Λυμένες ασκήσεις Μιγαδικές- Σάμαρης 2012

0.1. SQEDON LUMENES ASKHSEIS EPANALHYHS 1

0.1 Sqedìn lumènec ask seic epanlhyhc

1. ApodeÐxete ìti |z| ≤ |z|2 + |z − 1| z ∈ C gia kje z ∈ C.

LÔsh. Gia |z| ≤ 1 èqoume |z|2 + |z − 1| ≥ |z|2 + |z| − 1 ≥ |z|.Gia |z| > 1 èqoume |z| < |z|2 ≤ |z|2 + |z − 1|.(ExhgeÐste giatÐ èqoume isìthta mìno sthn perÐptwsh z = 1.)

2. An ρ rÐza tou poluwnÔmou zn + an−1zn−1 + an−2z

n−2 + . . .+ a0,

apodeÐxte ìti |ρ| < 1 + |an−1|+ |an−2|+ . . .+ |a0|.

LÔsh.

An |ρ| > 1 èqoume

|ρ| = |an−1

ρ+an−2

ρ2+ . . .+

a0

ρn−1| ≤ |an−1|+ |an−2|+ . . .+ |a0|.

3. UpologÐste ta

max|z|≤1

|z3 − 2||z3 + 2|

kai min|z|≤1

|z3 − 2||z3 + 2|

.

LÔsh.

Me bsh tic idiìthtec twn mètrwn èqoume

|z3 − 2||z3 + 2|

≤ |z|3 + 2|

2− |z|3≤ 3

1.

Sto shmeÐo autì kpoioi stamatoÔn nomÐzontac ìti èqoun telei¸sei me to na broun nwfrgma qwrÐc na apodeÐxoun ìti apoteleÐ kai tim thc sunrthshc. Sthn perÐptws macto nw frgma 3 apoteleÐ tim thc sunrthshc gia z = −1 opìte eÐnai h megÐsth tim .Doulèyte anloga gia to elqisto.

4. JewroÔme z, w ∈ C me Im(z)Im(w) > 0. BreÐte thn elaqÐsth tim thc sunrthshcf(x) = |x−z|+|x−w|, x ∈ R kai thn pragmatik rÐza thc exÐswshc |x−z|+|x−w| = |z−w|.

LÔsh.

Kpoioc me idiaÐterec euaisjhsÐec sthn eukleÐdeia gewmetrÐa ja anagn¸rize sth gl¸ssathc to akìloujo isodÔnamo prìblhma.

2

An mac dojoÔn dÔo shmeÐa Z,W sto Ðdio hmiepÐpedo miac eujeÐac na eurejeÐ shmeÐo X thceujeÐac pou kajist to jroisma apostsewn apì ta Z,W elqisto.

Gia thn kataskeu jewroÔme to summetrikì W ∗ tou W wc proc thn eujeÐa opìte gia kjeshmeÐo X thc eujeÐac isqÔei XW = XW ∗ kai XW + XZ = XW ∗ + XZ ≤ W ∗Z. EÐnait¸ra profanèc ìti èqoume isìthta an jewr soume X to shmeÐo tom c tou eujugrmmoutm matoc W ∗Z me thn eujeÐa.

<<Metafrzoume >> ta parapnw se migadik gl¸ssa. Gia x ∈ R èqoume x = x kai

|x− w|+ |x− z| = |x− w|+ |z − x| ≥ |z − w|.

An mporoÔsame gia katllhlo x na epitÔqoume isìthta sthn parapnw anisìthta tìte hzhtoÔmenh elaqÐsth tim ja tan |z − w|. EÐnai gnwstì ìti h trigwnik anisìthta gÐnetaiisìthta sthn perÐptwsh pou z − x = t(x − w), t ≥ 0 dhlad Rez − x = t(x − Rew),Imz = tImw me t ≥ 0, opìte ikanopoieÐtai h sqèsh me t = Imz

Imw> 0.

Telik paÐrnoume wc elaqÐsth tim to |z − w| gia x = Rez+tRewt+1

me t = ImzImw

. H parapnwtim tou x apoteleÐ profan¸c kai thn zhtoÔmenh lÔsh thc exÐswshc tou probl matoc.

5. An a, b ∈ C− 0 kai n fusikìc arijmìc apodeÐxte ìti gia kje

r > 0 uprqei z ∈ C me |z| = r ¸ste |a+ bzn| > |a|.

LÔsh.

Jètoume a = |a|eıt1 , t1 = arga, omoÐwc b = |b|eıt2 , z = reıt opìte

|a+ bzn| = ||a|eıt1 + |b|eıt2rneınt| = ||a|+ |b|rneı(nt+t2−t1)|An proume t me nt+t2−t1 = 0 paÐrnoume thn tim |a|+|b|rn pou eÐnai h megÐsth dunat tim gia |z| = r kai ikanopoieÐ thn apaÐthsh tou probl matoc . (An proume t me nt+t2−t1 = πpaÐrnoume thn elaqÐsth tim ||a| − |b|rn| ).

6. UpologÐste ta

w = (1 + ı√

3)24, Rew, Imw, X(w,∞).

LÔsh.

An z = 1 + ı√

3, eÔkola diapist¸noume ìti z2 = −2z opìte z3 = −2zz = −2|z|2 = −8,

w = (z3)8 = 88 kai X(w,∞) = 2√1+|w|2

= . . ..

(EpilÔste to arqikì er¸thma me thn qr sh trigwnometrik c èkfrashc)

7. Perigryete gewmetrik ta sÔnola:

a. A = z ∈ C : Re(z2) = 1,b. B = 1

z: Re(z) ≤ −1


c. C = z ∈ C : Re(z) = Re( ¯(ız),

LÔsh.

Gia to sÔnolo A anatrèxte sthn analutik gewmetrÐa gia thn gramm x2 − y2 = 1.

Gia to sÔnolo B qrhsimopoi¸ntac tic idiìthtec tou s¸matoc twn migadik¸n arijm¸n èqoumediadoqik tic sqèseic

Z = 1z, z = 1

Z, Rez = Re 1

Z= Re Z

|Z|2 ,

Re Z|Z|2 ≤ 1,|Z|2 − 2Re(1

2Z) ≤ 0, |Z − 1

2|2 ≤ 1

4.

EÐnai t¸ra profanèc ìti so sÔnolo apartÐzetai apì ton kleistì dÐsko D[12, 1

2].

(XanalÔste to prìblhma metatrèpontc to se prìblhma analutik c gewmetrÐac.)

To sÔnolo C perigrfetai apì thn eujeÐa x+ y = 0.

8. Perigryte gewmetrik to sÔnolo pou dÐdetai apì thn exÐswsh

|z|2 + 5Re(z) + 3Im(z) + d = 0, gia tic diforec timèc tou d ∈ R.

LÔsh.

Metatrèpontac to prìblhma se analutik c gewmetrÐac èqoume thn exÐswsh

x2 + y2 + 5x+ 3y + d = 0 (x+ 52)2 + (y + 3

2)2 = 34

4− d.

Apì thn parapnw morf prokÔptoun oi paraktw peript¸seic.

An 344− d < 0 to sÔnolo eÐnai kenì.

An 344− d < 0 èqoume monosÔnolo me stoiqeÐo to z = (−5

2,−3

2) = −5

2,−ı3

2

kai gia 344− d > 0 èqoume ton kÔklo kèntrou K = (−5

2,−3

2) = −5

2,−ı3

2kai aktÐnac

R =√

344− d.

H antimet¸pish mporeÐ na gÐnei me thn lgebra twn migadik¸n. Pio sugkekrimèna h arqik exÐswsh gÐnetai

|z|2 +Re(z(5 + 3ı)) +|5 + 3ı|2

4=|5 + 3ı|2

4− d

|z + 5+3ı2|2 = 34

4− d kai suneqÐzoume ìpwc prohgoumènwc.

9. BreÐte tic rÐzec twn exis¸sewn

z4 = 24, z2001 = 1, 1 + z + z2 + . . . z2000 = 0, ez = 1 + ı.

LÔsh.

4

Gia thn pr¸th exÐswsh èqoume lÔseic

zk =4√

24e2kπı

4 , k = 0, 1, 2, 3

z0 = 2, z1 = 2eıπ4 =√

2 + ı

√2

2, z2 = 2eıπ = 2ı, z3 = 2eı3π = −z1.

Gia thn deÔterh èqoume rÐzec zk = e2πık2001 , k = 0, 1, 2 . . . 2000.

Gia thn trÐth an pollaplasisoume me z−1 paÐrnoume thn exÐswsh z2001 = 1. thn lÔnoumeìpwc thn prohgoÔmenh all exairoÔme thn lÔsh z0 = 1. Gia thn tètarth oi lÔseic dÐnontaiapì to sÔnolo

log(1 + ı) = Log|1 + ı|+ ıArg(1 + ı) + 2kπı, k = 0± 1,±2,± . . . =

Log|2|2

+ıπ

4+ 2kπı, k = 0± 1,±2,± . . . .

