Math 715 Homework 1 Solutions

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1 . [arrier, Krook and Pearson Section 2- Exercise ] Show that no purely real function can be analytic, unless it is a constant. onsider a function f(z) = u(x, y) + iv(x, y) where z = x + iy and where u and v are real functions. For f to be purely real means v(x, y) =. For f to be analytic means u x = v y and u y = v x hold at every point in the complex plane. Therefore, if f is a purely-real analytic function, it follows that u x = and u y =. Now u x = implies for y fixed that the function x u(x, y ) is constant and similarly u y = implies for x fixed that the function y u(x, y) is constant. Let c = u(x, y ). onsider any point x +iy in the complex plane. Since x u(x, y ) is constant holding y fixed, then u(x, y ) = u(x, y ) = c. Since y u(x, y) is constant holding x fixed, then u(x, y ) = u(x, y ) = c. It follows that u is identically equal to c throughout the entire complex plane. Therefore f is constant.

2 2. [arrier, Krook and Pearson Section 2-2 Exercise ] Evaluate 3 2i +i sin z dz in two ways. First by choosing any path between the two end points and using real integrals as in f(z) dz = (u dx v dy) + i (u dy + v dx) (2-5) and second by use of an indefinite integral. Show that the inequality f(z) dz f(z) dz ML (2-6) where f M and L is the length of is satisfied. The trigonometric identities will be used in this exercise. First, consider the path given by sin z = sin(x + iy) = sin x cosh y + i cos x sinh y cos z = cos(x + iy) = cos x cosh y i sin x sinh y +i 3 2i which may be written as the sum of the paths [γ ] and [γ 2 ] where γ (t) = + i( 3t) and γ 2 (t) = ( + 2t) 2i. Since sin z dz = (sin x cosh y dx cos x sinh y dy) + i (sin x cosh y dy + cos x sinh y dx), compute the real path integrals [γ ] 2 (sin x cosh y dx cos x sinh y dy) = cos sinh y dy = cos (cosh 2 cosh ) [γ 2 ] (sin x cosh y dx cos x sinh y dy) = 3 2 sin x cosh 2 dx = (cos cos 3) cosh 2

3 2 i (sin x cosh y dy + cos x sinh y dx) = i sin cosh y dy = i sin (sinh 2 + sinh ) [γ ] 3 i (sin x cosh y dy + cos x sinh y dx) = i cos x sinh 2 dx = i(sin sin 3) sinh 2 [γ 2 ] and add them to obtain sin z dz = cos cosh cos 3 cosh 2 i ( sin sinh + sin 3 sinh 2 ). This finishes the computation using real path integrals. Second, compute using an indefinite integral to obtain 3 2i +i sin z dz = cos(3 2i) + cos( + i) = cos 3 cosh 2 i sin 3 sinh 2 + cos cosh i sin sinh. This answer is the same as the answer found using the path integrals. We now show inequality (2-6) is satisfied. The term of the left hand side is To approximate the integral sin z dz = cos( + i) cos(3 2i) use numerical techniques. The Maple script i sin z dz # ompute integral f(z) dz along the path gamma+gamma2 2 restart; 3 f:=z->sin(z); 4 gamma:=t->+i*(-3*t); 5 gamma2:=t->(+2*t)-2*i; 6 I:=Integrate(abs(f(gamma(t))*diff(gamma(t),t)),t=..); 7 F:=evalf(I); 8 I2:=Integrate(abs(f(gamma2(t))*diff(gamma2(t),t)),t=..); 9 F2:=evalf(I2); print("the integral f(z) dz is approximately", F+F2); gives the output f := z -> sin(z) gamma := t -> + ( - 3 t) I gamma2 := t -> 2 t I 3

4 / I := 3 sin( + ( - 3 t) I) dt / F := / I2 := 2 sin(2 t I) dt / F2 := "The integral f(z) dz is approximately", Therefore Finally we compute M using Maple sin z dz # Find the maximum value of f(z) on gamma and gamma2 2 restart; 3 _EnvAllSolutions:=true; 4 5 f:=z->sin(z); 6 g:=t->+i*(-3*t); 7 ff:=f(g(t))*conjugate(f(g(t))) assuming t::real; 8 dff:=diff(ff,t); 9 cp:=solve(dff=,t); k:=; for j from to nops([cp]) 2 do 3 for i from -3 to 3 4 do 5 c:=evalf(subs(_z=i,cp[j])); 6 if(abs(im(c))<. and Re(c)< and Re(c)>) 7 then 8 k:=k+; 9 cpn[k]:=abs(c); 2 end 2 end 4

