Mat 311 - Spring 013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, 013 Question 1. [p 56, #10 (a)] 4z Use te teorem of Sec. 17 to sow tat z (z 1) = 4. We ave z 4z (z 1) = z 0 4 (1/z) (1/z 1) = z 0 4 (1 z) = 4. Question. [p 56, #13] Sow tat a set is unbounded (Sec. 11) if only if every neigborood of te point at infinity contains at least one point in S. If S C is unbounded, ten for eac n 1, tere is a point z n S wit z n n. Now, given ɛ > 0, coose n 0 1 wit 0 < 1 n 0 < ɛ, ten z n0 n 0 > 1 ɛ, z n0 is in te ɛ-neigborood {z : z > 1/ɛ} of te point of infinity. Tus, every neigborood of te point at infinity contains at least one point in S. Conversely, if every neigborood of te point at infinity contains at least one point in S, ten for eac n 1, we can coose a point z n S wit z n n (tat is, z n is in te 1/n-neigborood of te point at infinity). Ten S cannot be bounded, since z n S for all n 1, z n = +, tere is no n M > 0 suc tat z M for all z S. Question 3. [p 6, #1] Use te results in Sec. 0 to find f (z) wen (a) f(z) = 3z z + 4; (b) f(z) = ( 1 4z ) 3 ; (c) f(z) = z 1 ( ) 1 + z 4 (z 1/); (d) f(z) = z + 1 z (z 0). (a) f (z) = 6z.
(b) f (z) = 3(1 4z ) ( 8z) = 4z(1 4z ). (c) f (z) = 1 (z + 1) (z 1) 3 (z + 1) =, for z 1/. (z + 1) (d) f (z) = 4(1 + z ) 3 z z (1 + z ) 4 z z 4 = (1 + z ) 3 z 3 (3z 1), for z 0. Question 4. [p 6, #3] Apply definition (3), Sec. 19, of derivative to give a direct proof tat f (z) = 1 z wen f(z) = 1 z (z 0). If f(z) = 1 z for z 0, ten f f(z + ) f(z) (z) = = 1 z + 1 z = z (z + ) z(z + ), tat is, for z 0. f (z) = z(z + ) = 1 z(z + ) = 1 z Question 5. [p 63, #8 (b)] Use te metod in Example, Sec. 19, to sow tat f (z) does not exist at any point z wen f(z) = Im z. Let f(z) = Im(z), ten for z, C, wit 0, we ave Im(z + ) Im(z) Now, if 0 troug real values, Im() = 0, = Im(z) + Im() Im(z) = Im(). Im(z + ) Im(z) real = real However, if 0 troug imaginary values, say = it were t R t 0, ten 0 = 0. (1) Im() Im(z + ) Im(z) imag = t it = i, = ( i) = i. () imag Terefore, (1) () imply tat Im(z + ) Im(z) doesn t exist, tat is, f (z) does not exist for any z C.
Question 6. [p 71, #1] Use te teorem in Sec. 1 to sow tat f (z) does not exist at any point if (a) f(z) = z; (b) f(z) = z z; (c) f(z) = x + i xy ; (d) f(z) = e x e i y. (a) If f(z) = z = x iy, ten u(x, y) = x v(x, y) = y, so tat = 1 1 =, te Caucy-Riemann equations do not old at any point z C. Terefore f (z) does not exist for any z C. (b) If f(z) = z z = iy, ten u(x, y) = 0 v(x, y) = y, so tat = 0 =, again te Caucy-Riemann equations do not old at any point z C. Terefore f (z) does not exist for any z C. (c) If f(z) = x + ixy, ten u(x, y) = x v(x, y) = xy, so tat =, = 0, = xy = y. Now te Caucy-Riemann equations are = xy 0 = y tese equations ave no solutions (x, y). Terefore, f (z) does not exist at any point z C. (d) If f(z) = e x e i y = e x cos y ie x sin y, ten = ex cos y, = ex sin y, = ex cos y = ex sin y. Now te Caucy-Riemann equations are since e x 0 for all x R, tese equations are e x cos y = 0 e x sin y = 0 cos y = 0 sin y = 0, but tis is impossible since cos y+sin y = 1. Terefore, tere are no solutions to te Caucy-Riemann equations, f (z) does not exist for any z C.