10. DeÐxete gewmetrik ìti ez+1z−1 ∈ D(0, 1) gia kje z ∈ D(0, 1).

LÔsh.

To sÔnolo tim¸n thc parapnw sunrthshc tautÐzetai me thc sunrthshc e−Z , ReZ ≥ 0to opoÐo ja broÔme. H exÐswsh e−Z = w,w 6= 0 dÐnei Z = −Log|w| − ıargw. H sqèshReZ ≥ 0 isodunameÐ me Log|w| < 0 w ∈ D(0, 1)− 0.

11. JewroÔme thn sunrthsh f(z) = eiz+1 sto sÔnoloΩ = z ∈ C : |Im(z)| < 1.(a) EÐnai h f 1-1 sto Ω?(b) BreÐte thn eikìna f(Ω).

LÔsh.

Apì thn exÐswsh w = eiz+1 prokÔptei sÔnolo lÔsewn

ız + 1 = log(w) = Log|w|+ ıArgw + 2kπı z = −ıLog|w|+ Argw + 2kπ − 1.

O periorismìc −1 < Imz < 1 isodunameÐ me −1 < −Log|w| < 1 e−1 < |w| < e1 poushmaÐnei ìti to sÔnolo tim¸n eÐnai o daktÔlioc ∆(0, e−1, e1). (

Epeid h exÐswsh èqei peirec lÔseic gia opoiod pote w ston daktÔlio h sunrthsh deneÐnai amfimonos manth.

12. BreÐte thn eikìna tou sunìlou mesw thc dosmènhc sunrthshc stic peript¸seic

A = z = x+ iy ∈ C : x ≤ −1, 0 ≤ y ≤ π, f(z) =1− z1 + z

.


B = z = x+ iy ∈ C : x ≤ −1, g(z) = e1+z1−z .

LÔsh.

H sqèsh

Z = X + ıY ) =1− z1 + z

met tic prxeic dÐdei

z = x+ ıy =1− Z1 + Z

x =1−X2 − Y 2

(1 +X)2 + Y 2, y =

−2Y

(1 +X)2 + Y 2.

O periorimìc x ≥ −1 antistoiqeÐ sto kleistì hmiepÐpedo X ≤ 1 , o periorismìc y ≥0 antistoiqeÐ sto kleistì hmiepÐpedo Y ≤ 0 kai o periorismìc y ≥ π antistoiqeÐ sto(1 + x)2 + (y − 1

pi)2 ≥ 1

pi2dhlad sto C−D(1 + 1

pi, 1pi. H zhtoÔmenh eikìna tautÐzetai me

thn tom twn parapnw sunìlwn.

To deÔtero er¸thma t¸ra diamorf¸netai sthn eÔresh tou Z = ew : Rew ≤ 1. Apì thlÔsh thc exÐswshc prokÔptei

w = Log|Z|+ ıargZ kai o periorismìc Rew ≤ 1 isodunameÐ me 0 < |Z| < e.

13. ApodeÐxte ìti |cosz|2 + | sin z|2 = 12(e2y + e−2y)

LÔsh.

| cos z|2 = |eız + e−ız

2|2 =

|eız|2 + |e−ız|2 + 2Re(eıze−ız)

4,

| sin z|2 = |eız − e−ız

2ı|2 =

|eız|2 + |e−ız|2 − 2Re(eıze−ız)

−4.

AjroÐzontac paÐrnoume thn posìthta |eız |2+|e−ız |2

2.

All |eız|2 = |eı(x+ıy)|2 = |eıx|2|e−y|2 = e−2y kai omoÐwc |e−ız|2 = e2y.

14. BreÐte ta sÔnola pou prokÔptoun apì tic ekfrseic ((1 + ı)3)ı, (1 + ı)3ı, (3ı)(1+ı).

LÔsh.

((1 + ı)3)ı = (−2 + 2ı)ı = eılog((−2+2ı).

Epeid

6

log((−2 + 2ı) = Log|(−2 + 2ı)|+ ıArg(−2 + 2ı) + 2kπı =3

2Log2 +

3

4πı+ 2kπı

èqoume

((1 + ı)3)ı = e32ıLog2− 3

4π−2kπ, k = 0± 1± 2 . . . .

Epeid

log(1 + ı) = Log|1 + ı|+ ıArg(1 + ı) + 2kπı =1

2Log2 +

1

4πı+ 2kπı

èqoume

(1 + ı)3ı = e3ılog(1+ı) =1

2Log2 +

1

4πı+ 6kπı, k = 0± 1± 2 . . . .

(ParathroÔme ìti apì tic ekfrseic den proèkuyan Ðdia sÔnola all to deÔtero eÐnaiuposÔnolo tou pr¸tou.)

Gia thn trÐth èkfrash akolouj¸ntac thn Ðdia diadikasÐa èqoume

log(3ı) = Log3 + π2

+ 2kπı opìte

(3ı)(1+ı) = e(1+ı)log(3ı) = eLog3−π2−2kπ+ı(Log3+π

2+2kπ), k = 0± 1± 2 . . . .

15. An z1 = 4 + 3ı, z2 = 3 + 4ı, z3 = −3 + 4ı, z4 = −4 + 3ı apant ste poièc apì tic paraktwisìthtec alhjeÔoun.

Logz1z2 = Logz1 + Logz2, Logz4z2 = Logz4 + Logz2, Logz2z4 = Logz2 + Logz4,

Logz41 = 4Logz1, Logz3

3 = 3Logz3.

LÔsh.

Apì to tetarthmìrio pou an kei to kje shmeÐo paÐrnoume tic paraktw anisìthtec.

0 < Argz1 <π4, π

4< Argz2 <

π2, π

2< Argz3 <

3π4kai 3π

4< Argz4 < π.

H pr¸th isìthta isqÔei en kai mìno en −π < Argz1 + Argz2 ≤ π. Axiopoi¸ntac ticprohgoÔmenec sqèseic èqoume π

4< Argz1 + Argz2 <

3π4opìte h apnthsh eÐnai jetik .

Sthn deÔterh perÐptwsh èqoume Argz4 + Argz2 > π opìte h apnthsh eÐnai arnhtik .

Sthn trÐth perÐptwsh epeid Argz4 = −Argz4, èqoume −π < Argz4 < −3π4kai

−3π4< Argz4 + Argz2 < −π

4opìte h apnthsh eÐnai jetik .

Gia thn tètarth perÐptwsh epeid 0 < 4Argz1 < π h apnthsh eÐnai jetik .

Gia thn teleutaÐa perÐptwsh epeid 3Argz3 >3π2h apnthsh eÐnai arnhtik .


16. An f eÐnai analutik sunrthsh se anoiktì sÔnolo apodeÐxte ìti∂2

∂x2 |f |2 + ∂2

∂y2|f |2 = |f ′|2

LÔsh. ∂∂x|f |2 = ∂

∂xff = fxf + ffx,

∂2

∂x2 |f |2 = fxxf + 2|fx|2 + fxxf ,

∂2

∂y2|f |2 = fyyf + 2|fy|2 + fyyf .

Lambnontac upìyei ìti fxx + fyy = 0, fxx + fyy = 0 kai |fx| = |fy| = |f ′| prosjètontactic dÔo teleutaÐec sqèseic paÐrnoume to zhtoÔmeno.

Dialèxame ènan sunoptikì gr goro trìpo gia thn apìdeixh. O anagn¸sthc kaleÐtai naorgan¸sei mia leptomer lÔsh jètontac f = u+ ıv.

Eidik perÐptwsh tou parapnw sumpersmatoc apoteleÐ h paraktw skhsh.

17. An f eÐnai analutik sunrthsh se tìpo me |f |2 armonik se autìn tìte h sunrthsh eÐnaistajer . (ExhgeÐste giatÐ h diaforopoÐhsh tou pedÐou orismoÔ se tìpo eÐnai ousiastik .)

18. An f(z) eÐnai analutik sunrthsh se sÔnolo O summetrikì wc proc ton xona twn x tìteh sunrthsh g(z) = f(z) eÐnai analutik sto O kai isqÔei g′(z) = f ′(z) gia kje z ∈ O

LÔsh.

Epeid z → z0 ìtan z → z0 èqoume

limz→z0

f(z)− f(z0)

z − z0

= f ′(z0)

kai paÐrnontac ta suzug

limz→z0

f(z)− f(z0)

z − z0

= f ′(z0),

dhlad gia kje z0 ∈ O uprqei h pargwgoc g′(z0) kai isqÔei g′(z0) = f ′(z0).

