Math Spring 2014 Solutions to Assignment # 4 Completion Date: Friday May 16, f(z) = 3x + y + i (3y x)

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1 Math Spring 2014 Solutions to Assignment # 4 Completion Date: Friday May 16, 2014 Question 1. [p 77, #1 (a)] Apply the theorem in Sec. 22 to verify that the function is entire. f(z) = 3x + y + i (3y x) Solution: If f(z) = 3x + y + i(3y x), then u(x, y) = 3x + y v(x, y) = 3y x, so that x = 3 = y y = 1 = x. Since the Cauchy-Riemann equations hold for all z C all partial derivatives are continuous everywhere, f (z) exists for all z C f(z) is analytic at each z C. Therefore f(z) is an entire function. Note that f(z) = 3(x + i y) + i( x i y) = 3z i z f (z) = 3 i. Question 2. [p 77, #1 (c)] Apply the theorem in Sec. 22 to verify that the function is entire. f(z) = e y sin x i e y cos x Solution: If f(z) = e y sin x i e y cos x, then u(x, y) = e y sin x v(x, y) = e y cos x, so that x = e y cos x = y y = e y sin x = x. Since all partial derivatives are defined continuous everywhere the Cauchy-Riemann equations hold for all z C, then f (z) exists for all z C, that is, f is entire. Question 3. [p 77, #4 (c)] For the function f(z) = z (z + 2)(z 2 + 2z + 2), determine the singular points of the function state why the function is analytic everywhere except at those points. Ans : z = 2, 1 ± i. Solution: Note that z 2 + 2z + 2 = (z + 1) = 0 when z = 1 ± i, f (z) doesn t exist for f is not analytic at any of these points. z 0 = 2, z 1 = 1 + i, z 2 = 1 i,

2 Note that f (z) exists except at each of these points, so that f is analytic everywhere except at these points. Therefore, given any one of these points, every ɛ-neighborhood of that point contains at least one point at which f is analytic, the points are singular points of f(z). z 0 = 2, z 1 = 1 + i, z 2 = 1 i, Question 4. [p 78, #6] Use the results in Sec. 23 to verify that the function g(z) = ln r + i θ (r > 0, 0 < θ < 2π) is analytic in the indicated domain of definition, with derivative g (z) = 1. Then show that the composite z function G(z) = g(z 2 + 1) is analytic in the quadrant x > 0, y > 0, with derivative G (z) = 2z z Suggestion : Observe that Im(z 2 + 1) > 0 when x > 0, y > 0. Solution: If g(z) = ln r + i θ (r > 0, 0 < θ < 2π), then u(r, θ) = ln r v(r, θ) = θ for r > 0, 0 < θ < 2π, r = 1 r = 1 r θ 1 r θ = 0 = r. Therefore, the Cauchy-Riemann equations hold at each point of the domain r > 0, 0 < θ < 2π, all the partial derivatives are continuous there, hence g is analytic in this domain, ( g (z) = e iθ r + i ) ( ) 1r = e iθ r + i 0 = 1 re iθ = 1 z, for r > 0, 0 < θ < 2π. Since the function z z is analytic everywhere, then the composition G(z) = g(z 2 + 1) is analytic wherever w = z = ρe iφ satisfies ρ > 0, 0 < φ < 2π. Now, if x > 0 y > 0, then Im(z 2 + 1) = 2xy > 0, so that ρ > 0 0 < φ < π < 2π, therefore G(z) = g(z 2 + 1) is analytic for z = x + i y with x > 0, y > 0, we can use the chain rule to differentiate it. Letting w = z 2 + 1, then G (z) = g (w) dw dz = 1 w 2z = 2z z 2 + 1, for x > 0, y > 0.

3 Question 5. [p 76, Example 4] Let f(z) be analytic in a domain D. Prove that f(z) must be constant throughout D if f(z) is constant throughout D. Suggestion : Observe that f(z) = c2 f(z) then use the main result in Example 3, Sec. 25. if f(z) = c (c 0); Solution: Suppose that f(z) = u(x, y) + i v(x, y) for z = x + i y D, then f(z) 2 = u(x, y) 2 + v(x, y) 2 = c 2 for z D, where c R is a constant. Differentiating with respect to x with respect to y, we get u x + v x = 0 u y + v y = 0 in D. Multiplying the first equation by u the second equation by v, we get using the Cauchy-Riemann equations, we have Adding these two equations, we have on D. In an entirely similar way, we find u 2 + uv x x = 0 uv + v2 y y = 0, u 2 + uv x x = 0 uv + v2 x x = 0. (u 2 + v 2 ) x = 0 (u 2 + v 2 ) y = 0, (u2 + v 2 ) x = 0, (u2 + v 2 ) y = 0, on D. Finally, since u 2 + v 2 = c 2, we have c 2 = c2 = c2 = c2 x y x y = 0 on D. Now, if c = 0, then f(z) = 0 therefore f(z) = 0 for all z D, however, if c 0, then therefore f(z) is a constant for z D. x = y = x = y = 0, Alternatively, using the suggestion, if f(z) = c for all z D, c = 0, then f(z) = 0 for all z D. On the other h, if f(z) = c for all z D, where c 0, then f(z) is never 0 in D, the function f(z) = c2 f(z) is also analytic on D, since f f are both analytic on D, then f(z) is a constant on D.

