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## Complex Variables

by: Mrs. Lizeth Bailey

24

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2

# Complex Variables MATH 3810

Mrs. Lizeth Bailey
TTU
GPA 3.97

Staff

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## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Mrs. Lizeth Bailey on Wednesday October 21, 2015. The Class Notes belongs to MATH 3810 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/225719/math-3810-tennessee-tech-university in Mathematics (M) at Tennessee Tech University.

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Date Created: 10/21/15
Math 3810 Spring 2009 Answer Key for final review sheet 1 25 71 Z gin20w 6i7r20k27r5 These points are equally spaced around the circle of radius 10 2 centered at the origin with the first point at an angle of 7r20 radians 2 2472231022 0 iff x22272z100 iffz0 or z 372640 1i3i x106iia cmn3 z 0 is a double root77 of this equation 3 a 37 3 27 78 so f is nowhere differentiable a a a a b 2x 4y3875y4 873202 The first equation becomes an 2243 The 2nd equation is 5244 73x2 732y32 712146 so 12246 5244 241212 5 0 and thus y 0 is the only possibility Thus f is differentiable only at the origin c 2x 27 2x 27 2y 2 7 1 717 2y The first equation is satisfied everywhere The second becomes 2y 7152 7 1 717 2y so f is differentiable on the parabola y 7x24 14 d Almost the same as f is differentiable on the parabola y 7334 7 14 Note c and d are more interesting if the real parts of f are changed to 2 7 242 Then for c the Cauchy Riemann equations reduce to 2 7 1 0 so f is differentiable on the points of the two horizontal lines at 1 and x 71 For d you get 2 1 0 so f is nowhere differentiable remember that x is a real number not complex 4 Parametrize O with 215 1 t2 7 2i for 156 01 Note that 215 1 215 1 7 225 So 1 C flt2gtd27 fltzlttgtgtzlttgtdt 1 31 225 517 2ti71 2t 7 817 2t2 7 2i dt 0 1 52tt14ilt68t718dt4016 0 6 Using the top part of the circle 1 7d2 0 OZ 7 a First find the Laurent series fz eZz5 1z51 z z22l 1z5 14lz 151 z6l so Resf 0 141 and f0 7 dz 271141 7ri12 b fz z9 cos1z z91 7 1z22l 1z44l z9 7z63l71z101 so Resf01101 and f0 72 dz 27ri101 c fz Zfij iz has poles at 0 and i Resfi limzniz7ifz 3 sini Resf 0 limzn0zfz 4 7 i 4i since limzno Sinjz 1 Both poles are inside the circle so f0 fz dz 27ri3 sini 717 4i d fz W dz has poles at 0 and the roots of z38 Which are 72 72w3 17x3i z74 Z i 1114 i37r2 7 1114 7 WW2 i7 z4 and 72w 1 It is easy to see that 0 is inside the circle 72 is outside You can also check that 11 lt 2 and 117 gt 2 So 0 and 1x3i are the poles inside the circle Resf0 18 and 7 1 3 1 1 71 Resf713i lim 2 3 1 11m 3 2 4 7 ZH1 i 22 8 z7gt13i 2 8 z3z 31 V303 24 Where the limit is computed using L7Hospital7s rule So f0 fz dz 27ri18 7 124 7ri6 8 2quot 1 1 1 1 2 O 73 sian6CmgdzC7iz ltz271dz 3z214iz73 The poles of the integrand are at i 7141 1 14 2 7 4373 i 77 i 6 3 But 7 gm 7444 so w is the only pole inside the unit circle Using L7Hospital7s rule again 77210i 772x10 2 2 3 gt 2 Res f nn lnn 7 3 ZS 772x 3z2 141z 7 3 ZH772 i 6z 14z 3 3 1 1 377 2 71 2amp1 2

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