CHAPTER THREE Basic Transcendental Functions of Complex Variables

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1 CHAPTER THREE Basc Transcndntal Functons of Compl Varabls 3- Th Eponntal Functon. If +, th compl ponntal s gvn b + (cos + sn ) 上述之定義中, 使用到 Eulr's Formula: θ cosθ + snθ, θ R 在此須先證明是否亦符合 Eulr's Formula, 使得 cos + sn () 所以, 底下我們將先舉出在實變函數中 Eponntal 函數的性質, 而後再證明之 * 在實變微積分中, 若定義 a f ( ) 則其有下列之性質 + ) ) f ( 0) ) f ( ) a a af ( ) Usng th abov mntond proprts, w want to prov that th Equaton () holds for an ral numbr. cos + sn Lt f ( ) g ( ) + h( ) whr both g() and h() ar ral functons and satsf th proprts mntond abov,.., f ( ) f ( ) f ( 0) Thus, whn f ( ) g ( ) + h ( ) f( ) [ g( ) + h( )] g( ) h( ) W know that whn th qualt holds an, thr must hav Ral part Ral part and Im Im So, w obtan th followng quatons: g ( ) h( ) () h ( ) g( ) (3) Dffrntatng quaton () wth rspct to, gvs g ( ) h ( ) g ( ) g ( ) + g ( ) 0 g ( ) c cos + c sn (4) From quaton (), w know that h( ) g ( ) csn ccos (5) And thn w us th proprt f ( 0) : f ( 0) g (0) + h(0) Ths mpls that 37

2 g (0) h (0) 0 c c 0 g ( ) cos h( ) sn f ( ) g( ) + h( ) cos + sn Som basc proprts of th ponntal functon ar dscussd as blow.. If, + and + ar two compl numbrs, thn + ( + ) ( + ) <pf.> [ cos + sn ] [ cos + sn ] + [(cos cos sn sn ) + (sn cos + sn cos ) ] + [ cos( + ) + sn( + ) ] ( + ) ( + ) ( + ) + ( + ) + 3. If f ( ) [cos + sn ] s analtc n th ntr compl plan, thn f ( ) * 為了證明上述結果, 我們必須先了解下列的性質 If f ( ) u(, ) + v(, ), 且其 u v v u, 而且 u v v u,,, 均為連續 f () s analtc <pf.> Snc f ( ) [cos + sn ] u(, ) cos v(, ) sn It s asl sn that u cos v cos u v () And v u sn, sn v u () u u v v Furthrmor,,, and ar contnuous vrwhr n th -plan, and satsf th Cauch-Rmann condtons for all fnt valus of vrwhr n ths plan. f () s analtc for all and s thrfor an ntr functon. 38

3 Th drvatv f ( ) s asl found as u v f ( ) + cos + sn [cos + sn ] Chan Rul: If g( ) s also an analtc functon, w hav d ( g( ) ) ( g( ) ) g ( ). d 4. 和之比較如下表所示 : ) > 0 沒有大小 ) 0 0 ( 0 ) 3) f ( ) s on-to-on functon *** f ( ) s a prodc functon. Its prod s *** [cos + sn ] [cos( + n ) + sn( + n )] ( + n ) + + n + n, n I T (magnar prod), 故知為週期函數 MATLAB Commands for plottng % Magntud of p() [-:0.05:]; [-6:0.05:6]; [X,Y]mshgrd(,); ZX+*Y; wp(z); wmabs(w); msh(x,y,wm);hold on ttl('magntud of p()'), R( ), and Im ( ) : % Ral part of p() [-:0.05:]; [-6:0.05:6]; [X,Y]mshgrd(,); ZX+*Y; wp(z); wmral(w); msh(x,y,wm);hold on surf(x,y,wm) ttl('ral Part of p()') % Imagnar part of p() [-:0.05:]; [-6:0.05:6]; [X,Y]mshgrd(,); ZX+*Y; 39