Question 7. [p 71, #3] From te results obtained in Secs. 1, determine were f (z) exists find its value wen (a) f(z) = 1 z ; (b) f(z) = x + i y ; (c) f(z) = z Im z; Ans : (a) f (z) = 1 z (z 0); (b) f (x + i x) = x; (c) f (0) = 0. (a) If f(z) = 1 z, ten so tat for z 0. Now, u(x, y) = f(z) = z z = x x + y i y x + y, x x + y v(x, y) = y x + y = y x (x + y ) = = xy (x + y ) =. Since te partial derivatives are all continuous at eac z C, z 0, te Caucy-Riemann equations old at eac z C, z 0, ten f (z) exists for all z 0, f (z) = + i = y x (x + y ) + ixy (x + y ), tat is, for z 0. f (z) = x y ixy (x + y ) = (z) z 4 = 1 z (b) If f(z) = x + i y, ten u(x, y) = x v(x, y) = y, = x, = 0, = y = 0, so te Caucy-Riemann equations old only for tose points z = x + i y wit x = y. Since all partial derivatives are continuous everywere, f (z) exists only for te points z = x + i x = x(1 + i), x R, f (x + ix) = (x, x) + i (x, x) = x. (c) If f(z) = zim(z) = (x + i y) y = xy + i y, ten u(x, y) = xy v(x, y) = y, so tat = y, = x, = y = 0, te Caucy-Riemann equations old if only if y = y x = 0,
tat is, if only if x = y = 0. Since all partial derivatives are continuous, ten f (z) exists only for z = 0, f (0) = (0, 0) + i (0, 0) = 0. Question 8. [p 71, #4 (b)] Use te teorem in Sec. 3 to sow tat te function f(z) = re i θ/ (r > 0, α < θ < α + π) is differentiable in te indicated domain of definition, ten use expression (7) in tat section to find f (z). Ans : f (z) = 1 f(z). If f(z) = re iθ/, r > 0, α < θ < α + π, ten so tat u(r, θ) = r cos θ/ v(r, θ) = r sin θ/. f(z) = r (cos θ/ + i sin θ/), Now, so tat for r > 0, α < θ < α + π. Also, so tat for r > 0, α < θ < α + π. r = 1 r cos θ/ r θ = cos θ/, r = 1 r cos θ/ = 1 r θ r θ = sin θ/ r = 1 sin θ/, r 1 r θ = 1 sin θ/ = r r Te partial derivatives are all continuous for eac (r, θ) wit r > 0, α < θ < α+π, te Caucy-Riemann equations old for eac suc point, so tat f (z) exists for all suc points, ( f (z) = e iθ r + i ) = 1 r r (cos θ/ + i sin θ/) e iθ = 1 r e iθ/ = 1 f(z), for r > 0, α < θ < α + π. Question 9. [p 7, #5] Sow tat wen f(z) = x 3 + i (1 y) 3, it is legitimate to write f (z) = u x + i v x = 3x only wen z = i.
If f(z) = x 3 + i(1 y) 3, ten u(x, y) = x 3 v(x, y) = (1 y) 3, so tat Te Caucy-Riemann equations become = 3x, = 3(1 y) = 0, = 0. 3x + 3(1 y) = 0 tese old only at te point z = (0, 1) = i. Since all partial derivatives are continuous everywere, ten f (i) exists, f (i) = (0, 1) + i (0, 1) = 0. Question 10. [p 7, #10] (a) Recall (Sec. 5) tat if z = x + i y ten x = z + z y = z z. i By formally applying te cain rule in calculus to a function F (x, y) of two real variables, derive te expression F z = F z + F z = 1 ( F + i F ). (b) Define te operator z = 1 ( + i ), suggested by part (a), to sow tat if te first-order partial derivatives of te real imaginary parts of a function f(z) = u(x, y) + i v(x, y) satisfy te Caucy-Riemann equations, ten z = 1 [(u x v y ) + i (v x + u y )] = 0. Tus derive te complex form z = 0 of te Caucy-Riemann equations. (a) Since z = x + i y z = x i y, ten x = z + z y = z z, i F z = F z + F z = 1 ( F + 1 ) F i = 1 ( F + i F ). (b) Now, if f(z) = u(x, y) + i v(x, y) te real-valued functions u v satisfy te Caucy-Riemann equations, ten from part (a), we ave tat is, since = z = 1 (u + i v) + i (u + i v) = 1 + i + i 1, z = 1 ( ) + i ( + ) = 0, =.