Complex Powers and Logs (5A) Young Won Lim 10/17/13

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1 Complx Powrs and Logs (5A)

2 Copyright (c) 202, 203 Young W. Lim. Prmission is grantd to copy, distribut and/or modify this documnt undr th trms of th GNU Fr Documntation Licns, Vrsion.2 or any latr vrsion publishd by th Fr Softwar Foundation; with no Invariant Sctions, no Front-Covr Txts, and no Back-Covr Txts. A copy of th licns is includd in th sction ntitld "GNU Fr Documntation Licns". Plas snd corrctions (or suggstions) to youngwlim@hotmail.com. This documnt was producd by using OpnOffic and Octav.

3 Powr and Taylor Sris Powr Sris Taylor Sris n=0 c n (z a) n n=0 f (n) (a) n! (z a) n = c 0 + c (z a) + c 2 ( z a) 2 + always convrgs if z a < R can also b diffrntiatd = f (a) + f '(a)(z a) + f ''(a) 2 ( z a)2 + f (z) = n=0 f (n) (a) n! (z a) n only valid if th sris convrgs Complx Logs (5A) 3

4 Powr and Taylor Sris Powr Sris Taylor Sris n=0 c n (z a) n = c 0 + c (z a) + c 2 ( z a) 2 + always convrgs if can also b diffrntiatd z a < R n=0 f (n) (a) n! (z a) n = f (a) + f '(a)(z a) + f ''(a) 2 ( z a)2 + f (z) = n=0 f (n) (a) n! (z a) n only valid if th sris convrgs z = a + (z a) = a + ( z a a ) f (z) = z f (n) (0) = = n=0 ( ) n a ( z a a )n z = n=0 z n n! Complx Logs (5A) 4

5 Complx Log : ln z Powr Sris Taylor Sris z = n=0 ( ) n a ( z a a )n z = n=0 z n n! convrgs if z a < a for any a 0 convrgs for all z for a sufficintly far away from 0 th siz of th disk can b mad big d d z (ln z) = z w = z w = ln z ln z : intgral of th powr sris ln z : invrs of z xpansion of at z = z Complx Logs (5A) 5

6 Polar Rprsntation z = r i θ y = r sin θ z 2 = r 2 iθ 2 θ x = r cosθ z z 2 = r iθ r 2 iθ 2 z = x + i y = r cosθ + i r sinθ = r (cosθ + isin θ) = r r 2 i(θ + θ 2 ) = r iθ i arg (z) = z z z 2 = r iθ r 2 iθ 2 = r r 2 i(θ θ 2) Complx Logs (5A) 6

7 Principal Argumnt z = x + i y = r cosθ + ir sinθ = r (cosθ + isinθ) = r i θ i arg( z) = z arg( z) : to b a function of z, it nds to b uniquly dfind for vry z, by choosing its rang by spcifying its rang an argumnt can b uniquly dfind uniqu angls Th spcial argumnt Principal Argumnt Arg(z ) uniqu angls θ = arg(z ) 0 arg( z) < 2π π Arg( z) < +π x) 0 arg( z) < 2π fr to choos a mor convnint dfinition for a particular problm. Argumnt many angls θ = arg(z ) = Arg(z ) + 2kπ Complx Logs (5A) 7

8 Complx Log z = w = u + iv = u+i v i arg(u+i v ) w = u + i v = u i v iarg(u i v ) = z i arg(z) = u iarg(i v ) u =, arg( u ) = 0 z = z i arg(z) = u iarg(i v ) z = ln z i(arg(z)+2k π) z = u arg(z) = arg( i v ) u = ln z v = arg(z) + 2kπ 2kπ is ncssary assuming arg is uniquly dfind with its rang spcifid Complx Logs (5A) 8

9 Complx Log : ln z z = z i arg(z) = u iarg(i v ) z = ln z i(arg(z)+2k π) many angls ln z + i(arg(z ) + 2k π) z = ln z = ln z + i(arg(z) + 2kπ) 2kπ is ndd : assuming arg is uniquly dfind with its rang spcifid uniqu argumnt ln z + i arg(z) z = ln z = ln z + iarg(z) z = z i arg(z) ln z = ln z + iarg(z) Complx Logs (5A) 9

10 A mapping of ln z ln 2 = ln(/2) ln(2) +π +3π/4 3π/4 +π/4 π/4 +3π/4 +π/4 π/4 3π/4 π Complx Logs (5A) 0

11 Complx Powrs z = w w = ln z a b = w a b = ln z z = w = ln z a b = w = ln ab = bln a z = ln z iarg(z) z b = b ln z ib arg( z) z b = w = ln z b = bln z z b = b(ln z + iarg( z)) = bln z ib arg(z) a z = z ln a i z arg(a) a z = w = ln az = z ln a a z = z (ln a + i arg(a)) = z ln a i z arg(a) ln z ral b ln z z ln a in gnral not ral Complx Logs (5A)