5 22 end; 23 cps:=[.,seq(cpn[n],n=..k),.]; 24 v:=seq(abs(f(g(cps[n]))),n=..nops(cps)); 25 print("the maximum of f(z) on gamma is",max(v)); g:=t->(+2*t)-2*i; 28 ff:=f(g(t))*conjugate(f(g(t))) assuming t::real; 29 dff:=diff(ff,t); 3 cp:=solve(dff=,t); 3 k:=; 32 for j from to nops([cp]) 33 do 34 for i from -3 to 3 35 do 36 c:=evalf(subs(_z2=i,cp[j])); 37 if(abs(im(c))<. and Re(c)< and Re(c)>) 38 then 39 k:=k+; 4 cpn[k]:=abs(c); 4 end 42 end 43 end; 44 cps:=[.,seq(cpn[n],n=..k),.]; 45 v2:=seq(abs(f(g(cps[n]))),n=..nops(cps)); 46 print("the maximum of f(z) on gamma2 is",max(v2)); print("the maximum of f(z) on is", max(v,v2)); 49 print("lm is",5*max(v,v2)); with the results _EnvAllSolutions := true f := z -> sin(z) g := t -> + ( - 3 t) I ff := sin( + ( - 3 t) I) sin( - ( - 3 t) I) dff := -3 I cos( + ( - 3 t) I) sin( - ( - 3 t) I) + 3 I sin( + ( - 3 t) I) cos( - ( - 3 t) I) / 2 /2 \ - + ( + tan(2) ) cp := -/3 I + I - arctan( ) - Pi _Z~, \ tan(2) / / 2 /2 \ 5

6 + ( + tan(2) ) -/3 I + I + arctan( ) - Pi _Z~ \ tan(2) / k := cps := [., ,.] v := , , "The maximum of f(z) on gamma is", g := t -> 2 t I ff := sin(2 t I) sin(2 t I) dff := 2 cos(2 t I) sin(2 t I) + 2 sin(2 t I) cos(2 t I) 2 /2 + ( - tanh(4) ) Pi _Z2~ cp := -/2 + I - /2 I arctanh( ) , tanh(4) 2 2 /2 - + ( - tanh(4) ) Pi _Z2~ -/2 + I + /2 I arctanh( ) tanh(4) 2 k := cps := [., ,.] v2 := , , "The maximum of f(z) on gamma2 is", "The maximum of f(z) on is", "LM is", Therefore M and L = = 5 so that LM Since we have shown that inequality (2-6) is satisfied for the curve. 6

7 3. [arrier, Krook and Pearson Section 2-2 Exercise 8a] Show that formal, term-by-term differentiation, or integration, of a power series yields a new power series with the same radius of convergence. Term-by-term Differentiation. Let f(z) = n= a nz n have radius of convergence R and g(z) = n= na nz n have radius of convergence r. laim R = r. First show r R. Let z be such that < z < R. hoose w such that z < w < R. Then by (-5) on page 9 we have that the series for f(w) converges absolutely. Thus, a n w n <. n= Now, let α = z / w. Then < α < and so nα n as n. It follows that nα n is bounded by some constant A so that nα n A for all n. Now, n= n a n z n = z n= n a n w n α n A z a n w n < shows that the series for g(z) converges for any z < R. It follows that r R. Second show that R r. Suppose z < r. Then by (-5) on page 9 we have that the series for g(z) converges absolutely. Thus, Now, n= n a n z n <. n= a n z n = a + z a n z n a + z n a n z n <. n= shows that the series for f(z) converges for any z < r. It follows that R r. Term-by-term Integration. Let f(z) = n= a nz n have radius of convergence R and h(z) = n= n+ a nz n+ have radius of convergence ρ. laim R = ρ. First show that R ρ. Suppose z < R. Then by (-5) on page 9 we have that the series for f(z) converges absolutely. Thus, Now, n= a n z n <. n= n + a n z n+ z n= n= a n z n <. shows that the series for h(z) converges for any z < R. It follows that R ρ. 7 n=