4 wp(z); wmmag(w); msh(x,y,wm);hold on ttl('imagnar Part of p()') 40

5 5. For +, w hav ) 0 ) and (Th magntud of s dtrmnd b th ral part of.) 3) If k, k constant. 4) If k whr k s an ntgr. 0 <pf.> ) From, snc th product s nvr ro, nthr factor can b ro. Thrfor, 0 C. ) cos + sn Snc [cos + sn ], thn cos + sn 3) Lt [cos + sn ] thus cos sn 0 Snc 0 R, w s that sn 0. Hnc n, n I Bcaus > 0 cos ( 不合 ) k, k I Thus, w sk that n k. If w tak 0 因此, n k 此為必要條件現在, 尚須證明其充分條件亦成立, 故 Suppos that k, whr k I. Thn, w obtan that k cosk + sn k 4) Snc Hnc from th proprt 3), w hav k, whr k I. Th thorm s thus stablshd. Snc 0 C, and W hav that 6. Lt us show that <pf.> + (cos sn ) 4

6 (cos + sn ) (cos sn ) Thus, w s that Som usful dntts: ),,, ) / 3 / / n θ + + or ( ) n 3) DMovr s Thorm: (cosθ sn θ) cos nθ sn nθ nθ. 4) Compl functons of a ral varabl: f () t u() t + v() t, whr ut () and vt () ar ral. f () t u () t + v () t Eampl Fnd 7 t d t 7 dt cos( ). <Sol.>. Drct dffrntaton laborous! t cos( ) R ( + ) t t.. Not that ( ) Lt Thus, ( + t ) f() t. Thn, w hav (7) 7 7 ( + t ) 7 ( + t ) f () t ( + ) ( 8)( + ) 7 t d cos( t) 7 ( + t ) 0 R ( 8 )( + ) R 7 dt snt+ cost + snt cost 0 t (snt+ cos t) H.W. (a) Suppos w want th n-th drvatv, wth rspct to t, of f() t t/ ( t ) t { ( ) ( ) } +. Notc that f() t R and that th n-th drvatv of th functon n th brackts s actl takn. Usng t th mthod of Eampl, as wll as th bnomal thorm (whch prhaps should b rvwd), show that ( n+ )/ k n+ k ( n) ( ) n!( n+ )! ( ) t f () t n+, for n odd, ( t + ) k 0 ( k)!( n+ k)! n / k n+ k ( n) n!( n+ )! ( ) t f () t n+, for n vn. ( t + ) k 0 ( k)!( n+ k)! f() t / t +. (b) Usng th mthod of part (a), fnd smlar prssons for th nth drvatv of ( ) Not that ths functon s dntcal to Im ( /( t ) ). 本題摘自 :A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3., Problm 4, Parson Educaton, Inc., 005. H.W. Th absolut magntud of th prsson ψ ψ ( N ) ψ N nψ n 0 P s of ntrst n man problms nvolvng radaton from N dntcal phscal lmnts (.g., antnnas, loudspakrs). Hr ψ s a ral quantt that dpnds on th sparaton of th lmnts and th poston of an obsrvr of th radaton. P can tll us th strngth of th radaton obsrvd. (a) Usng th formula for th sum of a fnt gomtrc srs (s H.W. 8 n pag 6, lctur not of ch_0), show that 4

7 sn Nψ / P( ψ ) sn ψ / (b) Fnd lm ψ 0 P( ψ ). (c) Us a calculator or a smpl computr program to plot P( ψ ) for 0 ψ whn N 3. 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3., Problm 5, Parson Educaton, Inc., 005. <Ans.> (a) Lt ψ N + N N, thn P Nψ Nψ Nψ / Nψ / Nψ / P( ψ ) ψ ψ ψ / ψ / ψ / Nψ / ( cos( Nψ / ) + sn( Nψ / ) ) ( cos( Nψ / ) sn( Nψ / ) ) ψ / ( cos( ψ / ) + sn( ψ / ) ) ( cos( ψ / ) + sn( ψ / ) ) Nψ / sn( Nψ /) ψ / sn( ψ /) Thus, w hav Nψ / sn( Nψ / ) sn( Nψ / ) P( ψ ) ψ / sn( ψ / ) sn( ψ / ) Q.E.D. (b) Consdr sn( Nψ / ) N / cos( Nψ / ) lm N ψ 0 sn( ψ /) / cos( ψ /) Thus, w hav lm P( ψ ) N ψ 0 (c) MATLAB commands: % H.W. part (c) N3; ph[0:0.:*p]; sn(n*ph)./sn(ph); Yabs(); plot(ph,y)

