Algebra - Solutions

1. Evaluate the following integrals. If you are using a particular theorem or formula make sure you indicate it. Z (a) sin (1/z) dz, where γ is a circle of radius 7 centered at 5 +5i traversed once in counterγ clockwise direction. √ The function f (z) = sin (1/z) is analytic in C\{0}, and since |5 + 5i| = 50 > 7 then 0 is outside the circle γ. R Then γ is contractible to a point in C\{0} and then by Cauchy’s Homotopy Theorem γ sin (1/z) dz = 0. √ Or, we can say f is analytic in D(5+5i, 49.5) and γ isR inside this disk, thus by Cauchy’s Theorem for Disks (or for simply connected regions), γ sin (1/z) dz = 0. Z cos z (b) 3 dz, where γ is the circle with radius 3 centered at π traversed once in counterγ (z − π) clockwise direction. Let f (z) = cos z, z0 = π. Since f is entire then by Cauchy’s Integral formula for derivatives we have that Z Z cos z cos z 2! 2! 00 f (z0 ) · I(γ, z0 ) = dz. 3 dz = 2πi γ (z − z0 ) 2πi γ (z − π)3 00 But f (z0 ) = − cos π = 1 and I(γ, z0 ) = 1 according to the problem. Then Z 2 cos z 1= dz 2πi γ (z − π)3 R cos z and then γ (z−π) 3 dz = πi. 2. Let f : {z ∈ C : |z| < 1} → C be analytic on its domain such that f (0) = f 0 (0) = 0 and f 00 (0) = 2i. In addition, |f (z)| ≤ |z| for all z such that |z| < 1. Prove that f (z) = iz 2 . Hint: Consider the function G(z) defined as G(0) = 0 and G(z) = f (z)/z whenever z 6= 0. Prove that G is analytic on {z ∈ C : |z| < 1}, determine G0 (0), and apply Schwarz Lemma. Let G(z) be defined as in the hint. Clearly G is analytic on D(0, 1)\{0}. However, f (z) − f (0) f (z) = lim = lim G(z). z→0 z→0 z z→0 z−0 G(0) = 0 = f 0 (0) = lim Thus G is continuous at z = 0. Thus by a Corollary to Morera’s Theorem, G is analytic on D(0, 1). Note that for z 6= 0, f (z) = z · G(z). Differentiating gives f 0 (z) = G(z) + zG0 (z) and f 00 (z) = G0 (z) + G0 (z) + zG00 (z) = 2G0 (z) + zG00 (z). Thus 2i = f 00 (0) = lim f 00 (z) = lim(2G0 (z) + zG00 (z)) = 2G0 (0), z→0 z→0 and then G0 (0) = i. Finally, note that |G(z)| = |f (z)|/|z| ≤ 1 for all z ∈ D(0, 1) and G(0) = 0. Thus by Schwarz Lemma, |G(z)| ≤ |z| and |G0 (0)| ≤ 1. But |G0 (0)| = |i| = 1, so equality in Schwarz Lemma is achieved and thus G(z) = cz for some c with |c| = 1. Finally, G0 (z) = c and thus c = G0 (0) = i and f (z) = z · (iz) = iz 2 . 1 3. Evaluate the following integral, where γ is the figure “8” below traversed once according to the arrows in the figure. Hint: z 2 + 1 = (z − i) (z + i). Z ez dz. 2 γ z +1 Solution 1: Let γ 1 be the upper part of the figure 8 and γ 2 the lower part. Let f (z) = ez ez and g(z) = . (z + i) (z − i) Observe that f is analytic in C\{−i} and g is analytic in C\{i}. Thus f is analytic in D(i, 1.1) and γ 1 ⊆ D(i, 1.1); also g is analytic in D(−i, 1.1) and γ 2 ⊆ D(−i, 1.1). Thus by Cauchy’s Integral Formula we have that Z Z ez f (z) dz = dz = 2πif (i) · I(γ 1 , i) 2 γ1 z + 1 γ1 z − i µ i¶ e (1) = πei = 2πi 2i and γ2 ez dz = z2 + 1 Z Z ez dz = z4 − 1 Z Z g(z) dz = 2πig(−i) · I(γ 2 , −i) γ2 z + i µ −i ¶ e (−1) = πe−i . = 2πi −2i Then γ ez dz + z4 − 1 Z ez dz 4 γ1 γ2 z − 1 ¡ ¢ = πei + πe−i = π ei + e−i = π (cos 1 + i sin 1 + cos 1 − i sin 1) = 2π cos 1. 