midtermSol non ponce

# midtermSol non ponce - Name gﬂlm'10[A Directions Piease...

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Unformatted text preview: Name: gﬂlm ['10 [A # Directions: Piease be sure to answer each question carefully in the space previded. Piease present your work neatly so yea can receive maximum credit If you have any questions please feel free to ask Midterm Exam: November 8th, 2007 Professor: Peal J. Atzberger Scoring: Probiernl: ProblemZ: Problem3: Problemd: Final Score: Problem 1: a) “711.31; are the real and imaginary parts of Jmtlnﬁi‘: l l“? t3~+la— :1 Rumio Iw‘mzla 1)) Compute the modulus |zl when z = (Hint: Use properties of the modulus to compute the solution eﬂiciently.) I ‘4, I ( l‘l’l I ‘i‘ Lf 11L 1 (a l(\ P} "‘2 l 1-1 :Ull‘ﬂ‘") its—“f ’"3‘0 ll"3ﬂl} 0) Sketch the region in the complex plane corresponding to the fol», lowing set being sure to Show where it intersects the x~axis or y—axis: {z=\$+iy]|z—-1—i|2 \$2}. (X*H}+IY4T*EQ~Jdisk wahed M (m) an; X=03 Il+wwn*sl :3 CV-I)“&} :7 “LEV—14] ‘3? 0,4352}, v10: (an)}+13‘é3~ up (have | :7 ~léx-lil 2:) oaks-M skews lm‘tcmsecllew WHM 1”” “cm” {or l°.)0)1(0,3~);(3~10)- d) Compute the polar form 2 = “rem when the complex number 2 = -~2 ~— 21" Give r' encl all the possible values which can be used £016 r)“ :_ [2,)“: 21:7: ‘2. 333933” :: ‘3 ~57 r: L f; W T I QBZW+JqEEiL is one POSSJbli ﬁll/19,3: All angles awe 0* HA2 {Wt/‘4 eueo+mrl<..:§-\FH~TFK, l4-=025'11%M~o Problem 2: a) State sufﬁcient conditions for a complex—mined function of a com— plex variabie f(z) to be complex differentiable“ wt 20 e: (3.) [nus lCom‘Hl/unaws ‘WCI‘VS'II Pﬁw‘lftrak om ox (Item/Layla (OmHHVHMQ ch (H) Ha): U\(X_,l{)+} \r-(XWU/ 22%!” (M5 “HF Winkbe saiisﬂ “Hag, (audth Kremth €0[V\U\.+£\0V\_§, “dc Zn, b) Show that f (z) = :2 + :5 satisﬁes the Cauchymﬁiemann equations. 2::- X-Jr‘Hf Hz) = (x+1\pa‘+(x+§\;) :(X3~__\13~+X Hum HI) Gleam”) 4: [nus C0 [AJTI mwm_s- yaw ‘Hmfs (“4 +149, whiff; (MMWLX (mg/ML, 5} {E head OLE/\(‘L IWR9;HQ\,~-g (Davis!!! mfg Folrhohmfj ﬂak haircut shows 15(5) 3.9 (oi/Mfth d}Hew,\/d:lmb\e‘ c) Show that f (z) m In r—H‘H satisﬁes the Cauchy—Riemann equations on the region {2! — 7r < 9 < 7r,r > 0} when 2 = T810 with r = Izl and 0 = Arg(z) l 43(3) “5 {A‘va‘w-‘(r/M g“; m”? :” Vrg (1) Show that; f(z) = 111T ~43— i6 is discontinuous at z m —1 by comput— ing the limits limysa‘fb—l + ﬁg) and the limits limysa f'(-—1 — iy)‘ \ 11m HMHIY) : Hm UMWQMAW ("1749] V1401" \fw? 0+ ‘ l 9 ‘ I 2 ".W‘ “MVW’H‘I‘W‘JA’VSLIMQ V“70+ , ‘1“70 ‘ (Pruviafed +|ne lwm‘is vast) = {Ht/1(1) +°§ V W W BY (ome“IV\m{1‘\/ 9V PWPWNYS 0&- wx). 