8 3- Th Trgonomtrc Functon. Snc cos + sn cos sn thn w s that sn + cos 屬於實數 and snh + cosh 屬於實數 ) Gvn an compl numbr, w dfn sn + cos ) Th functons sn and cos ar analtc for all valus of. Morovr, d + sn d + cos d cos d ( ) sn. 底下是一些三角函數和 Hprbolc Functon 的性質 : ) cos + sn, cosh snh ) d sn + c, d snh + c + ω 3) L{sn ωt}, s + ω ω L{snh ωt} s ω s 4) L{cos ωt}, s + ω s L{cosh ωt} s ω 3. If +, thn ) sn sn cosh + cos snh ) cos cos cosh sn snh 44

9 <pf.> ) sn ( + ) ( + ) + [cos + sn ] [cos sn ] cos ( ) + sn ( + ) + cos + sn sn cosh + cos snh ) 同理可證在此另舉一證法 : 由 Talor's Srs, 知 3 5 n n+ ( ) sn + + 3! 5! n 0 (n + )! 3 5 n+ snh ! 5! n 0 (n + )! 比較上列二式可得 snh sn sn snh 同樣地, 由下二式 4 n n ( ) cos + +! 4! n 0 ( n )! 4 n cosh + + +! 4! n 0 ( n)! 比較可得 cosh cos cos cosh 由上述結果我們便可證明 ) 之結果 : Snc cos cos( + ) cos cos snsn cos cosh sn snh 4. sn, 在實變函數中成立 ; 但在複變函數中, sn, 不一定成立 Snc sn sn cosh + cos snh, thus sn sn cosh + cos snh sn ( + snh ) + ( sn sn + snh If w tak,,thn obtan sn, snh > 0 sn > sn, 不一定成立 5. If sn 0 n, n I. )snh 45

10 From sn 0 ( + ) [cos + sn ] sn 0 Snc > 0,whn sn 0, w know cos ± But n ths cas w must tak cos + ( 如此 > 0 才能配合題意之要求 ). n n 而因 cos, 故取故知 0 + n n, n I 6. If cos 0 n+, n I. + From cos 0, w s that. [cos + sn ] Thn, w hav sn + 0. Snc > 0,whn sn 0, w know that cos ±. But n ths cas w must tak cos (such that th satsf > 0, cos ). (n + ) n+ 又因 cos, 故取 0 0 故 + n+ + 0 n + 7. Gvn th compl numbr, w dfn ) sn tan, n+ cos ) cos cot, sn n 3) sc, + n cos 4) csc, sn n whr n I n all abov cass. 5) tan and cot both hav a fundamntal prod of. 6) sc and csc both hav a fundamntal prod of. 8. tan, cot, sc and csc ar analtc functons of cpt for th abov mntond

11 lmtng valus. W can dfn that d ) (tan ) sc, + n d d ) (cot ) csc, n d d 3) (sc ) sc tan, + n d d 4) (csc ) csc cot, n d whr n all cass, n I. 9. If +, thn ) sn sn, cos cos <pf.> sn sn( + ) sn( ) sn cos cos sn sn cosh cos snh sn sn sn sn cosh + cos snh sn sn cosh cos snh 同理 cos cos ) sn sn + snh 3) cos cos + snh 4) sn + cos 5) sn( ± ) sn cos ± cos sn 6) cos( ± ) cos cos sn sn 7) sn cos 8) sn sn cos 9) cos cos sn 以上各式之證明均非常簡單, 留待給讀者證明 0. Lt us show that tan + tan tan( + ) tan tan whr + n+, n I. sn( + ) <pf.> tan( + ) cos( + ) tan + tan tan tan sn cos cos cos + cos sn MATLAB Commands for plottng cos( ) : % Magntud of cos() [-6:0.05:6]; [-:0.05:]; [X,Y]mshgrd(,); ZX+*Y; wcos(z); sn sn 47