4 Question 6. [p 81, #1 (c)] Show that the function u(x, y) = sinh x sin y is harmonic in some domain find a harmonic conjugate v(x, y). Ans : v(x, y) = cosh x cos y. Solution: If u(x, y) = sinh x sin y, then 2 u x 2 (x, y) + 2 u (x, y) = sinh x sin y sinh x sin y = 0 y2 for all (x, y) C, u is harmonic on all of C. In order to find a harmonic conjugate v(x, y) of u, we use the Cauchy-Riemann equations to get y = = cosh x sin y x x = = sinh x cos y, y Integrating the first equation with respect to y holding x fixed, we have v(x, y) = cosh x cos y + h(x) where h(x) is an arbitrary function of x. Using the second equation, we have x = sinh x cos y + h (x) = sinh x cos y, that is, h (x) = 0 for all x, so that h(x) = C (constant) for all x. Therefore, for any real constant C, the function v(x, y) = cosh x cos y + C is a harmonic conjugate for u(x, y) = sinh x sin y on C. Question 7. [p 82, #6] Verify that the function u(r, θ) = ln r is harmonic in the domain r > 0, 0 < θ < 2π by showing that it satisfies the polar form of Laplace s equation, obtained in Exercise 5. Then use the technique in Example 5, Sec. 26, but involving the Cauchy-Riemann equations in polar form (Sec. 22), to derive the harmonic conjugate v(r, θ) = θ. (Compare Exercise 6, Sec. 24) Solution: If u(r, θ) = ln r, for r > 0, 0 < θ < 2π, then 2 u r 2 = ( ) (ln r) = r r r Therefore, ( ) 1 = 1 r r 2, r = 1 r, 2 u θ 2 = 0. 2 u r r r u ( r 2 θ 2 = 1r ) r for all r > 0, 0 < θ < 2π, u is harmonic in this region. ( ) = 0 r In order to find a harmonic conjugate v(r, θ) of u, we use the Cauchy-Riemann equations r = 1 r θ 1 r θ = r

5 we want ( ) 1 θ = r = 1 r r = 0 Integrating the first equation with respect to θ holding r fixed, we have v(r, θ) = θ + h(r) where h(r) is an arbitrary function of r. Using the second equation, we have r = h (r) = 0, that is, h (r) = 0 for all r > 0, so that h(r) = C (constant) for all r > 0. Therefore, for any real constant C, the function v(r, θ) = θ + C is a harmonic conjugate for u(r, θ) = ln r for r > 0, 0 < θ < 2π. So we may take v(r, θ) = θ, r > 0, 0 < θ < 2π, as the harmonic conjugate of u = ln r. The function f(z) = ln r + i θ, r > 0, 0 < θ < 2π, is analytic in this domain, ( f (z) = e iθ r + i ) = e iθ 1 r r = 1 re iθ = 1 z for r > 0, 0 < θ < 2π. Question 8. [p 88, #4] We know from Example 1, Sec. 22, that the function f(z) = e x e i y has a derivative everywhere in the finite plane. Point out how it follows from the reflection principle (Sec. 28) that f(z) = f (z) for each z. Then verify this directly. Solution: If f(z) = e x e i y, then u(x, y) = e x cos y v(x, y) = e x sin y, when y = 0, we have f(x) = e x cos 0 + ie x sin 0 = e x, which is real for all x R. Since the domain of f is symmetric about the real axis f is real-valued on the real axis, then from Schwarz s Reflection Principle, we have for all z C. If f(z) = e x e i y = e x cos y + ie x sin y, then again f(z) = f(z) for all z C. f(z) = f(z) f(z) = e x cos y ie x sin y, f(z) = e x e i y = e x cos y ie x sin y,