Skìpima sto s¸ma thc apìdeixhc den anafèrjhka sthn summetrikìthta tou sunìlou af -nontc h ston anagn¸sth na entopÐsei pou qrhsimopoi jhke. O anagn¸sthc kaleÐtaiepÐshc na d¸sei mia deÔterh lÔsh jewr¸ntac f(z) = u(x, y) + ıv(x, y) kai g(x, y) =u(x,−y)− ıv(x,−y).

19. Na exetastoÔn wc proc thn diaforisimìthta kai analutikìthta oi sunart seic,

f(z) = z2

z, z 6= 0, f(0) = 0, g(z) =

√|xy|, h(z) = |z|5 + 3ez + 1.

LÔsh. Gia z 6= 0 èqoume fz = 2zz6= 0, dhlad den isqÔoun oi sunj kec C-R opìte den

uprqei h pargwgoc. Sto shmeÐo z = (0, 0) èqoume f(x, 0) = x, f(0, y) = ıy opìte

8

fx(0, 0) = −ıfy(0, 0) = 1, dhlad isqÔoun oi sunj kec C-R. Parìla aut h pargwgoc

den uprqei sto mhdèn Prgmati h sunrthsh f(z)−f(0)z−0

= z2

z2ìtan to z → 0 kinoÔmeno ston

pragmatikì xona èqei orio 1 kai gia z = t(1 + ı), t ∈ R èqoume ìrio −1.

(Wc gnwstì den arkoÔn oi sunj kec C-R gia thn paragwgisimìthta all qreizetai epi-plèon kai h diaforisimìthta me thn pragmatik ènnoia pou sthn perÐptws mac den isqÔei.)

Sqetik me thn sunrthsh g = u+ıv me u(x, y) = 4√x2y2 kai v = 0 me aploÔc upologismoÔc

diapist¸noume ìti den isqÔoun oi sunj kec C-R sta shmeÐa pou isqÔei xy 6= 0.

Epeid g(0, y) = 0, g(x, 0) = 0 èqoume gx(0, 0) = −ıgy(0, 0) = 0 dhlad isqÔoun oisunj kec C-R.

H klÐsh g(z)−g(0)z−0

an kinhjoÔme sthn diqotìmo y = x dhlad z = x + ıx gÐnetai |x|(1+ı)x

pou

gia x > 0, x→ 0 dÐnei ìrio 11+ı

kaÐ gia x < 0, x→ 0 dÐnei ìrio − 11+ı

, epomènwc den uprqei

h pargwgoc sto mhdèn. An x0 6= 0 , f(x0, y) =√|x0|√|y| apì thn pragmatik anlush

eÐnai gnwstì ìti den uprqei h pargwgoc wc proc y sto mhdèn dhlad den uprqei tofy(x0, 0) gia tuqìn x0 6= 0.

Me thn Ðdia diadikasÐa mporoÔme na apodeÐxoume ìti den uprqei to fx(y0, 0) gia tuqìny0 6= 0. Telik den uprqei h pargwgoc se kanèna shmeÐo.

Sqetika me thn sunrthsh h. Epeid h sunsthsh 3ez + 1 eÐnai analutik se ìlo tomigadikì epÐpedo arkeÐ na melet soume thn sunrthsh q(z) = |z|5 = (

√x2 + y2)5.

Gia z 6= 0 èqoume fx = 2x(x2 + y2)32 ,fy = 2y(x2 + y2)

32 opìte fx(z) 6= −ıfy(z) .

Gia z = 0 èqoume q(z)−q(0)z−0

= z|z|3 → 0 ìtan z → 0, dhlad q′(0) = 0. Telik h sunrthshh(z) = |z|5 + 3ez + 1 èqei pargwgo mìno sto shmeÐo mhdèn thn tim h′(0) = 3.

Se kamÐa apì tic peript¸seic den èqoume shmeÐo pou h antÐstoiqh sunrthsh na eÐnaianalutik .

20. Na exetastoÔn wc proc thn diaforisimìthta kai thn analutikìthta oi sunart seic,

f(z) = 9x3 − x − ı(4y3 − y), g(z) = |z − 1|2 + |z − 2|2 + |z − 3|2, h(z) = ex(x cos y −y sin y) + ı(ex(y cos y + x sin y) z ∈ C.

LÔsh. Stic peript¸seic pou exetzoume ìlec oi sunart seic eÐnai diaforÐsimec me thnpragmatik ènnoia se ìlo to pedÐo orismoÔ touc diìti uprqoun oi merikèc pargwgoikai eÐnai suneqeÐc, epomènwc èqoun pargwgo akrib¸c sta shmeÐa ekeÐna pou isqÔoun oisunj kec C-R.

Oi sunj kec fx = 27x2− 1 = −ıfy = −12y2 + 1, dÐnoun ìti to sÔnolo paragwgisimìthtacthc f apartÐzetai apì ta shmeÐa thc èlleiyhc 27x2 + 12y2 = 2 kai se aut isqÔei f ′(z) =fx = 27x2 − 1

Se kanèna shmeÐo h sunrthsh den eÐnai analutik diìti den uprqei kanènac anoiktìcdÐskoc pou h sunrthsh na eÐnai paragwgÐsimh se ìla ta shmeÐa tou.

Sqetik me thn sunrthsh g h morf g(z) = (z−1))(z−1)+(z−2))(z−2)+(z−3))(z−3)mac <ÍpagoreÔei>> na qrhsimopi soume tic sunj kec


C-R sthn isodÔnamh ekdoq gz = 0 (z − 1) + (z − 2) + (z − 3) = 0 pou shmaÐnei ìtièqoume monadikì shmeÐo paragwgisimìthtac z0 = 2. H pargwgoc dÐnetai apì ton tÔpofz(z0) = (z0 − 1) + (z0 − 2) + (z0 − 3) = 0. Profan¸c h sunrthsh den eÐnai poujenanalutik .

Gia thn sunrthsh h met tic merikèc paragwgÐseic o anagn¸sthc mporeÐ na diapist¸seiìti isqÔoun oi sunj kec C-R se ìlo to migadikì epÐpedo kai epeid eÐmaste se anoiktìsÔnolo h sunrthsh ja eÐnai analutik . 'Ena exoikeiwmèno blèmma me thn lgebra twnmigadik¸n arijm¸n ja mporoÔse na apofÔgei touc upologismoÔc parathr¸ntac ìti f(z) =ex(x+ ıy)(cos y + ı sin y) = zez pou profan¸c eÐnai analutik .

21. Na breÐte sunrthsh orismènh se olìklhro to migadikì epÐpedo pragmatik diaforÐsimhpou to sÔnolo twn shmeÐwn pou paragwgÐzetai na apartÐzetai apì toÔc dÔo xonec kai ticdiqotìmouc twn gwni¸n tou.

LÔsh. Ta shmeÐa thc paragwgisimìthtac pou apaitoÔme perigrfontai me eniaÐa exÐswsh(x− y)(x+ y)xy = 0.

Jètoume x = z+z2, y = z−z

2ıkai met tic prxeic h parapnw exÐswsh paÐrnei thn ÐsodÔnamh

migadik morf z4 − z4 = 0.

AnazhtoÔme mÐa pargousa tou pr¸tou skèlouc wc proc z. MÐa tètoia eÐnai h sunrthshf(z) = zz4 − 1

5z5.

H parapnw sunrthsh èqei merikèc parag¸gouc suneqeÐc sto epÐpedo opìte eÐnai prag-matik diaforÐsimh kai oi sunj kec C-R fz = z4−z4 = 0 ikanopoioÔntai apì thn kataskeu thc apokleistik sto sÔnolo pou apaiteÐtai. Epiplèon sto sÔnolo autì h pargwgoc dÐde-tai apì ton tÔpo f ′(z) = fz(z) = z4. Epeid to sÔnolo twn shmeÐwn paragwgisimìthtacden èqei eswterik shmeÐa h sunrthsh den eÐnai analutik se kanèna shmeÐo.

22. Na exetasteÐ poièc sunart seic èqoun suzug armonik sto pedÐo orismoÔ touc kai sthnjetik perÐptwsh na eurejoÔn stic peript¸seic.

u1(x, y) = y3 − 3yx2, u2(x, y) = y3 − 3yx2, u3(x, y) = Log[(x− 1)2 + (y − 2)2],

u4(x, y) = x2 − y2 +x+ y

x2 + y2, u5(r, θ) = 5 + 3r cos θ + 7r sin θ + 8r2 sin θ cos θ.