8 Second show R ρ. Let z be such that < z < ρ. hoose w such that z < w < ρ. Then by (-5) on page 9 we have that the series for h(w) converges absolutely. Thus, n= n + a n w n+ <. Now, let α = z / w. Let A be the bound so that nα n A for all n. Now, n= a n z n = z n= n + a n w n+ (n + )α n+ A z n= n + a n w n+ < shows that the series for f(z) converges for any z < ρ. It follows that R ρ. 8

9 4. [arrier, Krook and Pearson Section 2-2 Exercise 8b] The uniform-convergence property of a power series implies that term-by-term integration yields the integral of the sum function. Show that the integrated sum function is single valued and analytic within the circle of convergence. Let R be the radius of convergence of the infinite series defining f and h in part a. Let n n f n (z) = a k z k and h n (z) = k + a kz k+. k= Suppose ξ and z are such that ξ < R and z < R. Let Γ = { γ(t) : t [, t] } be any path such that γ() = z, γ() = ξ and γ(t) < R for t [, ]. Since f n is a polynomial, it is analytic. Therefore the path integral along Γ is path independent and since h n = f n we obtain ξ f n (ζ) dζ = f n (ζ) dζ = h n (ξ) h n (z). Γ z Since γ(t) < R for t [, ] then there is η > such that γ(t) R η for t [, ]. From the results on page 9 we obtain that ( ) ( ) f n γ(t) f γ(t) uniformly in t as n. Therefore, It follows that lim f n (ζ) dζ = lim n Γ Γ = n k= f n (γ(t))γ (t) dt f(γ(t))γ (t) dt = Γ f(ζ) dζ. ( f(ζ) dζ = lim hn (ξ) h n (z) ) = h(ξ) h(z). n Since this equality holds for any Γ inside the radius of convergence, the integral is path independent. Therefore, we may write ξ z f(ζ) dζ = h(ξ) h(z) for any z and ξ such that ξ < R and z < R where the integral is to be interpreted as a path integral along along any path from z to ξ that lies strictly inside the radius of convergence. Since h is single valued then the integrated sum function is single valued. We now claim h is analytic and h (z) = f(z). Let ϵ >. Since f is continuous at z there is δ > so that ζ z < δ implies f(ζ) f(z) < ϵ. Define γ(t) = ξ( t) + zt so that γ (t) = z ξ. Now, ξ z < δ implies γ(t) z < δ for t [, ] and therefore h(ξ) h(z) f(ζ) f(z) f(γ(t)) f(z) f(z) = dζ = γ (t) dt ξ z Γ z ξ z ξ f(γ(t)) f(z) dt < ϵ dt = ϵ. onsequently h (z) = f(z) for every z < R. 9

10 5. [arrier, Krook and Pearson Section 2-2 Exercise 8c] Show that a power series converges to an analytic function within its circle of convergence. onsider the power series for f(z) defined above with radius of convergence equal to R. By part a the function g(z) also has radius of convergence equal to R. Since f(z) may be obtained from g(z) through term by term integration of g(z) we have by part b that f (z) = g(z) for every z such that z < R. Thus, f is differentiable and therefore analytic in its circle of convergence.

11 6. [arrier, Krook and Pearson Section 2-3 Exercise ] Use auchy s integral formula to evaluate the integral around the unit circle z = of sin z 2z + i, ln(z + 2) z + 2, z 3 + asinh(z/2) z 2 + iz + i and cot z. Let D = { z : z < } be the unit disk and Γ = D be its boundary oriented in the positive (counterclockwise) direction. Since sin z is analytic on D and i/z is contained within it, then auchy s formula implies Γ sin z 2z + i = 2 Γ sin z z + i/2 = πi sin( i/2) = πi sin(i/2). Since then sin z = exp(iz) exp( iz) 2i sin(i/2) = exp( /2) exp(/2) 2i exp(/2) exp( /2) = i 2 = i sinh(/2). It follows that Γ sin z 2z + i = π sinh(/2). For the next integral note that, by definition, the principle branch ln z of the logarithm is the inverse of the function z exp(z): { x + iy : x R and y ( π, π] }. Therefore ln z is analytic on the domain \ (, ] and consequently ln z + 2 is analytic on \ (, 2]. It follows that ln z + 2 z + 2 is analytic in an open set containing D. By auchy s theorem we have For the next integral note that Γ ln(z + 2) z + 2 =. sinh z = exp(z) exp( z) 2 Setting ψ = sinh z and ω = exp z yields 2ωψ = ω 2. ( )