2 Solution 2: Let F (z) = ez , this function is analytic on C and γ is homotopic to a point in C. Also from partial fraction decomposition we know that µ ¶ 1 i 1 1 = − . z2 + 1 2 z+i z−i Then by Cauchy’s integral formula we have that µZ ¶ Z Z ez F (z) F (z) i dz = dz − dz 4 2 γ z −1 γ z+i γ z−i i (2πiF (−i) · I(γ, −i) − 2πiF (i) · I(γ, i)) = 2 ¡ ¢ = −π e−i (−1) − ei (1) ¢ ¡ = π e−i + ei = 2π cos 1. z around z0 = 0 and determine its radius of convergence. −2 √ √ The function √ f is analytic on C \ { 2i, − 2i}. Thus the radius of convergence of its Taylor Series is 2, which is the radius of the largest open P diskn centered at z0 = 0 on which f is analytic. Using the geometric series 1/(1 − w) =√ ∞ n=0 w , which is valid for |w| < 1, we set 2 2 w = z /2. Then as long as |z /2| < 1, i.e., |z| < 2, we get that 4. Find the Taylor Series of f (z) = z2 ∞ µ 2 ¶n ∞ X X 2 z 2n z 1 = = = . 2 2 − z2 2 2n 1 − z2 n=0 n=0 Multiplying by −z/2 yields X ³ z ´ z 2n X −1 z 2 z =− · − f (z) = 2 = = z 2n+1 , n+1 z −2 2 2 − z 2 n=0 2 2n 2 n=0 ∞ which converges uniformly for any |z| ≤ r and r < ∞ √ 2. 5. Let f be an analytic function on a region containing a circle γ and its interior. Suppose that f (z) 6= 0 for all z except for a point z0 located inside the circle γ. Prove that if f has a zero of order one at z0 , then Z zf 0 (z) 1 z0 = dz. 2πi γ f (z) Hint: Let f (z) = (z − z0 )φ(z) and apply Cauchy’s Integral Formula. Let f be defined as in the hint. Note that on a neighborhood D(z0 , r) of z0 , φ is analytic and moreover φ(z) 6= 0 in that neighborhood. Let α be a circle centered at z0 with radius smaller than r. Because γ and α are homotopic in the region bounded by γ, it follos that Z Z zf 0 (z) zf 0 (z) dz = dz. γ f (z) α f (z) 3 For z ∈ D(z0 , r) we have that f 0 (z) = φ(z) + (z − z0 )φ0 (z). Thus Z α Z zφ(z) + z(z − z0 )φ0 (z) dz (z − z0 )φ(z) α Z Z z zφ0 (z) dz. dz + = α z − z0 α φ(z) zf 0 (z) dz = f (z) If F (z) = z, then Cauchy’s Integral Formula for F implies that Z Z F (z) z 1 1 dz = dz. z0 = F (z0 ) · I(α, z0 ) = 2πi α z − z0 2πi α z − z0 On the other hand, because φ never vanishes in D(z0 , r), the function zφ0 (z)/φ(z) is analytic there. Thus by Cauchy’s Theorem Z zφ0 (z) dz = 0. α φ(z) It follows that Z γ as we wanted to prove. zf 0 (z) dz = f (z) Z α z dz + z − z0 Z α zφ0 (z) dz = 2πiz0 , φ(z) 6. (Extra) Suppose f is an entire function and Im f (z) < 0 for all z. Prove that f (z) is a constant function. Let 1 . f (z) − i Notice that f (z) 6= i for all z, since Im f (z) < 0. Thus F is defined for all z and since f is entire then F is also entire. Now recall that |z| ≥ Im z for all z, thus F (z) = |f (z) − i| ≥ |Im (f (z) − i)| = |Im f (z) − 1| and since Im f (z) < 0 then Im f (z) − 1 < −1 and consequently |f (z) − i| > 1. Therefore |F (z)| = 1 < 1 for all z. |f (z) − i| Then by Liouville’s Theorem F (z) = c for all z and clearly c 6= 0. Then f (z) = constant. 1 c + i is also a Remark: The function g(z) = 1/ (z − i) maps the halfplane {z : Im z < 0} to the disk D(i/2, 1/2), that is the reason we can later apply Liouville’s Theorem to the image under this transformation. 4