − 2 − − and so c lim x dx =lim0=0. c c c 0 →∞ − → On the other hand, 0 x2 0 a2 b x2 b b2 x dx = = and x dx = = 2 − 2 2 2 a a 0 0 so that 0 a2 b b2 lim x dx = lim = and lim x dx =lim = . a − a 2 −∞ b b 2 ∞ →−∞ a →−∞ →∞ 0 →∞ Thus, b lim x dx = + = a ,b −∞ ∞ ∞−∞ →−∞ →∞ a so that ∞ x dx −∞ does not exist. 32–4 Supplement: Verification of ( ) from proof of Theorem 32.2 † Suppose that Q(z)=b + b z + + b zm is a polynomial of degree m. Without loss of 0 1 ··· m generality assume that bm =1.Therefore, m bm 1 b0 z− Q(z)=1+ − + + z ··· zm and so by the triangle inequality, m bm 1 b0 bm 1 b0 z− Q(z) = 1+ − + + 1 − + + . | || | z ··· zm ≥ − z ··· zm Let M =max 1, b0 ,..., bm 1 and note that 2mM > 1. This means that if z 2mM, − then { | | | |} | |≥ bm j M M 1 − . zj ≤ z j ≤ z ≤ 2m | | | | Therefore, since there are m terms in the following sum, bm 1 b0 1 1 1 1 − + + + + + = z ··· zm ≤ 2m 2m ··· 2m 2 which implies that m bm 1 b0 1 1 z− Q(z) 1 − + + 1 = . | || |≥ − z ··· zm ≥ − 2 2 Hence, we obtain, z m Q(z) | | | |≥ 2 n for z sufficiently large. On the other hand, suppose that P (z)=a0 + a1z + + anz is a polynomial| | of degree n so that ··· n an 1 a0 z− P (z)=a + − + + . n z ··· zn If z > 1, then by the triangle inequality, | | n an 1 a0 z− P (z) an + − + + an + an 1 + + a0 | |≤| | z ··· zn ≤| | | − | ··· | | and so with C = a + + a we obtain | 0| ··· | n| n z− P (z) C | |≤ for z > 1. Now suppose that | | P (z) f(z)= Q(z) is the ratio of polynomials with deg Q(z) deg P (z)+2. If deg P (z)=n and deg Q(z)=n+k with k 2, then we find that for z suffi≥ciently large, ≥ | | P (z) z nP (z) C 2C 2C K = − = = | n | z n+k k 2 2 Q(z) z− Q(z) ≤ n | | z ≤ z z z − 2 | | | | | | | | | | since k 2. ≥ 32–5 Mathematics312(Fall2012) November30,2012 Prof. Michael Kozdron Lecture #33: Cauchy Principal Value Definition. Suppose that f : R R is a continuous function on ( , ). If → −∞ ∞ R lim f(x)dx R R →∞ − exists, then we define the Cauchy principal value of the integral of f over ( , )tobe this value, and we write −∞ ∞ R ∞ p.v. f(x)dx =lim f(x)dx R R −∞ →∞ − for the value of this limit. Remark. If ∞ f(x)dx −∞ exists, then ∞ ∞ f(x)dx =p.v. f(x)dx. −∞ −∞ However, ∞ p.v. f(x)dx −∞ may exist, even though ∞ f(x)dx −∞ does not exist. For instance, ∞ ∞ p.v. x dx =0 whereas x dx does not exist. −∞ −∞ We can now finish verifying Claim 2 from Example 32.1 of the previous lecture. Example 32.1 (continued). Recall that we had deduced π R x2 z2 = 2 2 dx + 2 2 dz 3 R (x +1)(x +4) C (z +1)(z +4) − R where CR is that part of the circle of radius R in the upper half plane parametrized by z(t)=Reit,0 t π. Taking the limit as R and using Theorem 32.2, we obtained ≤ ≤ →∞ π R x2 =lim 2 2 dx. 3 R R (x +1)(x +4) →∞ − 33–1 By the definition of the Cauchy principal value, we have actually shown 2 π ∞ x =p.v. dx. 3 (x2 +1)(x2 +4) −∞ We now observe that x2 (x2 +1)(x2 +4) is an even function so that R x2 R x2 2 2 dx =2 2 2 dx R (x +1)(x +4) 0 (x +1)(x +4) − which implies that 2 R 2 ∞ x x p.v. 2 2 dx =2 lim 2 2 dx. (x +1)(x +4) R 0 (x +1)(x +4) −∞ →∞ In order to verify that the improper integral actually exists, note that R x2 R x2 R 1 dx = dx dx =arctanR (x2 +1)(x2 +4) (x2 +1)(x2 +4) ≤ x2 +1 0 0 0 using the inequality x2 (x2 +4).Sincearctan R π/2asR ,weconclude ≤ → →∞ R x2 lim dx R (x2 +1)(x2 +4) →∞ 0 exists by the integral comparison test. Thus, 2 2 2 π ∞ x ∞ x ∞ x =p.v. 2 2 dx = 2 2 dx =2 2 2 dx 3 (x +1)(x +4) (x +1)(x +4) 0 (x +1)(x +4) −∞ −∞ so that 2 ∞ x π dx = . (x2 +1)(x2 +4) 6 0 Example 33.1. Compute ∞ 1 p.v. dx. x2 +2x +1 −∞ Solution. Suppose that C = C [ R, R]denotestheclosedcontourorientedcounter- R ⊕ − clockwise obtained by concatenating CR,thatpartofthecircleofradiusR in the upper half plane parametrized by z(t)=Reit,0 t π,with[ R, R], the line segment along the real axis connecting the point R to the point≤ ≤R.Supposefurtherthat− − 1 f(z)= z2 +2z +2 33–2 so that f(z)hastwosimplepoles.Theseoccurwhere z2 +2z +2=z2 +2z +1+1=(z +1)2 +1=0, namely at z = √ 1 1=i 1andz = √ 1 1= i 1= (i + 1). Note that only 1 − − − 2 − − − − − − z1 is inside C,atleastforR sufficiently large. Therefore, since 1 1 f(z)= = , z2 +2z +2 (z z )(z z ) − 1 − 2 we conclude that 1 1 1 1 Res(f; z1)= = = = . z z2 z1 z2 i i +(i +1) 2i z=z1 − − − This implies 1 1 dz =2πi = π z2 +2z +2 2i C so that 1 1 1 π = 2 dz = 2 dz + 2 dz C z +2z +2 [ R,R] z +2z +2 C z +2z +2 − R R 1 1 = 2 dx + 2 dz. R x +2x +1 C z +2z +2 − R Taking R yields →∞ R 1 1 ∞ 1 π =lim 2 dx +lim 2 dz =p.v. 2 dx R R x +2x +1 R C z +2z +2 x +2x +1 →∞ − →∞ R −∞ using Theorem 32.2 to conclude that the second limit is 0. 33–3 Mathematics 312 (Fall 2012) December 3, 2012 Prof. Michael Kozdron Lecture #34: The Fundamental Theorem of Algebra Theorem 34.1 (Fundamental Theorem of Algebra). Every nonconstant polynomial with complex coefficients has at least one root. Proof. Suppose to the contrary that m 1 m P (z)=a0 + a1z + + am 1z − + amz ··· − is a nonconstant polynomial of degree m (so that m 1) having no roots. Without loss of ≥ generality, assume that am =1.Consequently, 1 Q(z)= P (z) must be an entire function. Let M =max1, a0 , a1 ,..., am 1 . Using an argument − similar to that given in the supplement to the{ proof| | of| Theorem| | 32.2,|} we find that if z 2mM,then | |≥ 1 1 2 Q(z) = . | | P (z) ≤ z m/2 ≤ (2mM)m | | On the other hand, if z 2mM,thenwehaveacontinuousreal-valuedfunction,namely Q(z) , on a closed disk.| |≤ From calculus, we conclude that the function must be bounded. |Hence,| 1 Q(z)= P (z) is bounded and entire. Thus, Q(z)mustbeconstantandthereforeP (z)mustitselfbe constant. However, we assumed that m 1sothiscontradictsourassumptionandwe conclude that P (z)musthaveatleastoneroot.≥ m 1 m Corollary 34.2. If P (z)=a0 + a1z + + am 1z − + amz is a polynomial of degree m, ··· − then P (z) has m complex roots. Moreover, if the coefficients aj R for all j, then the roots come in complex conjugate pairs. ∈ Proof. By Theorem 34.1 we know that P (z)hasatleastoneroot.Thus,supposethatz1 is arootofP (z). This means that we can write m 1 P (z)=(z z1)(b0 + b1z + + bm 1z − ) − ··· − m 1 for some coefficients b0,b1,...,bm 1.However,Q(z)=b0 + b1z + + bm 1z − is a poly- − − nomial of degree m 1, and so applying Theorem 34.1 to this polynomial,··· we conclude that Q(z)hasatleastoneroot.Thismeansthat− P (z)musthaveatleasttworoots.Continuing in this fashion yields m complex roots for P (z). 34–1 m 1 m Suppose that P (z)=a0 + a1z + + am 1z − + amz with aj R for all j.Supposethat ··· − ∈ z0 is a root of P (z)sothatP (z0)=0.Considerz0.Wefind m 1 m m 1 m P (z0)=a0 + a1z0 + + am 1(z0) − + am(z0) = a0 + a1z0 + + am 1z0 − + amz0 ··· − ··· − m 1 m = a0 + a1z0 + + am 1z0 − + amz0 ··· − = P (z0) = 0 =0. Thus, if z0 is a root of P (z), then so too is z0.Inotherwords,therootsofapolynomial with real coefficients come in conjugate pairs. Example 34.3. Show that the roots of P (z)=z3 1comeinconjugatepairs. − Solution. From Theorem 34.1 we know that P (z)hasthreeroots.Theyarez1 =1, 2πi/3 4πi/3 z2 = e ,andz3 = e .Observethatz1 is real so that it is self-conjugate. Furthermore, 2πi/3 4πi/3 z2 = e− = e = z3 so that z2 and z3 are a conjugate pair of roots. 34–2