AME525_Homework_11_Solutions_112107

# AME525_Homework_11_Solutions_112107 - HM Solull‘lh’ls...

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Unformatted text preview: HM/ Solull‘lh’ls Section 21.1 2- f(w +131) = (95 +iy)2 — We + 1'31) = (952 — 3/2 + y) + 1(2953/ - 95); Thus u(-"c,y) = \$2 - 3/2 + y,v(a:,y) = 29:3; — 9:. For the Cauchy-Riemann equations, compute 2—“ = 2x, 33 = 29:, g—u = 93 3/ 3/ —2y + 1, g: 2 2y — 1; so Cauchy-Riemann equations hold everywhere. Since u,v and all partials are continuous, f is differentiable everywhere. 6. f(a: +131): cc+iy + Im(:t +z'y) = (2r +y) + iy, so u(:r,y) = a: + y,v(m,y) = y. For , . Bu 61) (911 v the Cauchy-Riemann equations, compute —— = 1, — = 1, — = l, —— = 0. The Cauchy— 69: 6 6 81E Riemann equations hold nowhere, so is not differentiable at any point. 10. f(a:+iy) = i(w+iy)+lx+iyl = —y+\/ac2 +y2+gx so u(\$,y) = —y+\/ac2 + y§,v(a:,y) = 9:. For the Cauchy-Riemann equations, compute l = ~L,l = 0 g = _1 6x y 612 . . PF + 3/2 , a = 1. The Cauchy—Riemann equations hold nowhere, so f is not differentiable at any point. a:+i(y— 1) _ x2+y2 — 1—2921' 12. ‘ = _ “UH-1y) a:+i(y+1) m2+(y+1)2 ’50 x2 + 3/2 — 1 —2m u ‘Ti 2 my ) = ﬁ- ( y) x2+(y+1)2 “m y) 932+(y+1)2 For the Cauchy-Riemann equations, compute 93 2 2x022 + (y + 1)2) — 2m(x2 +y2 — 1) _ 4x(y +1) 335 [I2 + (y + 1)2l2 - [902 + (y +1)2l2’ 6v 493(y + 1) 6y [m2 + (y + 1)2}2’ an 2349:? + (y + 1)2) — 2(y + 1)(9:2 + y2 — 1) 2 2(y + 1)2 — 2m? 67/ z [m2 + <y+ 1W [m2 + <y+ 1)2]2’ 6v _ —2(ac2 + (y + 1)?) + 49:2 _ W a - lav2 + (y + 1)le - [I2 + (y + 1)212' Since u, v, and all partials are continuous at all points except a: = 0,3] = —1, or z = —i, and the Cauchy-Riemann equations holds at all such points, f is differentiable at all points except at z: —i. ‘14» Section 21.4 I _ . _ i ' . 7K 2. Smce 2 27.- _ 2\/-2—e77r/41we have log(2—21) .—. 111(2\/§)_H + 27m). and Maia—22:) 1mm) + I 4- Since 1 + 51' = V/2—691Tan-1(5): we have 10g(1 + 51;) = 1110/56) +£(Tan'1(5) 4: 2m). and Log“ + 5i) = 1n(\/2—6) + iTan_1(5\, Section 21.5 2- (1 +10% = e2i1°s<1+i> = €2i(‘“(‘/§)+i("/4+2"")) = e—(§+4n")[008(1n(2) +isin(ln(2))] 6_ (1_i)1/a = [ﬁe—iﬁ’z‘+2mr)]l/3 = we—qgﬂgl)” = 0,1,2 8. (165)1/4 = [16ei<2mr>]1/4 = 2e“T”,n = 0,1,2,3. These values are 2, 21‘, —2, —2i. I0 117? ~l ) _ 1—1 ,n—O,1,2. These values are ' ' 1/3 First : i, so +1) : : [ei(%+2n7r)]1/3 = ei(%+ 1 - 1 . . §(\/§+1),§(—\/9—>+2),—2. Section 22.2 3 -2 2/(22—1‘2)dz= iJi 1“ 3 2 37r/2 1 4. For zon 1",z=4eit,1r/25 t g37r/2, so /ldz:/ __(4ieit)dt:i7r 1‘2 7r /2 4e“ 2 1 3 8./(l+z)dz=(z+—z) r ‘ 3 2i 1 . 2 — §(—8 + 41.) 3i = —12i —3i _ 10. Parametrize F by z = —4 + t(i + 4),0 S t S 1 so on F, [2|2 = 16(t — 1)2 +152 1 - ' Then/|z|2dz=:/16t—12+t2i "d‘=w _ 3 31 17 . P 0[ ( ) ](_+4)z 3 [16(2: 1) +t]0:—§.(4+z)_ . 2 14. Parametrize I‘ by z = e“, O gt 3 27r, so /Im(z)dz = f 1r sin(t)ie”dt I‘ 0 = i/02"[_ sin2(t)' + isin(t) cos(t)]dt = _1r ...
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AME525_Homework_11_Solutions_112107 - HM Solull‘lh’ls...

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