12 wmabs(w); msh(x,y,wm);hold on ttl('magntud of cos()') H.W. Lt f ( ) sn(/ ). (a) Eprss ths functon n th form u (, ) + v (, ). Whr n th compl plan s ths functon analtc? (b) What s th drvatv of f ( )? Whr n th compl plan f ( ) s analtc? 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3., Problm, Parson Educaton, Inc., 005. <Ans.> (a) Snc f ( ) sn(/ ), whr / s analtc cpt for 0. But, sn s analtc for all. Thrfor, w hav sn(/ ) s analtc for all 0. Snc +, w hav (b) whr sn(/ ) sn sn sn cosh cos snh u (, ) sn cosh and + + d sn(/ ) cos(/ ) d analtc for all 0. v (, ) cos snh + + sn( ) + snh( ) H.W. Show that tan cos( ) + cosh( ) 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3., Problm 8, Parson Educaton, Inc., 005. <Ans.> Snc sn sn cosh + cos snh tan cos cos cosh sn snh (sn cosh + cos snh )(cos cosh sn snh ) N cos cosh + sn snh D Consdr th dnomnator D: 48

13 ( ) ( ) cos cosh + sn snh cosh sn + sn cosh cosh sn ( + cosh ) [ cos( )] [ cosh( ) + cos( ) ] D Now, consdr th numrator N: Th ral part of N s cosh sn cos snh sn cos sn cos ( cosh snh ) sn cos sn( ) Thn, consdr th magnar part of numrator: cos cosh snh + sn cosh snh snh cosh ( cos + sn ) snh cosh snh( ) Thus, w hav N sn( ) + snh( ) D ( cosh + cos) Ths mpls that sn( ) + snh( ) tan Q.E.D. cos( ) + cosh( ) H.W. 3 (a) Snc sn sn cosh + cos snh and snh cosh, show that snh sn cosh. (b) Drv a comparabl doubl nqualt for cos. 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3., Problm 30, Parson Educaton, Inc., 005. <Ans.> (a) Snc sn sn cosh + cos snh, w hav sn sn cosh + cos snh sn cosh + cos snh sn cosh + cos cosh snh cosh ( ) cosh sn + cos cosh cosh cosh (snc cosh s postv!) sn cosh () On th othr hand, w hav sn sn cosh + cos snh sn snh + cos snh snh cosh snh sn + cos ( ) snh snh snh sn () From Eqs.() and (), w conclud that snh sn cosh Q.E.D. (b) Snc cos cos cosh sn snh, w hav 49

14 cos cos cosh sn snh cos cosh + sn snh cos cosh + sn cosh snh cosh cosh sn + cos ( ) cosh cosh cosh (snc cosh s postv!) cos cosh (3) Smlarl, w hav cos cos cosh + sn snh sn snh + cos snh snh cosh snh sn + cos ( ) snh snh snh cos (4) From Eqs.(3) and (4), w conclud that snh cos cosh Q.E.D. H.W. 4 Fnd all roots of th quaton sn cosh 4. <Ans.> ( ) + n ± 4, n 0, ±, ±, 本題摘自 :Jams Ward Brown and Rul V. Churchll, Compl Varabls and Applcatons, 6 rd d., Ercs 4, Problm 6, McGraw-Hll, Inc., 005. H.W. 5 Fnd all roots of th quaton cos. n + cosh or n ± ln + 3, n 0, ±, ±, <Ans.> ( ) 本題摘自 :Jams Ward Brown and Rul V. Churchll, Compl Varabls and Applcatons, 6 rd d., Ercs 4, Problm 7, McGraw-Hll, Inc.,

15 3-3 Th Hprbolc Functons. Rcall that whn w stud n th ral varabl, th hprbolc functons hav bn dfnd as followng: and snh, ts graph s as shown n Fg. (a). + cosh, ts graph s as shown n Fg. (b) snh cosh Fg. (a) -60 Fg. (b). Gvn an compl numbr, w dfn snh, sn + + cosh, cos ) 若取為實數, 則上述定義與實變函數中的雙曲線函數之定義可相符合 ) snh and cosh both hav a fundamntal prod of. 3) snh( ) snh cosh( ) cosh 4) cosh snh 5) d d snh d d ( ) cosh d d + cosh ( ) d d snh 6) If snh 0 n, n I. <pf.> Snc snh 0 0 Thn, w hav + Ths mans that 0 3 5