65¢ Avg 2: 3 ’IY HIM H441) 2: Jill/x“) + NH?) __:Mn.\4'( Mush hu‘i: ann‘ J SE) “5 deLUW‘ylWUUmﬁ rift Ztmlﬁ Problem 3: a) Show that f(z) = z + i + 1 is a continuous function. De“ 'A wﬂmmcﬂan 39 {whﬂv‘wiﬂ “Kit 20 H: hm Ha): H20), '2"? 2 o Wham (1(2)::Z‘FE‘H We Kmart Show Cw Hm 43ft) :2 Hag) :ZU+}+I. 2""720 mam-(m 1: [Ha—+1 “(to—mo): 1mg} Whemvw Izaaké :7 1H2) -41130Hcﬁ 50 M: W9» ((40056. 3‘56 £19m U\V\) 670 {Me was” MW, H142) “HENLC‘ WMQHW Iz‘zokgv This [How's b) Show that 11111340? 2 Do. I Iégzdﬁ’z) 3‘- 3,20), “W” mvk (3131/14 2% :0 7H7” 1 > 2:70 (a) Hmn "I; :1va L10 ‘a-70 ( ‘2’) r70 '2‘“ ‘ WWW—E» {Quotas Sham \wa Mm! 620/ Izwgkﬁ 3/...“ l2+\\§12] LgL [e ‘COW 6:6! \mﬁ [ik—d) ' (3+312~:2 : 2 c) Show that 131mMoo \ 3F 3 . \ '" “L J“ , 3, 3,. 111m (4244) m Hm M“ 2~7o 2‘70 W ° 6.) Show that 11mg” = 00 I ‘ 2% w M W M 2:90 ‘2:\-\ '— A) 4”"? ,2,“er k > v" 0 I 1%,)“ \ WW . J. 4);“ in)" a: 111m (11*! 2 ,Wz}.(*§:+|) Hi)“ 7’70 (2)} 2% 2% % w)?” Problem 4: a) Show that f(z) = 22 + z is complex differentiable in Hz): 2M2 :(X’LWX High/+7) inng {omisimuuws PivS‘i; meii‘misj Sime mgr 0W9. H momimiS. (H) Hz) =m+i\r, has mu» sm‘Hxﬁihq +109 Cmmﬁingvﬂi‘owwn eqwxﬂ‘ms by Part 35b6, Tia-9.96 JFWQ FOMdiHOM (We, Swarl‘if’i‘imt 40w Ht) +0 M {ow‘oiEX Oiiiiewmtt‘mioig, b) Solve the following steady—state heat equation using the complex differentiable function ﬁzz) = 2:2 + z and pmperties of Harmonic functions: um+uyy = O,fOI‘0\$x\$ 1,03y51- u(:c,y) = y, for .1: = O“ u(:1:,y) = O,for y m E} u(:c,y) = 3y,for m = 1. u(:c,y) = 2m +1,f01 y m1” iii/it eii’izv/ 61:; IMH’IZ are HorwomiL Sit/HQ {4(3) {‘5 (@WV/U’ afr‘PiPrChi-r‘ab/e iii (XS: X20, TX==~ 3" (does mi‘ wink, bud. {mlhtfamg ) U“ “M I (wa it v?) (v Uwomxtlﬂumw 4, i:- :0} z: 7“ - D—XM ) . (duebSMS‘a 63:2,?"1’0, T'i/xewimw wring mini?“ ‘i 3 mzﬁ’lmuia HAM—Va 0) Sketch the region obtained by mapping the rectangle {z m :1: + iyiO S :13 S 1n(2),0 g y S n/2,} under f(z) = e”. d) Sketch the Iegion obtained by mapping the semi~circle z = r‘e‘GIO 5 7” 51,0 5 t9 3 n/Z} uncier f(z) = 22‘ ...
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• Spring '08
• Ponce
• Complex number, Peal J. Atzberger, nus lCom‘Hl/unaws ‘WCI‘VS'II, differentiable function ﬁzz, State sufﬁcient conditions, X3~__\13~+X Hum HI

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midtermSol non ponce - Name gﬂlm'10[A Directions Piease...

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