12 Complx Logs Exampl fr to choos mor convnint dfinition for a problm iπ+i 2k π ( ) = ln( ) = ln + i arg( ).5 2 i(π 2k π) i π (Principal Argumnt) ( ) i = i ln( ) i( i( π 2kπ)) = (+) (+i) ( ) ( i) π 2 kπ = +π = 0 +iπ /2 = = iπ iπ /2 = +π 2 kπ (Principal Argumnt) (+) i (+i) i ( ) i ( i) i = 0 = π/2 = +π = +π/ ln( ) +3 π +π (i) i () i ( i) i π 3 π ( ) i +π Complx Logs (5A) 2

13 Complx Powr Exampls : z i +i(0 2k π) (+) = (+) i i ln (+) = = 0 +2 kπ (+) i = 0 +i(π /2 2 kπ) (+i) = (+i) i = i ln(+i) = π/2 +2 kπ (+i) i = π/2 i(π 2 kπ) ( ) = ( ) i = i ln( ) = +π 2 kπ ( ) i = +π ( i) i i( π/2 2kπ) = ( i) i = i ln( i) = +π/2 2 kπ ( i) i = +π/2 (+2) ln2 + i (0 2 kπ) = (+2) i i(ln2+i(0 2 kπ)) = = 0 +2 kπ i ln2 (+2) i = 0 i ln2 (+2i) ln2 + i (π/2 2kπ) = (+2i) i i(ln2+i( π/2 2kπ)) = = π/2 +2 kπ i ln2 (+2i) i = π/2 i ln2 ( 2) ln2 i (π 2k π) = ( 2) i i(ln2 i( π 2k π)) = = +π 2 kπ i ln2 ( 2) i = +π i ln2 ( 2i) ln2 i (π/2 2kπ) = ( 2i) i i(ln2 i( π/2 2kπ)) = = +π/2 2 kπ i ln2 ( 2i) i = +π/2 i ln2 ln 2 = cos(ln 2) = sin(ln 2) = π = π/2 = 4.80 Complx Logs (5A) 3

14 Complx Roots z n a = 0 z 4 = z = a /n i(arg(a) / n+2kπ / n) z n = r n inθ = a = a iarg(a) z = π 4 + 2k π 4 r n = a nθ = arg(a) + 2kπ k=2 k= n r = a θ = arg(a) n + 2k π n k=3 k=0 Complx Logs (5A) 4

15 Complx Roots: z 4 = - z 4 = π 2kπ 4 = π k=2 k= 2kπ 4 = π 2 k=0 2kπ 4 = 0 π 4 2kπ 4 = 3 π 2 k=3 z n a = 0 z = a /n i(arg(a) / n+2kπ / n) n r = a θ = arg(a) n + 2k π n z = π 4 + 2k π 4 k=2 k=3 k= k=0 Complx Logs (5A) 5

16 Complx Roots : z 4 = 4 r = a θ = ( π + 2k π) z 4 4 = θ i θ i 2 θ i 3 θ i 4 0 arg(z) < 2π θ 0 = 4 π : π θ 0 4 = π 2π arg(z) < 4 π θ = + 4 π : +π θ 4 = π 4π arg(z) < 6π θ 2 = π : +3π θ 2 4 = π 6π arg(z) < 8π θ 3 = π : 5π θ 3 4 = π Complx Logs (5A) 6

17 Complx Powrs: z ½ θ 0 +2 π z dz = θ0 iθ/2 i iθ d θ z = θ 0 +2 π i i3 θ/2 dθ = [ 2 3 i3 θ/2 ]θ0 θ 0 +2 π z = z /2 +i arg( z)/2 θ 0 arg( z) < θ 0 + 2π = θ0 θ 0 < arg( z) θ 0 + 2π = 2 3 (i 3(θ 0+2 π)/2 i3 θ 0/2 ) = 2 3 i3 θ 0/2 ( i3 π ) = 4 3 i3 θ 0/2 th dfinition of z also dpnds on θ 0 th circular intgration dpnds on whr th closd circl bgins θ 0 implicitly assumd dfinition of z dpnds on θ 0 Complx Logs (5A) 7

18 Complx Powrs: z ½ ral x 0 complx z y 2 = x w 2 = z y = + x y = x w = + z w = z Th MATLAB's dfintion z = z /2 +i arg( z)/2 π < arg( z) +π a discontinuity in z whn th ngativ ral axis x 0 R{ z } 0 if w would want this, thn z = z /2 +i arg( z)/2 π < arg( z) < +π z = z /2 +i arg( z)/2 π arg( z) < +π a discontinuity in z whn th positiv ral axis π/2 < arg( z)/2 < +π /2 but w would hav xcludd all th pur imaginary numbrs also. Complx Logs (5A) 8

19 Branch Cut z = z /2 +i arg( z)/2 π < arg( z) +π z = z /2 +i arg( z)/2 0 < arg( z) +2π jump jump branch cut branch cut branch point branch point if w go around a branch point, thn thr is always a discontinuity somwhr Complx Logs (5A) 9