16 (cos + sn ) 利用實部實部, 虛部虛部, 我們可得知 sn and cos Snc > 0 cos ( 不合 ) From, w can obtan that n n Agan, w nd 0 So, f snh 0, w conclud that n n 7) If cosh 0 n+, n I <pf.> + Snc cosh 0, [cos + sn ] sn 0 Snc > 0 cos (n + ) 故 n+, n I. Agan w nd 0 So, f snh 0, w conclud that + n+ * 由 (6) 及 (7) 知, 只有純虛數才能使 snh 和 cosh 等於 Gvn th compl numbr, w dfn th othr hprbolc functons as followngs: ) snh tanh, n+. cosh ) cosh coth, snh n. 3) sc h, n+. cosh 4) csc h, snh n. whr n I n all cass. 5) Both tanh and coth hav a fundamntal prod of. 6) Both sc h and csc h hav a fundamntal prod of. 4. tanh, coth, sc h and csc h ar analtc functons of, 其中值不可為上述重點 3 所限制的各不允許值, 則 d ) (tanh ) sch, n+ d d ) (coth ) csch, n d 5

17 3) d (sch ) sch tanh, n+ d 4) d (csch ) csch coth, d n whr n I n all cass. 5. If +, thn ) snh cos snh + sn cosh ) cosh cos cosh + sn snh <pf.> ) snh ( + ) ( + ) [ (cos sn ) + (cos sn )] + cos + sn cos snh + sn cosh ) cosh 同理可證, 亦可仿 3-3. 之 ) 證之 6. If +, thn a) snh ( ) sn, sn ( ) snh b) cosh ( ) cos, cos ( ) cosh c) snh snh, cosh cosh d) snh sn + snh cosh cos + snh H.W. (a) Whr on th ln s th quaton sn + snh 0 satsfd? (b) Usng MATLAB, obtan a thr-dmnsonal plot of sn + snh and vrf that th surfac obtand has ro hght at ponts n Part (a). Includ 0 and at last on othr soluton, on th ln, of th gvn quaton. 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3.3, Problm 0, Parson Educaton, Inc., 005. <Ans.> (a) Snc sn + snh 0, w hav sn cosh + cos snh + [snh cos + cosh sn ] (A) Usng R R and Im Im, gvs sn cosh cosh sn 0 Put n th prcdng quaton, lds sn cosh cosh sn satsfd! Equatng th magnar part n Eq.(A), gvs cos snh + snh cos 0 Put n th prcdng quaton, lds snh cos 0 0 or (b) MATLAB commands: [0:0.05:]; [0:0.05:] ; [X,Y]mshgrd(,); ZX+*Y; n ± +, 0,,, 3, n. Also, n ths cas. 53

18 wsn(z)+*snh(z); wmabs(w); msh(x,y,wm);vw(50,70) H.W. Fnd all roots of th quaton (a) cosh, (b) snh, and (c) cosh. 本題摘自:Jams Ward Brown and Rul V. Churchll, Compl Varabls and Applcatons, 6 rd d., Ercs 5, Problm 4, Parson Educaton, Inc., 005. <Ans.> (a) ( 3 ) (b) ( ) n±, n 0, ±, ±, n+, n 0, ±, ±, (c) ( ) ± ln (n+ ), n 0, ±, ±, 54

19 3-4 Th Logarthmc Functon. For an compl numbr 0, thr sts compl numbrs w such that w 0. In partcular, on of such w s s th compl numbr as shown blow: w ln + θ θ <pf.> Lt r, whr r. Hnc, w hav w ln + θ ln r+ θ ln r θ θ r Thus, w can dfn that ln ln + θ ln + arg ) Dfn Ln ln + θ, < θ. Ln s th prncpal valu of ln so that th Ln s a sngl-valud functon and ln s a multpl-valud functon. Hnc, ln Ln + n, n I. ln + θ + n, ) If < 0, 0 arg s not contnuous. Snc lm arg + 0 lm arg 0 lm arg dos not st. 0 arg s not contnuous for < 0, 0. f () s not analtc for < 0, 0. 3) Th functon Ln s analtc n th doman D consstng of all pont of compl plan cpt thos lng on th ngatv ral as. f ( ) Ln ln + θ f ( ) <pf.> W can us two was to show th abov thorm. ) Rctangular coordnat mthod: Suppos f ( ) Ln ln + θ, and lt f ( ) u(, ) + v(, ) u ln ln + v θ tan Not that f () s analtc n th doman D consstng of all pont of compl plan cpt for thos lng on th ngatv ral as, thn u v f ( )