(PedÐa orismoÔ jewreÐste ta antÐstoiqa uposÔnola tou migadikoÔ epipèdou pou èqoun nìhmaoi ekfrseic touc)

LÔsh. Sqetik me thn u1 Me stoiqei¸deic paragwgÐseic diapist¸noume ìti ∂2u1

∂x2 + ∂2u1

∂x2 = 0,dhlad h sunrthsh eÐnai armonik sto pedÐo orismoÔ thc. EpÐ plèon to pedÐo orismoÔ eÐnaiolìklhro to epÐpedo pou eÐnai apl sunektikì opìte sÐgoura uprqei suzug c armonik .

Gia thn eÔresh èqoume ux = −6y = vy opìte v(x, y) = −3y2 + c(x). Apì thn sqèshvx = −uy = −3y2 + 3x2 èqoume c′(x) = 3x2 c(x) = x3 + c kai v(x, y) = −3y2 + x3 + c.

10

Sthn perÐptws mac kai genik stic rhtèc sunart seic mporoÔme na efarmìsoume ènanligìtero epÐpono trìpo brÐskontac pr¸ta thn analutik sunrthsh kai sth sunèqeia napaÐrnoume thn suzug armonik apì to fantastikì thc mèroc. Pio sugkekrimèna an jèsoumex = z+z

2, y = z−z

2ımet tic prxeic h parapnw sunrthsh paÐrnei thn migadik morf

ız3 − ız3 = Re(2ız3). EÐnai t¸ra profanèc ìti v = Im(2ız3) + c.

Sqetik me thn sunrthsh u2 met tic stoiqei¸deic paragwgÐseic eÔkola diapist¸noume ìtiden eÐnai armonik opìte den èqei suzug armonik .

Sqetik me thn u3 eÔkola diapist¸noume ìti eÐnai armonik . Ed¸ prèpei ìmwc na eÐ-maste prosektikoÐ giatÐ to pedÐo orismoÔ C − 1 + 2ı den eÐnai apl sunektikì (èqeitrÔpa sto(1, 2)) kai den mporoÔme akÐnduna na suneqÐsoume thn prohgoÔmenh diadikasÐ-a. EÐmaste anagkasmènoi na perioristoÔme se katllhlo uposÔnolo apla sunektikìtìpo kai sth sunèqeia na doÔme an epekteÐnetai to apotèlesma. Sthn perÐptws mac hsunrthsh grfetai 2Log|z − 1 − 2ı| = Re

(2Log(z − 1 − 2ı)

)An perioristoÔme sto

sÔnolo OC = C − z : Im(z − 1 − 2ı) = 0, Re(z − 1 − 2ı) ≤ 0 Dhlad na afairè-soume apì to migadikì epÐpedo thn hmieujeÐa (y = 2, x ≤ 1) gnwrÐzoume ìti h sunrthsh2Log(z−1−2ı) eÐnai analutik ston parapnw tìpo opìte se autìn oi suzugeÐc armonikècthc arqik c sunrthshc eÐnai Im(2Log(z − 1 − 2ı) = 2Arg(z − 1 − 2ı) + c. EÐnai gnw-stì ìmwc ìti h prohgoÔmenh sunrthsh den epekteÐnetai suneq¸c se kanèna shmeÐo thcapomènousac hmieujeÐac pou shmaÐnei ìti h arqik den èqei armonik suzug . Profan¸c operiorismìc pou qrhsimopoi same eÐnai o megalÔteroc dunatìc pou dÐnei armonik suzug .

Sqetik me thn u4 ja diapist¸soume ìti , par to gegonìc ìti o tìpoc orismoÔ èqei trÔpasto mhdèn h sunrthsh èqei suzug armonik .

Prgmati an jèsoume x = z+z2, y = z−z

2ıèqoume

x2 − y2 = Rez2,x

x2 + y2= Re

z

zz= Re

1

z= Re

1

z,

y

x2 + y2= Re(

−ızzz

) = Reı

z

opìte u4(x, y) = Re(z2 + 1+ız

).

EÐnai t¸ra profanèc ìti h u4 èqei suzug armonik thn v4(x, y) = Im(z2 + 1+ız

).

Sqetik me thn u5 qrhsimopoi¸ntac tic isìthtec cos θ = Re(eıθ)

sin θ = Re(−ıeıθ) paÐrnoume

u5(r, θ) = 5 + 3r cos θ + 7r sin θ + 8r2 sin θ cos θ =

Re(5 + 3reıθ − 7ıreıθ + 4r2e2ıθ) = Re(5 + (3− 7ı)z + 4z2).

EÐnai t¸ra profanèc ìti h sunrthsh èqei armonik suzugeÐc tic Im(5+(3−7ı)z+4z2)+c.Af netai ston anagn¸sth na tic ekfrsei se morf polik¸n suntetagmènwn.


23. An R h aktÐna sÔgklishc dunamoseirc∑anz

n kai k fisikìc arijmìc breÐte tic aktÐnecsÔgklishc twn dunamoseir¸n

∑nkanz

n,∑anz

kn kai∑aknz

n.

LÔsh. Sqetik me thn pr¸th dunamoseir èqoume

1

R1

= lim n√nk|an| = (lim n

√n)klim n

√|an| = lim n

√|an| =

1

R.

Sqetik me thn deÔterh, an jèsoume zk = Z h sÔgklish thc seirc∑anZ

n gia |Z| < RdÐnei |z| < k

√R.

DeÔteroc trìpoc eÐnai na proume thn sqèsh

1

R2

= lim kn√|an| =

k

√lim n√|an| =

1k√R.

Sqetik me thn trÐth dunamoseir èqoume

1

R3

= lim n√|akn| = (lim n

√|an|)k =

1

Rk

.

24. An p, q polu¸numa, k,m akèraioi kai h(n) gnhsÐwc aÔxousa akoloujÐa fusik¸n arijm¸napodeÐxte ìti h seir

∑pk(n)(Log(q(n))mzh(n) èqei aktÐna sÔgklishc èna.

LÔsh.

Gia kje sugkekrimèno z ja exetsoume wc proc thn sÔgklish thn antÐstoiqh seir mi-gadik¸n arijm¸n

∑fn(z) me to krit rio tou lìgou dhlad ja broÔme ta z gia ta opoÐa

isqÔei,

|fn+1(z)||fn(z)|

= |p(n+ 1)

p(n)|k|Log(q(n+ 1))

Loq(q(n))|m|z|h(n+1)−h(n) < 1giagian > n0.

Epeid ta polu¸numa p(x), p(x + 1) èqoun ton Ðdio suntelest sto megistobjmio ìro

èqoume lim p(n+1)p(n)

= 1. EpÐshc isqÔei lim Log(q(n+1))Loq(q(n))

= 1. Gia thn apìdeixh ja qrhsi-mopoi soume ènan aplì kai idiaÐtera apotelesmatikì trìpo pou efarmìzetai gia akolou-jÐec pou proèrqontai apì periorismoÔc pragmatik¸n sunart sewn stic opoÐec efarmìzetaio kanìnac Hospital. (Ed¸ qreizetai prosoq diìti merikoÐ, esfalmèna profan¸c, efar-mìzoun ton kanìna qwrÐc na jewr soun sunrthsh paÐrnontac pargwgo akoloujÐac)

JewroÔme thn sunrthsh Log(q(x+1))Log(q(x))

kai ja apodeÐxoume ìti gia x→ èqei ìrio 1. Prgmati

èqoume thn perÐptwsh +∞+∞ opìte

(Log(q(x+ 1)))′

(Loq(q(x)))′=q′(x+ 1)q(x)

q′(x)q(x+ 1)→ 1 ìtanx→ +∞.

12

EpÐshc

|zh(n+1)−h(n))| ≤ |z| ìtan |z| < 1 kai |zh(n+1)−h(n)) ≥ |z| ìtan |z| > 1.

Gia |z| < 1 èqoume

|fn+1(z)||fn(z)|

≤ |p(n+ 1)

p(n)|k||Log(q(n+ 1))

Log(q(n))|m|z| → |z| < 1

pou shmaÐnei ìti telik gia ìlouc touc deÐktec isqÔei |fn+1(z)||fn(z)| < 1 opìte èqoume sÔgklish

sto dÐsko |z| < 1.

Gia |z| > 1 me ton Ðdio trìpo paÐrnoume |fn+1(z)||fn(z)| > 1 pou shmaÐnei ìti den sugklinei h seir

sto exwterikì tou monadiaÐou dÐskou opìte h aktÐna sÔgklishc thc dunamoseirc eÐnai èna.

25. Na upologistoÔn oi aktÐnec sÔgklishc twn dunamoseir¸n

1.∑

(an + bn)zn, |a| < |b|,2.∑

(n+ bn)zn,

3.∑an

2zkn, k fusikìc,

4.∑aknzn

2, k fusikìc,

5.∑

(1 + ı)4nz32n,

6.∑

(1 + ı)32nz4n,

7.∑

32nz4n!,

8.∑

34n!z2n,

9.∑

n!nnzkn,k fusikìc.