12 ompleting the square and factoring yields (ω ψ) 2 = ψ 2 +. We would like to take square roots to solve for ω. Recall that the principle branch of the square root function is defined as re iθ re iθ/2 for θ ( π, π]. This function is analytic on \ (, ]. The only time ψ 2 + (, ] is when Therefore the map ψ i(, ] or ψ i[, ). ψ ψ + ψ 2 + is analytic on \ ( i(, ] i[, ) ). From ( ) observe that if ω then ψ R. However, ψ R implies ω = ψ + ψ 2 + >, which is a contradiction. Therefore, it can t happen that ω. onsequently the function ψ log(ψ + ψ 2 + ) is analytic on \ ( i(, ] i[, ) ). As a result asinh(z) = log(z + z 2 + ) is an inverse to sinh(z) which is analytic on \ ( i(, ] i[, ) ). In particular sinh(z/2) is analytic on a neighborhood of the unit disk D. Note that z 2 + iz + i = (z z )(z z 2 ) where z = i + 4i > and z 2 = i 4i <. 2 2 Therefore z 3 + asinh(z/2) z z is analytic on D. It follows from auchy s formula that z 3 + asinh(z/2) Γ z 2 dz = 2πi z3 2 + asinh(z 2 /2) i. + iz + i z 2 z For the final integral note that cot z = cos z sin z and that sin z = ( ) k ( (2k + )! z2k+ = z ( ) k (2k + )! z2k) = z sinc z, k= where sinc z is defined by the absolutely convergent series in parenthesis shown above. Since the only zeros of sin z are z = kπ where k Z then the only zeros of sinc z are for z = kπ where k Z \ {}. It follows that cos z sinc z is analytic on D. onsequently ( cos ) cot z dz = 2πi = 2πi. sinc Γ k= 2

13 7. [arrier, Krook and Pearson Section 2-3 Exercise 2] If Φ(z) is analytic in a simply connected region in which a closed contour is drawn, obtain all possible values of where z is not a point on itself. ζ 2 z 2 Let = Ω be the boundary of an open set and S = Ω be a set on which the Green s theorem holds. onsider separately two cases. The first where z = and the second where z. If z =, the possible values of the integral in question are given by { 2πiΦ ζ 2 = () for Ω for Ω. If z, we note that ζ 2 z 2 = (ζ z)(ζ + z). Therefore, the possible values are ζ 2 z 2 = πiφ(z)z πiφ( z)z for z Ω and z Ω πiφ(z)z for z Ω and z Ω πiφ( z)z for z Ω and z Ω for z Ω and z Ω. If one also considers closed contours that do not necessarily bound a domain, then the contours may wind around the singularities multiple times. In this case we obtain the following possibilities where k, k, k 2 Z. 2πikΦ () and πik Φ(z)z πik 2 Φ( z)z 3

14 8. [arrier, Krook and Pearson Section 2-3 Exercise 3] If n is an integer, positive or negative, and if is a closed contour around the origin, use f (n) (z) = n! 2πi f(ζ) dζ (2 4) (ζ z) n+ to show that dz = unless n =. zn If n then /z n is analytic. Therefore, auchy s theorem implies dz z n =. If n >, we write f(z) = so that f (n ) (z) =. Then (2-4) implies z n dz = 2πi (n )! f (n ) () =. Finally, if n = then 2π z dz = 2π e it ieit dt = idt = 2πi which is non-zero. 4