20 ) # Polar coordnat mthod: Lt w f () and + r(cosθ + sn θ ), thn r cosθ r snθ In Chaptr, w hav drvd th formula as shown blow: d w w (cosθ snθ ) d r Not that w f ( ) Ln ln + θ ln r + θ d w w (cosθ snθ ) d r (cosθ sn θ) r r (cosθ + sn θ ) Q.E.D.. In th ral varabl analss, w know that f 0 and 0, thn ln ln + ln Howvr, n th compl varabl analss, th followng Ln Ln + Ln ma b not tru. For ampl: Lt, + thn Frst, w hav Ln Ln ln + and Ln Ln( + ) 3 ln ln + 4 But Ln( ) Ln ( ) 3 ln + ( ) 4 3 ln 4 3 Hnc, Ln + Ln + ln ln + 4 > > 56

21 Ln( ) # But, th followng thorms ar stll satsfd. 3. If th compl numbrs, 3 ar dffrnt from ro, thn th prncpal valus of th argumnts and logarthms of th product, quotnt, and powrs among ths compl numbrs ar gvn b ) Arg ( ) arg + arg + n (, ) -( a) Log ( ) Ln + Ln + n (, ) -( b) whr n s assumd to b th valus of, 0, as followng:, f < arg + arg n (, ) 0, f < arg + arg, f < arg + arg <pf.> Lt θ and θ, whr < θ, < θ Ths gvs < θ + θ Not that ( θ+ θ) ( θ+ θ+ n) whr n s an ntgr such that < θ + θ + n () ) Suppos < θ + θ, thn < θ + θ 0 Thus, < θ + θ Comparng ths nqualt wth quaton (), w obtan n (, ) ) Suppos < θ + θ, comparng ths nqualt wth Eq. (), gvs n (, ) 0 ) Suppos < θ + θ, w hav 0 θ + θ + < So, < θ + θ + Comparng ths nqualt wth Eq. (), w hav n (, ) # 已證得 (a) Lt, thn Ln Ln ln ln[ + arg( ] + [arg ) + arg + n ( + ln + arg + arg + n ( ) + arg ] + [ln + arg ] + n ( + Ln + n ( ) ln [ln ) Ln ) 基本自然對數函數尚具有底下之各項性質 : arg arg arg + n(, ) Ln Ln Ln + n(, ) ) 57

22 whr n s assumd to b th valus of, 0, as followngs: n (,, ) 0,, 3) arg ( ) n arg + k (, n) n f f f < arg < arg < arg arg arg arg Ln ( ) n Ln + k (, n) whr n s an ntgr, and k s th ntgr gvn b th brackt functon: n k(, n) arg( ) n 4. If 0 and w s an compl numbr, w dfn w w ln wln w whr ln s a multpl-valud functon. That s, wln w[ln + arg + n ] whr n 0, ±, ±,. Ths mans that valus. Hr, w dfn th prncpal valu of w ln w as followng: w wln s calld sngl-valud functon. ) If 0, 0, and w s an compl numbr, thn w w w wn (, ) ( ) () w w wn (, ) () w whr n (, ) and (, ) <pf.> has nfnt numbr of computd n ar th ntgrs dfnd n prcdng dscussons. Us th dfntons and thorms that w hav dscussd: w w Ln( ) ( ) w[ln + Ln + n ( )] wln wln wn(, ) w w wn (, ) Eampl Fnd th prncpal valu of <Sol.> Ln (ln + ) R. Eampl Fnd th prncpal valu of ( + ). <Sol.> Ln( + ) ( + ) (ln + 3 /4) 3 /4+ ln 3 /4 [cos ln + sn ln ] 3 /4 3 /4 cosln + sn ln 58

23 Eampl 3 Fnd Ln ( + ). <Sol.> Ln ( + ) ln + + arg( + ) ln + 4 Eampl 4 Fnd th prncpal valu of <Sol.> 3 ( 3). 3 ( 3) 3 3 ( ) ln + arg [cos3 + sn 3 ] 3 ln+ 3 ( ) If 0, w and λ ar an compl numbr, thn Ln ( w ) wln + k () <pf.> Dnot thn w λ wλ λ k ( ) () whr k s th ntgr gvn b brackt functon Im( w)ln + R( w) Arg k w b α, α w wln u ln vθ arg b θ, and lt w u + v, [ u+ v][ln + θ ] [ vln + uθ ] Thus, u ln vθ α arg v ln + uθ + k (3) whr k s th ntgr such that < v ln + uθ + k (4) Solvng for k n Eq. (4), w obtan t < k t + whr v Ln + uθ t Thrfor, 59

