MTH3101 Spring 2017 HW Assignment 4: Sec. 26: #6,7; Sec. 33: #5,7; Sec. 38: #8; Sec. 40: #2 The due date for this assignment is 2/23/17.

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1 MTH0 Spring 07 HW Assignment : Sec. 6: #6,7; Sec. : #5,7; Sec. 8: #8; Sec. 0: # The due date for this assignment is //7. Sec. 6: #6. Use results in Sec. to verify that the function g z = ln r + iθ r > 0, 0 < θ < π is analytic in the indicated domain of definition, with derivative g z =. Then show that the composite function G z = g z + is analytic in the quadrant > 0, y > 0, with derivative G z = z z +. Proof. We will use the theorem in Sec.. First, g z = ln r+iθ r > 0, 0 < θ < π is defined throughout some ε > 0 neighborhood of a nonzero point z 0 = r 0 e iθ0 with r 0 > 0 and 0 < θ 0 < π. Furthermore, the first order partial derivatives of the functions u r, θ = ln r, v r, θ = θ with respect to r and θ eist everywhere in the neighborhood. Moreover, those partial derivatives are continuous at r 0, θ 0 and satisfy the polar form of the Cauchy-Riemann equations at r 0, θ 0 since ru r r0,θ 0 = r 0 r 0 = = v θ r0,θ 0, u θ r0,θ 0 = 0 = rv r0,θ 0. Therefore, by the theorem in Sec. the derivative g z 0 eists and is equal to Net, we have implying g z 0 = e iθ0 u r r0,θ 0 + iv r r0,θ 0 = e iθ 0 r 0 = r 0 e =. iθ0 z 0 z + = + iy + = y + + iy Im z + = y > 0 if > 0, y > 0. Thus, since z + is an entire function and the domain { + iy C : > 0, y > 0} { re iθ : r > 0, 0 < θ < π }, then this implies the composition G z = g z + is analytic in { + iy C : > 0, y > 0} and its derivative is, by the chain rule, This completes the proof. G z = g z + d z + = z dz z +.

2 Sec. 6: #7. Let a function f be analytic everywhere in a domain D. Prove that if f z is real-valued for all z in D, then f z must be constant throughout D. Proof. Let a function f be analytic everywhere in a domain D and suppose that f z is real-valued for all z in D. Then f z = f z for all z in D, in particular, the conjugate of f z and f z are analytic in the domain D. By Eample in Sec. 6, f z is constant throughout D. This completes the proof. Sec. : #5. a Show that the two square roots of i are Then show that log e iπ/ = e iπ/ and e i5π/. n + πi n = 0, ±, ±,... and log e i5π/ = n + + πi n = 0, ±, ±,.... Conclude that b Show that log i / = n + πi n = 0, ±, ±,.... log i / = log i as stated in Eample 5, Sec., by finding the values on the right-hand side of this equation and then comparing them with the final result in part a. Proof. a First, e iπ/ = e iπ/ = e iπ/ = i, e i5π/ = e i5π/ = e i5π/ = e iπ/ e iπ/ = i which proves that the two square roots of i are Net, log e iπ/ π = ln + i + nπ = log e i5π/ = ln + i 5π/ + nπ = e iπ/ and e i5π/. n + πi n = 0, ±, ±,..., n + + πi n = 0, ±, ±,....

3 Thus, as n and n + range over all the even and odd integers, the collection of them all being the integers, and we conclude that log i / = n + πi n = 0, ±, ±,.... b On the other hand, we have log i = π ln i + i + nπ = Therefore, from this and part a we conclude that log i = log i /. Sec. : #7. Show that a branch Sec. n + πi n = 0, ±, ±,.... log z = ln r + iθ r > 0, α < θ < α + π of the logarithm function can be written log z = ln + y + i tan y in rectangular coordinates. Then, using the theorem in Sec., show that the given branch is analytic in its domain of definition and that d dz log z = z there. Proof. Let α > 0 be given. On the domain D = { re iθ : r > 0, α < θ < α + π } we have for any z = + iy D, z = + y so that ln z = ln z = ln + y and with z = re iθ, where r > 0 and α < θ < α + π, we have which implies that y = Im z Re z = r sin θ r cos θ = tan θ θ tan y α, α + π.

4 In particular, for any z 0 = 0 + iy 0 D there is a choice of the real branch of tan t on an open real interval of y0 0 such that for any z = + iy D suffi ciently near z 0 we have tan y = θ, d dt tan t = + t, log z = ln + y + i tan y We will now use the theorem in Sec.. First, g z = ln + y + i tan y is defined throughout some ε > 0 neighborhood of a nonzero point z 0 = 0 + iy 0 D such that the first order partial derivatives of the functions u, y = ln + y, v, y = tan y with respect to and y eist everywhere in the neighborhood and those partial derivatives are continuous at, y and satisfy the Cauchy-Riemann equations at 0, y 0 since u 0,y 0 = y 0 u y 0,y 0 = y y 0 = = y + y0 y 0,y 0 = v y 0,y 0, 0 + y0 0 y 0,y 0 = v 0,y 0. Therefore, by the theorem in Sec. the derivative g z 0 eists and is equal to g z 0 = u y 0,y 0 + iv y 0,y 0 = y 0 = 0 iy y 0 = z 0 z 0 = z 0 z 0 z 0 = z 0. i y y 0 This completes the proof. Sec. 8: #8. Point out how it follows from epressions 5 and 6 in Sec. 7 for sin z and cos z that a sin z sin ; b cos z cos. Proof. a We have for z = + iy, sin z = sin + sinh y sin which implies by taking square roots that b We have for z = + iy, sin z sin. cos z = cos + sinh y cos which implies by taking square roots that cos z cos.

5 Sec. 0: #. Solve the equation sin z = for z by a equating the real parts and then imaginary parts in that equation; b using the epression in, Sec. 0, for sin z. Solution. a Let z = + iy. We have sin z = if and only if Re sin z = and Im sin z = 0. By Sec. 8,, Re sin z = and Im sin z = 0 if and only if sin cosh y = and cos sinh y = 0. Now sinh y = 0 if and only if e y = if and only if y = 0 implying that if sin cosh y = and sinh y = 0 then sin =, a contradiction. Also, cos = 0 if and only if = π + nπ n = 0, ±, ±,..., in which case sin = sin π + nπ = n. Thus, sin z = if and only if = π + nπ and cosh y = n n = 0, ±, ±,... if and only if = π + nπ and cosh y = n = 0, ±, ±,.... But = cosh y = ey + e y if and only if e y e y + = 0 if and only if e y = ± [ ] = ± if and only y = ln [ ± ] = ± ln +. Hence, sin z = if and only if = π + nπ and y = ± ln [ + ] n = 0, ±, ±,.... Therefore, z = n + [ π ± i ln + ] n = 0, ±, ±,.... b On the other hand, using the epression in, Sec. 0, for sin z we have z = sin = i log [i + ] / = i log [i + /] = i log i ± i [ ] [ = i ln ± + arg ± ] i [ ] = i ln ± + arg i = n + [ π ± i ln + ] n = 0, ±, ±,.... 5