LÔsh.

1. Apì thn isìthta

|an+1 + bn+1

an + bn| = |

(ab)n+1 + 1

1b(ab)n + 1

b

|

lambnontac upìyh ìti (ab)n → 0 paÐrnoume

|an+1 + bn+1

an + bn| → |b| = 1

R1

.

Enac deÔteroc trìpoc eÐnai na apodeÐxoume ìti oi seirèc∑anzn kai

∑bnzn èqoun aktÐnec

sÔgklishc 1|a| kai

1|b| antÐstoiqa pou den eÐnai Ðsec metaxÔ touc opìte h mikrìterh eÐnai aktÐna

sÔgklishc tou ajroÐsmatoc.

2. Gia |a| ≤ 1 èqoume an

n→ 0 opìte

|n+ 1 + an+1

n+ an| = |

1 + 1n

+ aan

n

1 + an

n

| → 1 =1

R2

.


Gia a| > 1 èqoume nan→ 0 (breÐte giatÐ) opìte

|n+ 1 + an+1

n+ an| = |

n+1an+1 + 11a

+ nan

1a

| → |a| = 1

R2

.

3. DiereunoÔme thn uparxh toÔ orÐou thc akoloujÐac n√|a|kn2 = (|a| 1k )n. Wc gnwstì to

ìrio thc parapnw akoloujÐac sthn perÐptwsh |a| < 1 eÐnai mhdèn opìte R3 = +∞. SthnperÐptwsh |a| = 1 èqoume R3 = 1 kai sthn perÐptwsh |a| > 1 to ìrio eÐnai +∞ opìteR3 = 0.

4. lim n2√|a|kn lim |a|

knn2 = lim |a| kn = 1 = 1

R4.

5. DiereunoÔme wc proc to ìrio thn akoloujÐa |1 + ı|4n32n . To ìrio thc akoloujÐac 4n

32n eÐnaimhdèn (ìpoioc jèlei na apofÔgei mia apeujeÐac anamètrhsh mazÐ tou mporeÐ na katafÔgeisthn bo jeia tou kanìna Hospital jewr¸ntac thn sunrthsh 4x

32x ). Telik èqoume 1R5

=1 = R5.

6. Apì thn prohgoÔmenh perÐptwsh èqoume 32n

4n→ +∞ , opìte

|1 + ı|32n

4n = (√

2)32n

4n → +∞

pou shmaÐnei ìti R6 = 0.

7. Epeid 32n4n! = 3

2(n−1)! → 1 èqoume R7 = 1.

8. Epeid 4n!2n→ +∞, 3

4n!2n → +∞ èqoume R8 = 0.

9. An fn(z) = n!nnzkn tìte

|fn+1(z)||fn(z)|

=n+ 2

n(1 +

1

n)n|z| → e|z|k.

Epeid e|z|k < 1 ìtan |z| < k

√1eèqoume R9 = k

√1e.

26. UpologÐste tic aktÐnec sÔgklishc twn paraktw dunamoseir¸n qrhsimopoi¸ntac apotelès-mata thc prohgoÔmenhc skhshc.

1.∑

(an + bn)nzn+8, |a| < |b|.2.∑

(n+ bn)n(n− 1)(n− 2) . . . (n− k + 1)zn.

3.∑

(1 + ı)32nn(4n− 1)(4n− 2)z4n.

LÔsh

1. Gia kje z h antÐstoiqh seir gÐnetai z9∑

(an+bn)n)zn−1, epomènwc h zhtoÔmenh aktÐnatautÐzetai me thn aktÐna sÔgklishc thc dunamoseirc∑

(an + bn)nzn−1 =∑

(an + bn)(zn)′.

14

Apì to gnwstì je¸rhma parag¸gishc dunamoseir¸n h aktÐna thc teleutaÐac sumpÐptei methn aktÐna sÔgklishc thc dunamoseirc

∑(an + bn)zn pou melet same sto pardeigma 1

thc prohgoÔmenhc skhshc.

2. H zhtoÔmenh aktÐna sumpÐptei me thn aktÐna sÔgklishc thc dunamoseirc∑(n+ bn)n(n− 1)(n− 2) . . . (n− k + 1)zn−k =

∑(n+ bn)(zn)(k)

pou me thn seir tou tautÐzetai me thn aktÐna tou deÔterou paradeÐgmatoc thc prohgoÔmenhcskhshc,

3. Apì thn èkfrash thc antÐstoiqhc seirc sth morf 14z3∑

(1 + ı)32n4n(4n − 1)(4n −

2)z4n−3 prokÔptei ìti h zhtoÔmenh aktÐna tautÐzetai me thn aktÐna thc dunamoseirc∑(1 + ı)32n

(z4n)(3)

pou me th seir thc tautÐzetai me thn aktÐna tou paradeÐgmatoc 6 thc prohgoÔmenhc skhshc.

27. Na upologisteÐ to jroisma∑+∞n=0(an3 + bn2 + cn+ d)(

zn+k1

zn+m2

),ìtan |z1| < |z2|.

LÔsh.

Apì tic isìthtec

n2 = n(n− 1) + n, n3 = n(n− 1)(n− 2) + 3n(n− 1) + n

h seir paÐrnei thn morf tou ajroÐsmatoc

+∞∑n=0

an(n− 1)(n− 2)zn+k

1

zn+m2

++∞∑n=0

(3a+ b)n(n− 1)zn+k

1

zn+m2

+

+∞∑n=0

(a+ b+ c)nzn+k

1

zn+m2

++∞∑n=0

dzn+k

1

zn+m2

.

Epeid f(z) = 11−z =

∑+∞n=0 z

n ìtan |z| < 1 o pr¸toc prosjetèoc grfetai

azk+31

zm+32

f (3)( z1z2

), O deÔteroc (3a+ b)zk+21

zm+22

f (2)( z1z2

),

o trÐtoc (a+ b+ c)zk+11

zm+12

f ′( z1z2

) kai o tètartoc d zk1

zm2f( z1

z2).

28. QwrÐc na efarmìsete ton tÔpo Taylor all thn bo jeia thc gewmetrik c seirc anaptÔxtese dunamoseir me kèntro to mhdèn thn sunrthsh Log(1 − z), |z| < 1. Sth sunèqeia me


thn bo jeia tou anaptÔgmatoc pou ja breÐte d¸ste se morf seirc rht¸n migadik¸n toucarijmoÔc Log(1+ı

2), π. LÔsh.

Apì thn sqèsh

(Log(1− z))′ = − 1

1− z= −

+∞∑n=0

zn

paÐrnoume

Log(1− z) =+∞∑n=0

zn+1

n+ 1+ c

kai jetontac z = 0 paÐrnoume c = 0.

Epeid 1+ı2

= 1− 1−ı2

kai |1−ı2| < 1 èqoume

Log(1 + ı

2) = −−

+∞∑n=0

(1− ı)n+1

2n+1(n+ 1).

Apì tic sqèseic (1−ı)2 = −2ı, (1−ı)3 = −2(1+ı), (1−ı)4 = −4, an jèsoume n = 4k+m,m = 0, 1, 2, 3 èqoume

(1− ı)n+1

2n+1(n+ 1)=

(−1)k(1− ı)m+1

(4k +m+ 1)22k+m+1,

opìte

Log(1 + ı

2) = −

3∑m=0

(1− ı)m+1

+∞∑k=0

(−1)k

(4k +m+ 1)22k+m+1=

−(1− ı)+∞∑k=0

(−1)k

(4k + 1)22k+1+ 2ı

+∞∑k=0

(−1)k

(4k + 2)22k+2+

2(1 + ı)+∞∑k=0

(−1)k

(4k + 3)22k+3+ 4

+∞∑k=0

(−1)k

(4k + 4)22k+4

Epeid ImLog(1+ı2

) = Arg(1+ı2

) = π4, apì ta fantastik mèrh sto prohgoÔmeno anptugma

èqoume

π = +4+∞∑k=0

(−1)k

(4k + 1)22k+1+ 8

+∞∑k=0

(−1)k

(4k + 2)22k+2+ 8

+∞∑k=0

(−1)k

(4k + 3)22k+3

16

29. Na upologistoÔn ta ajroÐsmata

1.∑+∞

n=025n+1

n!33n+2 , 2.∑+∞

n=025n+1

(2n+1)!33n+2 , 3.∑+∞

n=025n+1

(2n)!33n+2 , 4.∑+∞

n=0(−1)n25n+1

(2n)!33n+2 ,

5.∑+∞

n=0(−1)n25n+1

(2n+1)!33n+2 .

LÔsh.