24 v Ln + uθ k [ t + ] Im( w) Ln + R( w)arg Now, usng Eq. (3), w s that Ln ( w ) Lnα ln α + argα Ln[ u ln vθ ( u + v)(ln + θ ) + k wln + k 故 () 式已被證得同理, 可證得 () 式 ] + [ v ln + uθ + k ] u ln vθ + [ v ln + uθ + k ] n H.W. Consdr th dntt ln nln, whr n s an ntgr, whch s vald for approprat choc of th logarthms on ach sd of th quaton. Lt + and n 5. (a) Fnd th valus of ln n n and ln that satsf ln nln. n (b) For th gvn and n, s ln nln satsfd? n (c) Suppos n and s unchangd. Is ln nln thn satsfd? 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3.4, Problm 6, Parson Educaton, Inc., 005. <Ans.> 5 (a) 5ln ln 5 Ln + 4 (b) No. (c) Ys. 6. Dfnton (Branch): A branch of a multvalud functon s a sngl-valud functon analtc n som doman. At vr pont of th doman, th sngl-valud functon must assum actl on of th varous possbl valus that th multvalud functon can assum. Dfnton (Branch cut) A ln usd to crat a doman of analtct s calld a branch ln or branch cut. Dfnton (Branch pont) An pont that must l on a branch cut no mattr what branch usd s calld a branch pont of a multvalud functon. E. () Ln prncpal valu of ln, whch s dfnd for all cpt 0. () Ln s also usd to dnot th prncpal branch of th logarthm functon, whch s dfnd for all cpt 0 and valus of on th ngatv ral as. (3) f ( ) Lnr+ θ, whr 3 / < θ /. It s dscontnuous at th orgn and at all ponts on th postv magnar as. (4) f ( ) Lnr+ θ, whr 3 /+ k < θ /+ k, k 0, ±, ±,, ar, for ach k, analtc branchs, provdd s confnd to th doman D. Eampl 5 (a) Fnd th largst doman of analtct of ( ) Ln[ (3 4 )] f +. (b) Fnd th numrcal valu of (0) f. <Sol.> (a) For th gvn functon, th non-analtc ponts ar locatd at: 60

25 Doman of analtc of Ln (shadd) j Branch cut j Doman D. Im w 0, R w 0 If w (3+ 4 ), ths two condtons can b rwrttn as [ ] [ ] Im ( + ) (3+ 4 ) 0 4 R ( + ) (3+ 4 ) 0 3 So, th full doman of analtct s shown n th followng Fg. (a). Branch cut 4, 3 j j (appro.) Fg. (a) 4 Fg. (b) (b) f (0) Ln( 3 4 ) Ln(5) + arg( 3 4 ), whr < arg( 3 4 ) <. f(0) Ln(5).4 Fg. (b). H.W. (a) Show that Ln ( Ln( )) s analtc n th doman consstng of th -plan wth a branch cut along th ln 0,. (S blow) (b) Fnd d Ln ( Ln( ) ) dwthn th doman of analtct found n part (a). (c) What branch cut should b usd to crat th mamum doman of analtct for Ln Ln ( Ln( ) )? 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3-5, Problm 5, Parson Educaton, Inc., 005. Branch cut j Fg. HW- H.W. 3 Th compl lctrostatc potntal Φ (, ) φ + ψ Ln(/ ), whr 0, can b cratd b an lctrc ln charg locatd at 0 and lng prpndcular to th -plan. (a) Sktch th stramlns for ths potntal. (b) Sktch th qupotntals for φ, 0,,. 6

26 (c) Fnd th componnts of th lctrc fld at an arbtrar pont (, ). 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3-5, Problm 6, Parson Educaton, Inc., 005. <Ans.> (a) Snc Φ (, ) φ + ψ Ln(/ ) Ln φ(, ) Ln +, ψ (, ) tan ( ) (b) Snc Ln + φ, thrfor + f φ Also, whn 0 Ln + φ 0, thrfor + f φ 0 Whn Ln + φ, thrfor / + f φ Th lctrc fld s dφ E + E d E and E + +, whr ψ s ar ramp manatng from orgn. φ ψ 3/4 ψ /4 φ φ 0 ψ /4 (c) MATLAB commands: % HW (c) [-.5: :.5]; ; [X,Y]mshgrd(,); ZX+*Y; wlog(z--*); %wmral(w); % for ral part wmmag(w); % for mag part Msh(X,Y,wm); % for mag part and ral Vw(45,45); 3 ( ) H.W. 4 Lt f( ) 0. Ths functon s valuatd such that f ( ) s ral whn. Fnd f ( + ). Whr n th compl plan s f ( ) analtc? 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3-6, Problm 4, Parson Educaton, Inc., 005. <Ans.> 6

f ( + ) 0.37 0.048 ; f ( ) s an ntr functon. (a) ral part H.W.