15 9. [arrier, Krook and Pearson Section 2-3 Exercise 6] Show, that if Φ(z) is any function continuous on and if a function f(z) is defined for z include by f(z) = 2πi ζ z dζ then f(z) is analytic inside. Invent an example to illustrate the fact that, as an interior point z approaches a boundary point z, then f(z) need not approach Φ(z ). To show that f(z) is analytic, it is enough to show it has a derivative. By definition, f f(ξ) f(z) (z) = lim. ξ z ξ z Therefore, it remains to show that this limit exists. Now, f(ξ) f(z) ξ z = ξ z ( 2πi ζ ξ ) dζ ζ z = ( ) dζ 2πi (ζ ξ)(ζ z) = ( 2πi (ζ z) 2 + (ζ ξ)(ζ z) ) (ζ z) 2 dζ = ( 2πi (ζ z) 2 + ξ z ) (ζ ξ)(ζ z) 2 dζ = ( 2πi (ζ z) 2 + ξ z ) dζ. ζ ξ It follows that f(ξ) f(z) ξ z 2πi (ζ z) 2 dζ 2π ( ξ z ) (ζ z) 2 dζ ζ ξ Now since is continuous on it s maximum exists and we may define Since z then M = max { : ζ }. µ = dist(z, ) = min { z ζ : ζ } >. Finally, if ξ is close enough to z then also ξ. onsequently, ν = dist(ξ, ) >. By assumption = { γ(t) : t [, ] } where γ(t) is a differentiable function such that L = γ (t) dt <. 5

16 Therefore, follows that 2π ( ξ z ) (ζ z) 2 dζ ζ ξ 2π 2π Φ(γ(t)) ξ z (γ(t) z) 2 γ(t) ξ γ (t) dt ML µ 2 ξ z ν as ξ z. Therefore f(z) is analytic at each point inside. For an example where f(z) does not approach Φ(z ) as the interior point z approaches a boundary point z, consider Φ(z) = Re(z), z = and = { exp(it) : t [ π, π] }. Now However, 2πi Φ(z) = Re(z) as z f(ζ) ζ z dζ = π 2πi π cos t e it z ieit dt = 4πi = ζ + ζ 4πi ζ z dζ = 4πi = ( z 2 + ) = z 2 z z 2 2 π π e it + e it e it z ieit dt ζ 2 + (ζ z)ζ dζ as z. 6

17 . [arrier, Krook and Pearson Section 2-3 Exercise 7] A function Φ(z) is known to be continuous throughout a simply connected region and to have the property that Φ(z)dz = for any closed contour in that region. Show that Φ(z) must be analytic in that region. Let Ω be the simple region mentioned above. For z Ω fixed, define F (ξ) = Γ dζ for ξ Ω where Γ = { γ(t) : t [, ] } is any path that remains inside Ω and such that γ() = z and γ() = ξ. laim that F (ξ) is independent of the path Γ. onsider two different paths Γ i = { γ i (t) : t [, ] } such that γ i () = z and γ i () = ξ for i =, 2. Let = { c(t) : t [, 2] } where c(t) = { γ (t) for t [, ] γ 2 (2 t) for t (, 2]. Then is a closed contour and c(t) is pathwise differentiable. It follows that = dζ = = 2 Φ ( c(t) ) c (t)dt Φ ( γ (t) ) γ (t)dt + Setting s = 2 t so that ds = dt yields Therefore 2 2 Φ ( γ 2 (2 t) )( γ 2(2 t) ) dt. Φ ( γ 2 (2 t) )( γ 2(2 t) ) dt = Φ ( γ 2 (s) ) γ 2(s)ds. dζ = dζ, Γ Γ 2 which shows the integral defining F does not depend on the choice of path. laim F is differentiable on Ω. Let z Ω be fixed. Since Ω is open, there is α > such that B α (z) Ω. Therefore, for ξ such that ξ z < α, the straight line path Γ 3 = { γ 3 (t) : t [, ] } where γ 3 (t) = z + t(ξ z) lies entirely within Ω. It follows that F (ξ) F (z) ξ z = ( ξ z ) dζ dζ = ξ dζ. ξ z z z ξ z z 7

18 Therefore F (ξ) F (z) ξ z Φ(z) = ξ z ( Φ(γ3 (t)) Φ(z) ) γ 3(t)dt Φ(γ 3 (t)) Φ(z)) dt Given ε >, since Φ is continuous, there exists δ > with δ < α such that ξ z < δ implies Φ(ξ) Φ(z) < ε. It follows that F (ξ) F (z) ξ z Φ(z) ε whenever ξ z < δ. Therefore the limit exists and F (z) = Φ(z) for every z Ω. It follows that F is analytic and consequently that Φ is also analytic. 8