6 Question 9. [p 92, #1] Show that (a) exp(2 ± 3π i) = e 2 ; ( ) 2 + π i (b) exp = 4 e (1 + i); (c) exp(z + π i) = exp z. 2 Solution: (a) From the definition of the exponential function, we have exp(2 ± 3π i) = e 2 (cos 3π ± i sin 3π) = e 2 ( 1 + i 0) = e 2. (b) From the definition of the exponential function, we have ( ) 2 + π i exp = e 1 2 e π i/4 = e (cos π/4 + i sin π/4) = 4 (c) From the definition of the exponential function, we have e (1 + i). 2 exp(z + π i) = e x+i(y+π) = e x (cos(y + π) + i sin(y + π)) = e x (cos y + i sin y) = e x+iy = e z. Question 10. [p 92, #3] Use the Cauchy-Riemann equations the theorem in Sec. 21 to show that the function f(z) = exp z is not analytic anywhere. Solution: If f(z) = exp z = e x (cos y i sin y), then u(x, y) = e x cos y v(x, y) = e x sin y, x = ex cos y, y = ex sin y, y = ex cos y x = ex sin y. The Cauchy-Riemann equations hold if only if 2e x cos y = 0 2e x sin y = 0, that is, if only if sin y = cos y = 0. However, this is impossible, since sin 2 y + cos 2 y = 1. Therefore, there are no points z C for which f(z) = e z is differentiable, so no points z C at which f is analytic. Question 11. [p 92, #5] Write exp(2z + i) exp(iz 2 ) in terms of x y. Then show that exp(2z + i) + exp(iz 2 ) e 2x + e 2xy. Solution: We have therefore exp(2z + i) = e 2x+i(2y+1) = e 2x e i(2y+1) = e 2x, exp(iz 2 ) = e i(x2 y 2 +2ixy) = e 2xy e i(x2 y 2) = e 2xy, e 2z+i + e iz2 e 2z+i + e iz2 = e 2x + e 2xy.

7 Question 12. [p 92, #6] Show that exp(z 2 ) exp( z 2 ). Solution: We have e z 2 = e x 2 y 2 e 2ixy = e x 2 y 2 e 2ixy = e x 2 y 2, since x 2 y 2 x 2 + y 2, the (real) exponential function is increasing, then e x2 y 2 e x2 +y 2, so that e z 2 = e x 2 y 2 e x2 +y 2 = e z 2, that is, for all z C. e z2 e z 2 Question 13. [p 92, #8] Find all values of z such that (a) e z = 2; (b) e z = i; (c) exp(2z 1) = 1. Ans : (a) z = ln 2 + (2n + 1)π i (n = 0, ±1, ±2,... ). ( (b) z = ln 2 + 2n + 1 ) π i (n = 0, ±1, ±2,... ). 3 (c) z = 1 + nπ i (n = 0, ±1, ±2,... ). 2 Solution: (a) Note that e z = 2 if only if e x e iy = 2 e iπ if only if e x = 2 e iy = e iπ. This last condition is true if only if x = ln 2 y = π + 2kπ for k = 0, ±1, ±2,..., that is, if only if z = ln 2 + (2k + 1)πi for k = 0, ±1, ±2,.... (b) Note that e z = i if only if e z = 2 ( ) 3 i 2 if only if e z = 2 e iπ/3. This last condition is true if only if x = ln 2 y = π 3 + 2πk for k = 0, ±1, ±2,..., that is, if only if ( π ) z = ln 2 + i 3 + 2πk for k = 0, ±1, ±2,....

8 (c) Note that e (2z 1) = 1 if only if e 2x 1 e 2i y = 1 e i 0 this last condition is true if only if e 2x 1 = 1 2y = 2πk, for k = 0, ±1, ±2,..., that is, if only if 2x 1 = ln 1 = 0 y = πk, for k = 0, ±1, ±2,..., that is, if only if z = πki for k = 0, ±1, ±2,.... Question 14. [p 92, #13] Let the function f(z) = u(x, y) + i v(x, y) be analytic in some domain D. State why the functions U(x, y) = e u(x,y) cos v(x, y), V (x, y) = e u(x,y) sin v(x, y) are harmonic in D why V (x, y) is, in fact, a harmonic conjugate of U(x, y). Solution: If f(z) = u(x, y) + i v(x, y) is analytic in a domain D, let then writing we have so that g(z) = e f(z), e f(z) = U(x, y) + i V (x, y), e u(x,y) [cos v(x, y) + i sin v(x, y)] = U(x, y) + i V (x, y) U(x, y) = e u(x,y) cos v(x, y) V (x, y) = e u(x,y) sin v(x, y). Now, since f(z) is analytic in D, then g(z) is also analytic in D, therefore U V are harmonic in D the Cauchy-Riemann equations U x = V U y y = V x hold in D, that is, V is a harmonic conjugate of U in D.