To ousiastikì prìblhma me ask seic tic parapnw morf c eÐnai na tic susqetÐsoume me tokatllhlo gnwstì anptugma dunamoseirc. Anmesa sta lÐga prgmata pou qreizetaikaneÐc na èqei apomnhmoneÔsei afoÔ pr¸ta ta èqei melet sei eÐnai ta anaptÔgmata twnorismènwn klassik¸n sunart sewn pou onomsame jemeli¸deic .

Sqetik me to 1. To anptugma pou mac prodiajètei na susqetÐsoume to jroisma eÐnai to

ez =+∞∑n=0

zn

n!.

Prgmati+∞∑n=0

25n+1

n!33n+2=

2

9

+∞∑n=0

(32

27)n

1

n!=

2

9e

3227 .

Sqetik me to 2. To anptugma pou ja qreiastoÔme eÐnai

sinh z =1

2(ez − e−z) =

+∞∑n=0

z2n+1

(2n+ 1)!

.

Prgmati an jèsoume 5n+ 1 = 52(2n+ 1)− 3

2, 3n+ 1 = 3

2(2n+ 1)− 1

2èqoume

+∞∑n=0

25n+1

(2n+ 1)!33n+2=

2−32

3−12

+∞∑n=0

(2

52

332

)2n+1 1

(2n+ 1)!=

√3

8sinh

(√32

27

).


cosh z =1

2(ez + e−z) =

+∞∑n=0

z2n

(2n)!.

.

Prgmati,

+∞∑n=0

25n+1

(2n)!33n+2=

2

9

+∞∑n=0

(2

52

332

)2n 1

(2n!)=

2

9cosh

(√32

27

).

Sqetik me to 4.


To anptugma pou ja qreiastoÔme eÐnai

cos z =+∞∑n=0

(−1)nz2n

(2n)!

kai ergazìmenoi ìpwc sto 3. (H diaforopoÐhsh uprqei sto (−1)n) paÐrnoume

+∞∑n=0

(−1)n25n+1

(2n)!33n+2=

2

9cos(√32

27

).


sin z =+∞∑n=0

(−1)nz2n+1

(2n+ 1)!

kai me th diadikasÐa tou 2. brÐskoume

+∞∑n=0

(−1)n25n+1

(2n+ 1)!33n+2=

√3

8sin(√32

27

).

.

30. Na anaptuqjeÐ se seir Laurent h sunrthsh 11+2z+4z2+8z3+16z4+32z5+64z6

stouc megalÔter-ouc dunatoÔc daktulÐouc.

LÔsh.

Epeid

1

1 + 2z + 4z2 + 8z3 + 16z4 + 32z5 + 64z6=

1− 2z

1− 26z6

oi megalÔteroi dunatoÐ daktÔlioi kajorÐzontai apì tic sqèseic |26z6| < 1 |z| < 12kai

|26z6| > 1 |z| > 12.

Sthn pr¸th perÐptwsh apì ton gnwstì tÔpo thc gewmetrik c seirc èqoume

1− 2z

1− 26z6= (1− 2z)

+∞∑n=0

26nz6n =+∞∑n=0

26nz6n −+∞∑n=0

26n+z6n+1.

Sthn deÔterh perÐptwsh èqoume | 126z6| < 1 opìte

2z − 1

26z6 − 1=

2z − 1

26z6

1

1− 126z6

=2z − 1

26z6

+∞∑n=0

1

26nz6n=

+∞∑n=0

1

26n+5z6n+5−

+∞∑n=0

1

26n+6z6n+6.

18

31. An∑+∞

n=0 anzn to anptugma Taylor thc sunrthshc 1

a+bz+cz2,a 6= 0, apodeÐxte ìti

aan + ban−1 + can−2 = 0 ìtan n = 2, 3, . . ..

LÔsh. Gia praktikoÔc lìgouc orÐzoume ak = 0 ìtan k < 0. Apì thn sqèsh

1

a+ bz + cz2=

+∞∑n=0

anzn

èqoume

1 =+∞∑n=0

aanzn +

+∞∑n=0

banzn+1 +

+∞∑n=0

canzn+2 =

+∞∑n=0

azn ++∞∑n=0

ban−1zn +

+∞∑n=0

can−2zn =

+∞∑n=0

(aan + ban−1 + can−2)zn.

Apì thn taÔtish tou teleutaÐou anaptÔgmatoc me thn monda prokÔptei h zhtoÔmenh sqèsh.

32. BreÐte ta sÔnola twn akeraÐwn sunart sewn gia tic opoÐec isqÔei

1. Ref ≥ 0, 2. |f (4)(z)| ≤ 32.

LÔsh. Sqetik me to 1. An F = e−f tote profan¸c F akeraÐa kai |F | = e−Ref ≤ 1.

Efarmìzontac to Je¸rhma tou Liouville sthn sunrthsh F prokÔptei ìti eÐnai stajer epomènwc F ′ = eff ′ = 0 f ′ = 0 opìte f stajer .

(Ena genikìtero sumpèrasma pou t¸ra mporeÐ na apodeÐxei kaneÐc eÐnai ìti an se akeraÐasunrthsh isqÔei opoiad pote sqèsh thc morf c aRef + bImf + c ≥ 0,a, b, c ∈ R, (a, b) 6=(0, 0), tìte f stajer . Epeid aRef+bImf+c = Re((a−ıb)f+c) mporoÔme na ergastoÔmeìpwc prohgoumènwc qrhsimopoi¸ntac antÐ thc f thn sunrthsh (a− ıb)f .)Sqetik me to 2. Epeid f akeraÐa ìpwc eÐnai gnwstì ja èqei kje txhc pargwgo stopedÐo orismoÔ thc dhlad se olìklhro to migadikì epÐpedo opìte kai h f (4) ja eÐnai akeraÐa.Efarmìzontac to Je¸rhma tou Liouville sthn sunrthsh f (4) = c prokÔptei ìti eÐnaistajer kai autì isodumameÐ me thn prìtash ìti sumpÐptei me polu¸numo to polÔ bajmoÔ 4me suntelest thc tetrthc dÔnamhc c

24, |c| ≤ 32. (Genikìtera se mia analutik sunrthsh

se tìpo isqÔei f (k) stajer en kai mìno en tautÐzetai me polu¸numo to polÔ bajmoÔ k.Prgmati h sqèsh f (k) = c1 = (c1z)′ dÐnei f (k−1) = c1z + c2,f (k−2) = c1

2z2 + c2z + c3 kai

suneqÐzontac prokÔptei to zhtoÔmeno. apodeÐxte to Ðdio anaptÔssontac se dunamoseir)

33. JewroÔme to sÔnolo twn sunart sewn pou eÐnai analutikèc se sÔnolo C − A, kai seìla ta shmeÐa tou A parousizei memonwmènh anwmalÐa. BreÐte thn morf twn parapnwsunart sewn gia tic opoÐec isqÔei


|f(z)| ≤ |z21−z3 | se ìla ta z pou èqei nìhma h anisìthta kai exetste poièc èqoun akèraia

epèktash.

Sth sunèqeia epilÔste to Ðdio prìblhma stic peript¸seic

1. |f(z)| ≥ |z2|1−z3 |, 2. |f

(3)(z)| ≤ |z|3,

3. |f (2)(z)| ≥ 2|z|2 4. |f ′(z)| ≤ 5|z3 + z2 + 1|, 5. |f ′(z)| ≥ 5|z3 + z2 + 1|

6. |f(z) ≤ |e 1z − 1| 7. |f(z) ≥ |e 1

z − 1| .'Olec oi parapnw peript¸seic enopoioÔntai apì thn paraktw skhsh

34. An dÔo sunart seic eÐnai analutikèc se sÔnolo C−A, se ìla ta shmeÐa tou A parousizounmemonomènh anwmalÐa kai epipleon isqÔei |f | ≤ |g| tìte uprqei stajer c me f = cg, |c| ≤1.

LÔseic.

Sthn genik perÐptwsh epeid h sunrthsh g den tautÐzetai me thn mhdenik tote kajeshmeÐo tou epipèdou pou pou den h sunrthsh f

gden eÐnai analutik ja eÐnai memonomèno

an¸malo shmeÐo thc. Epeid ìmwc |fg| ≤ 1, dhlad eÐnai topik fragmènh se kje mem-

onomeno an¸malo shmeÐo me bsh to je¸rhma epektashc tou Riemman ja epekteÐnetaianalutik se autì. Telik h sunrthsh f

gja epekteÐnetai se akeraÐa kai epeid eÐnai

fragmènh apì to je¸rhma Liouville prokÔptei ìti eÐnai stajer ,opìte f = cg me |c| ≤ 1.

Efarmìzontac to parapnw sthn arqik prokÔptei ìti to sÔnolo twn sunart sewn pouzhtme perigrfetai apì thn sqèsh

f(z) = c( z2

1−z3 ) me |c| ≤ 1.