20 Winding Numbr (z ½ ) Considr computing th squar root of points along th diffrnt paths that ar continuous +i π/4 z = 2 z = 2 +i π/4 z = 2 +i π/4 +i π/4 z = 2 r : 2 r : 2 r : 2 r : 2 θ :0 π/4 θ :0 π/4 + 2π θ :0 π/4 + 4π θ :0 π/4 6π two classs of paths z = r +i θ = r +iθ/2 z = + 2 z = r +i(θ+2 π) = r +iθ/2 i π +i π/8 +i π/8 z = 2 vn numbr of rounding odd numbr of rounding Complx Logs (5A) 20

21 Rang of Argumnts (z ½ ) 2 r = 2 θ = 2 ( π 4 + 2kπ) z 2 = 2 + j π /4 θ i θ i 2 vn numbr of rounding 8 π 4 π 0 arg(z) < 2π : θ 0 = + 2 ( 4 π) π 4 θ 0 2 = 4 π 9 4 π 2π arg(z) < 4 π : θ = + 2 ( 9 4 π) 9 π 4 θ 2 = 9 4 π 9 8 π odd numbr of rounding a closd loop in a Rimann surfac = + Complx Logs (5A) 2

22 Domain & Rang Complx Plans (z ½ ) z = r +iθ = r +iθ/2 + r cos θ 2 + i r sin θ 2 z = r +i θ+2 π = r +i θ/2 i π r cos θ 2 i r sin θ 2 R{ z} = + r cos θ 2 R{ z} = r cos θ 2 I{ z} = + r sin θ 2 I{ z} = r sin θ 2 z z /2 +cosθ +cos θ 2 π θ < +π 2 π θ 2 < + 2 π π θ < +π π θ < +π +π θ < +3π + 2 π θ 2 < π +π θ < +3π +π θ < +3π cosθ cos θ 2 Domain Complx Plan Rang Complx Plan Complx Logs (5A) 22

23 Rimann Surfac (z ½ ) - () R{ z} = + r cos θ 2 R{ z} = r cos θ 2 +cos θ 2 +cos θ 2 cos θ 2 cos θ 2 Rimann surfac for th function ƒ(z) = z. Th two horizontal axs rprsnt th ral and imaginary parts of z, whil th vrtical axis rprsnts th ral part of z. For th imaginary part of z, rotat th plot 80 around th vrtical axis. [wikipdia.org] For visualizing a multivalud function. a gomtric construction that prmits surfacs to b th domain or rang of a multivalud function. Complx Logs (5A) 23

24 Rimann Surfac (z ½ ) - (2) y R( z) x +cos θ 2 I{ z} π θ < +π π θ < +π R{ z} cos θ 2 Domain R. surfac Rang Complx Logs (5A) 24

25 Rimann Surfac (z ½ ) - (3) R{ z} = +cos θ 2 y R( z) x z z /2 c π θ < +π a +π θ < +3π b a b c Domain Rang R{ z} = cos θ 2 Complx Logs (5A) 25

26 Rimann Surfac (z ½ ) - (4) This Rimann surfac maks full squar root function continuous for all z 0. btwn th point a and th point b, th domain switchs from th uppr half-plan to th lowr half-plan. no othr way to gt from th point a to th point b xcpt going countrclockwis. y R( z) x c π θ < +π a b a a b c +π θ < +3π c b Domain Rang Complx Logs (5A) 26

27 Rimann Surfac A on-dimnsional complx manifold. can b thought of as "dformd vrsions" of th complx plan: locally nar vry point thy look lik patchs of th complx plan, but th global topology can b quit diffrnt. For xampl, thy can look lik a sphr or a torus or a coupl of shts glud togthr. Holomorphic functions (conformal maps, rgular function) Rimann surfacs ar nowadays considrd th natural stting for studying th global bhavior of ths functions, spcially multivalud functions such as th squar root and othr algbraic functions, or th logarithm. Th phras "holomorphic at a point z0" mans not just diffrntiabl at z0, but diffrntiabl vrywhr within som nighborhood of z0. Th xistnc of a complx drivativ in a nighborhood is a vry strong condition, for it implis that any holomorphic function is actually infinitly diffrntiabl and qual to its own Taylor sris.. Complx Logs (5A) 27

28 Rfrncs [] [2] [3] M.L. Boas, Mathmatical Mthods in th Physical Scincs [4] E. Kryszig, Advancd Enginring Mathmatics [5] D. G. Zill, W. S. Wright, Advancd Enginring Mathmatics [6] T. J. Cavicchi, Digital Signal Procssing [7] F. Walff, Math 32 Nots, UW 202/2/ [8] J. Naring, Univrsity of Miami [9]

SECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero.

SECTION where P (cos θ, sin θ) and Q(cos θ, sin θ) are polynomials in cos θ and sin θ, provided Q is never equal to zero. SETION 6. 57 6. Evaluation of Dfinit Intgrals Exampl 6.6 W hav usd dfinit intgrals to valuat contour intgrals. It may com as a surpris to larn that contour intgrals and rsidus can b usd to valuat crtain