Fnd f ( ) and f ( /). 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d.

27 f ( + ) ; f ( ) s an ntr functon. (a) ral part H.W. 5 Lt ( ) f( ) 0 (b) magnar part. Ths functon s valuatd such that f( /). Fnd f ( ) and f ( /). 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3-6, Problm 4, Parson Educaton, Inc., 005. <Ans.> [Ln(0) ] f + ( ) Ln(0) + and f ( / ) [ ] 63

28 3-5 Invrs Trgonomtrc and Hprbolc Functons, thn sn ln + ( ) /. ) Lt sn w <pf.> w w Snc, assum that w p and / p / p p p p p 0 Solv ths quaton for p: p + ( ) / w or + ( ) / ( ) / w ln + Snc w sn, w hav ( ) / sn ln + ) Othr nvrs trgonomtrc functons: cos ln + ( ) / tan ln +. Invrs of Hprbolc functon ) snh ln + ( + ) / tanh ln + ) cosh ln + ( ) / 3) 3. Drvatv of nvrs trgonomtrc functons: d d / ) sn ln ( ) d d + ) 3) d cos d d tan d / ( ) ( + ) 4. Drvatv of nvrs hprbolc functons: d ) snh d / + ) d cosh d ( ) ( ) / ( ) p w, thn w hav / 64

29 3) d tanh d ( ) Eampl Fnd <Sol.> sn (/ ). / sn (/ ) ln Takng ( 3/4) / 3/, th abov prsson bcoms 3 sn (/ ) ln + ln ( ( / 6) ) + k, k 0, ±, ±, 6 Takng ( 3/4) / 3/, th abov prsson bcoms 3 5 sn (/ ) ln ln ( (5 / 6) ) + k, k 0, ±, ±, 6 Eampl Fnd all th numbrs whos sn s. <Sol.> Ths quston s to solv sn. Thus, w hav sn () ln + ( 3) / Takng ( 3) / + 3, th abov prsson bcoms ( ) ( ) sn () ln 3 Ln 3 k ( + k ).37, k 0, ±, ±, Takng ( 3) / 3, th abov prsson bcoms ( ) ( ) sn () ln 3 Ln 3 k + + ( + k ) +.37, k 0, ±, ±, Eampl 3 Fnd all th numbrs whos sn s. <Sol.> Ths quston s to solv sn. Thus, w hav sn ( ) ln + ( ) / ln( ± ) Takng ( 3) / + 3, th abov prsson bcoms Ln ( ) k k Ln ( ) ( ) ( ) ( ) + sn ( ) ln ± Ln k + k Ln + whr k 0, ±, ±,. H.W. Show that sn n ( ) n ( ) + ln ( ) + +. 本題摘自 :Jams Ward Brown and Rul V. Churchll, Compl Varabls and Applcatons, 6 rd d., Scton 9, Eampl, McGraw-Hll, Inc., 005. θ H.W. Show that tanh ( ) (/)ln cot ( θ /). 本題摘自:A. Davd Wunsch, Compl Varabl wth Applcatons, 3 rd d., Ercs 3-7, Problm 6, 65

30 Parson Educaton, Inc.,

0 0 = 1 0 = 0 1 = = 1 1 = 0 0 = 1

0 0 = 1 0 = 0 1 = = 1 1 = 0 0 = 1 0 0 = 1 0 = 0 1 = 0 1 1 = 1 1 = 0 0 = 1 : = {0, 1} : 3 (,, ) = + (,, ) = + + (, ) = + (,,, ) = ( + )( + ) + ( + )( + ) + = + = = + + = + = ( + ) + = + ( + ) () = () ( + ) = + + = ( + )( + ) + = = + 0