Apì autèc mìno h mhdenik eÐnai akeraÐa diìti oi upìloipec den epekteÐnontai analutik stoèna (To ìrio sto èna eÐnai peiro kai èqoume pìlo).

Sthn perÐptwsh 1. antstrèfontai oi rìloi kai èqoume thn sqèsh

z2

1−z3 = cf(z) me |c| ≤ 1. Profan¸c c 6= 0 opìte

f(z) = 1c( z2

1−z3 ) me 0 < |c| ≤ 1. Sthn perÐptwsh aut den uprqei kamÐa akèraia.

Sthn perÐptwsh 2. Meme ton Ðdio trìpo paÐrnoume f (2)(z) = cz2 me |c| ≤ 1 kai diadoqik

f ′(z) = c3z3 + c1 , f(z) = c

12z4 + c1z + c2 me |c| ≤ 1.

Sthn perÐptwsh 3. ergazìmenoi ìpwc sthn dÔo èqoume

f (2)(z) = 1cz2 kai f(z) = 1

12cz4 + c1z + c2 me 0 < |c| ≤ 1.

Sthn perÐptwsh 4. èqoume

f ′(z) = 5c(z3 + z2 + 1) f(z) = 5c( z4

4+ z3

3+ z) + c1 me |c| ≤ 1.


f ′(z) = 5c(z3 + z2 + 1) f(z) = 5

c( z

4

4+ z3

32+ z) + c1 me 0 < |c| ≤ 1.

20

Kai stic tèsserec prohgoÔmenec peript¸seic èqoume mìno akèraiec sunart seic.


f(z) = c(e1z − 1) me |c| ≤ 1 me akeraÐa mìno thn mhdenik .


f(z) = 1c(e

1z − 1) me 0 < |c| ≤ 1 pou kamÐa den eÐnai akeraÐa.

35. An f akeraÐa sunrthsh meleteÐste thn morf thc stic paraktw peript¸seic

1. |f(z)| ≤ a|z|√

2 ìtan |z| ≥ R. 2. |f(z)| ≤ a|z|√

172 ìtan |z| ≥ a. 3. |f ′(z)| ≤

1 + a|z| 12 + |z|√

226 ìtan |z| ≥ a. 4. |f ′(z)| ≤ Log3(1 + 3|z|) + |z| 212 .Oi parapnw peript¸seic enopoioÔntai sto paraktw .

36. An λ0 < λ1 < . . . < λn, an 6= 0 , k akèraioc me 0 ≤ k ≤ λn < k + 1

kai f akeraÐa sunrthsh me |f(z)| ≤ a0 + a1|z|λ1 + . . .+ an|z|λn| ìtan |z| > a

tìte h f tautÐzetai me polu¸numo to polÔ bajmoÔ k.

LÔseic.

Apì thn anisìthta Cauchy gia |z| ≤ r èqoume

|f (n)(0)| ≤ n!|a0|+ |a1|rλ1 + . . .+ |an|rλn

rn=|a0|rn

+|a1|rn−λ1

+ . . .+|an|rn−λk

.

. Otan n > k gia r → +∞ to deÔtero skèloc èqei ìrio to mhdèn opìte f (n)(0) = 0 ìtann > k kai h sunrthsh grfetai

f(z) =+∞∑n=0

f (n)(0)

n!zn =

k∑n=0

f (n)(0)

n!zn.

Efarmìzontac to parapnw sthn perÐptwsh 1 epeid 1 <√

2 < 2 prokÔptei ìti h f eÐnaipolu¸numo to polÔ bajmoÔ èna dhlad èqei thn morf f(z) = cz + d.

Sthn deÔterh perÐptwsh epeid 2 <√

172< 3 h sunrthsh tautÐzetai me polu¸numo to polÔ

bajmoÔ dÔo.

Sthn trÐth perÐptwsh epeid h sunrthsh tautÐzetai me polu¸numo to polÔ bajmoÔ dekapènte

kai sthn tètarth perÐptwsh. epeid Log(3(1+3|z| 12 ))+|z| 32 ≤ 9|z| 12 )+3+|z| 212 h sunrthshtautÐzetai me polu¸numo to polÔ bajmoÔ 10.

37. BreÐte an uprqoun tic analutikèc sunart seic se tìpo OC stic paraktw peript¸seic

1.f( 1n) = 2n+1

2n, 0 ∈ OC.

2. f( 1n) = (−1)n 2n+1

2n, 0 ∈ OC.


3. f( 1+1√3n

) = 8n3+18n3+2

, 1 ∈ OC.

4. f(1 + 1√3n

) = (−1)n 8n3+18n3+2

, 1 ∈ OC.

5. f( 12n

) = n sin 1n− n cos 1

n+ ne

1n , 0 ∈ OC.

6. f(n+1n+2

) = Log 3n+13n+2

27(zn−1)6+154(zn−1)6+1

, 1 ∈ OC.

7. f(1 + 1n) = (1 + 1

n)

1n , 1 ∈ OC

(n = 1, 2, 3 . . .)

LÔsh. Sqetik me to pr¸to er¸thma jewroÔme thn akoloujÐa zn = 1n. H akoloujÐ-

a èqei ìrio to mhdèn pou an kei ston tìpo OC kti pou apoteleÐ proôpìjesh gia naqrhsimopoi soume to je¸rhma tautismoÔ. H upìjesh f(zn) = 3

2+ zn mac upagoreÔei na

qrhsimopoi soume thn sunrthsh g(z) = 32

+ z, z ∈ OC. me lla lìgia oi sunart seic f, geÐnai analutikèc se tìpo OC kai to sÔnolo tautismoÔ touc èqei shmeÐo susswreÔsewc pouan kei ston tìpo , opìte me bsh to je¸rhma tautismoÔ ja isqÔei f = g.

Sqetik me to deÔtero er¸thma. An jewr soume thn akoloujÐa z2n → 0 èqoume f(z2n) =g(z2n) kai me bsh to je¸rhma tautismoÔ f = g.

An jewr soume thn akoloujÐa z2n+1 → 0 èqoume f(z2n+1) = −g(z2n+1) kai me bshto je¸rhma tautismoÔ f = −g. Apì ton sunduasmì twn dÔo prohgoumènwn prokÔpteif = 0, topo pou shmaÐnei ìti den uprqei sunrthsh pou na ikanopoieÐ tic upojèseic touerwt matoc.

Sqetik me to trÐto er¸thma

An jewr soume thn akoloujÐa zn = 1+1√3n

)1 ∈ OC tìte n = 13(zn−1)2

kai f(zn) = 27(zn−1)6+154(zn−1)6+1

.

Efarmìzontac to je¸rhma tautismoÔ èqoume f(z) = 27(z−1)6+154(z−1)6+1

.

Sqetik me to tètarto er¸thma akolouj¸ntac pist th diadikasÐa tou deÔterou apodeÐq-noume ìti den uprqei sunrthsh.

Sqetik me to pèmpto er¸thma .

An zn = 12n→ 0 ∈ OC tìte f(zn) = sin 2zn−cos zn+e2zn

2zn.

JewroÔme t¸ra thn sunrthsh sin 2z−cos z+e2z

2z, z 6= 0 pou faÐnetai na mhn eparkeÐ gia na

efarmìsoume to je¸rhma tautismoÔ. ParathroÔme ìmwc ìti to ìriì thc sto mhdèn uprqeikai isoÔtai me 4. (ParakaloÔme na to upologÐste qwrÐc qr sh kanìna Hospital). Mebsh to je¸rhma epèktashc tou Riemman h sunrthsh epekteÐnetai analutik sto mhdènme thn tim 4. T¸ra mporoÔme na efarmìsoume to je¸rhma tautismoÔ opìte f(z)) =sin 2z−cos zn+e2z

2zìtan z 6= 0 kai f(0) = 4.

Sqetik me to èkto er¸thma èqoume,

zn = n+1n+2→ 1 ∈ OC, f(zn) = Log(5zn−2

4zn−1)

opìte f(z) = Log( 5z−24z−1

) upì thn proôpìjesh ìti o tìpoc den perilambnei kanèna shmeÐo

tou diast matoc [14, 2

5] pou den epekteÐnetai analutik h prohgoÔmenh sunrthsh.

22

Sqetik me to teleutaÐo er¸thma èqoume, zn = 1 + 1n→ 1 ∈ OC , n = 1

zn−1,f(zn) =

z1

zn−1n = e

1zn−1

Logzn kai efarmìzontac to je¸rhma tautismoÔ paÐrnoume f(z) = e1zLogz.

38. An f, g analutikèc sunart seic se tìpo OC pou h f den eÐnai h mhdenik kai h g èqeipargousa thn G, zn akoloujÐa stoiqeÐwn tou tìpou me zn → z0 ∈ OC kai epÐ plèonisqÔei f ′(zn) = f(zn)g(zn), n = 1, 2 . . . tìte f = ceG me c 6= 0.

LÔsh.

Epeid ikanopoioÔntai oi proôpojèseic efarmog c tou jewr matoc tautismoÔ sthn sunrthshf ′ − fg èqoume f ′ − fg = f ′ − fG′ = 0. SuneqÐzoume me thn gnwst diadikasÐa lÔshc thcstoiqei¸douc diaforik c exÐswshc, dhlad e−Gf ′−e−GfG′ = (e−Gf)′ = 0 opìte e−Gf = ckai f = ceG me c 6= 0.

Sthn genikìterh perÐptwsh pou h sunrthsh g den èqei pargousa ston tìpo mporoÔme nathn periorÐsoume se opoiod pote dÐsko me kèntro to z0 pou epeid eÐnai kurtìc tìpoc èqeipargousa se autìn kai epilÔoume to prìblhma se autìn. Epeid kje lÔsh èqei to polÔmÐa monadik epèktash se ìlo ton tìpo exetzoume an uprqei tètoia. An den uprqei tìteden èqoume lÔsh se olìklhro ton tìpo.

'Ola ta parapnw den eÐnai anagkaÐa sthn perÐptwsh pou mporoÔme thn g na thn jèsoumesthn morf g = h′

hopìte apì thn sqèsh f ′ − fg = f ′h − fh′ = 0 paÐrnoume (f

h)′ = 0kai

f = ch.

39. Na eurejoÔn oi mh mhdenikèc sunart seic analutikèc sunart seic se tìpoOC stic paraktwpeript¸seic.

1. n2f ′( 1√n) = f( 1√

n)(2n2 + 1 + 1

n3 + 1n3 + 3

n4 , 0 ∈ OC

2.f ′( 12n

) = f( 12n

)(cos(1+nn

) + sin 13n− e 5

n + Log(2n+12n

)),0 ∈ OC.

3. f ′( 1n) = f( 1

n)( n2+2n+3n3+n2+n+1

),0 ∈ OC.(n = 1, 2, 3 . . .) .

LÔsh. Sqetik me to er¸thma 1.

Jètoume zn = 1√n→ 0 ∈ OC, opìte f ′(zn) = f(zn)(2 + z4

n + z10n + 3z16

n ). JewroÔme thn

sunrthsh g(z) = (2 + z4 + z10 + 3z16) pou èqei pargousa se ìlo to migadikì epÐpedothn G(z) = 2z + z5

5+ z11

11+ 3z17

17. Antigrfontac sth sunèqeia thn prohgoÔmenh apìdeixh

kai paÐrnoume f = ceG me c 6= 0.

Sqetik me to deÔtero er¸thma .

Jètoume zn = 12n→ 0 ∈ OC opìte

f ′(zn) = f(zn)(cos(1 + 2zn) + sin(2zn3

)− e10zn + Log(zn + 1), 0 ∈ OC.


JewroÔme thn sunrthsh

g(z) = (cos(1 + 2z) + sin(2z

3)− e10z + Log(z + 1).

Gia thn pargousa efarmìzontai kont sto mhdèn oi klasikèc mèjodoi eÔreshc tou aorÐstouoloklhr¸matoc opìte

G(z) =sin(1 + 2z)

2− 3

2cos

2z

3− 1

10e10z + (z + 1)Log(z + 1)− Log(z + 1).

kai f = ceG me c 6= 0 upì thn proôpìjesh ìti o tìpoc den perièqei kanèna shmeÐo thcpragmatik c hmieujeÐac x ≤ −1 pou den epekteÐnetai analutik h sunrthsh Log(z + 1).

Sqetik me to trÐto er¸thma. Jètoume zn = 1n→ 0 ∈ OC, opìte

f ′(zn) = f(zn)(3z2

n + 2zn + 1

1 + zn + z2n + z3

n

)

kai me to je¸rhma tautismoÔ

f ′(z) = f(z)(3z2 + 2z + 1

1 + z + z2 + z3) = f(z)

h′(z)

h(z)

ìpou h(z) = 1 + z+ z2 + z3. Apì thn teleutaÐa sqèsh mesa prokÔptei (fh)′ = 0 f = ch.

40. An se mh stajer sunrthsh analutik f orismènh se tìpo OC isqÔei Ref ≥ 0 apodeÐxteìti Ref(z) > 0 gia kaje z ∈ OC.

LÔsh. An sunèbaine Ref(z0) = 0 epeid prìkeitai gia m stajer analutik sunrthshse tìpo apì thn arq tou elaqÐstou gia pragmatik èqoume ìti uprqei z1 ∈ OC meRef(z1) < Ref(z0) = 0 pou sugkroÔetai me thn upìjesh Ref ≥ 0,

41. An f analutik sunrthsh sto dÐsko D(0, 4) kai epiplèon isqÔei |f(z)| < 6 − Log(5 −|z|)− |z| 12 − |z| gia kje z ∈ D(0, 4) apodeÐxte ìti f = 0.

LÔsh.

JewroÔme tuqìn ζ ∈ D(0, 4) me r > |ζ| kai efarmìzontac thn arq tou megÐstou èqoume

|f(ζ)| ≤ max|z|=r |f(z)| ≤ 6− Log(5− r)− r 12 − r opìte gia r → 4 paÐrnoume |f(ζ)| ≤ 0.

42. An f analutik sunrthsh se tìpo OC kai uprqei a ∈ OC ¸ste |f(z)| ≥ |f(a)| > 0 seanoikt perioq tou a na apodeiqjeÐ ìti h f eÐnai stajer ston tìpo

LÔsh.

24

An h sunrthsh den eÐnai stajer sthn perioq tou a epeid den èqei rÐzec se aut epitrèpetai h efarmog thc arq c tou elaqÐstou sÔmfwna me thn opoÐa apokleÐetai tona paÐrnei h sunrthsh elaqÐsth tim sto a diìti eÐnai eswterikì shmeÐo. Epeid h upì-jesh mac to dÐnei san jèsh elaqÐstou prokÔptei ìti h sunrthsh ja eÐnai stajer sthnperioq . Apì to je¸rhma tautismoÔ prokÔptei ìti eÐnai stajer ston tìpo.

43. An f analutik sunrthsh se tìpo OC kai uprqei kleist kampÔlh γ ston tìpo tètoia¸ste Ref(z) ≥ 0 gia kje z ∈ γ∗ na apodeiqjeÐ ìti h f eÐnai stajer ston tìpo.

LÔsh.

An G to eswterikì thc kampÔlhc tìte γ∗ = ∂G kai h lÔsh dÐnetai sto sqetikì pardeigmatou edafÐou thc arq c tou megÐstou.

(Thn upìjesh Ref(z) ≥ 0 gia kje z ∈ γ∗ mporoÔme na thn antikatast soume me aRef +bImf + c ≥ 0 gia kje z ∈ γ∗ me a + ıb 6= 0. EpÐshc sthn perÐptwsh pou h sunrthshden èqei rÐzec ston tìpo mporoÔme na thn antikatast soume me thn upìjesh ìti to mètroparamènei stajerì epnw sto Ðqnoc thc kampÔlhc.)

44. An γ1, γ2 jetik prosanatolismènec kampÔlec pou ta Ðqnh touc perigrfontai apì tic sqè-seic

γ1 : x2 + y2

4= 1 me x ≥ 0, y ≥ 0 , γ2 : x2 + y2 = 1, me x ≤ 0, y ≤ 0 kai γ3 = [−ı, 1]

na upologÐste ta oloklhr¸mata∫γ1fk(z)dz,

∫γ1+∗γ2

fk(z)dz,∫γ1+∗γ2+∗γ3

fk(z)dz me f1(z) = z kai f2(z) = sin2 z

LÔsh.

Jètoume

z = γ1(t) = cos t + 2ı sin t me t ∈ [0, π], dz = (− sin t + 2ı cos t)dt kai z = cos t − 2ı sin topìte ∫

γ1

f1(z)dz =

∫γ1

zdz =

∫ π

0

(2ı+3 cos 2t

2) = 2πı.

Jètoume

z = γ2(t) = eıt, z = e−ıt me t ∈ [π, 3π2

] kai dz = ıeitdt opìte

∫γ2

f1(z)dz =

∫γ2

zdz =

∫ 3π2

π

ıdt =πı

2.

Jètoume

z = γ3(t) = t(1 + ı)− ı me t ∈ [0, 1] z = t(1− ı) + ı kai dz = (1 + ı)dt opìte


∫γ3f1(z)dz =

∫γ3zdz =

∫ 1

0(2t+ ı− 1)dt = ı.

Sth sunèqeia paÐrnoume ta antÐstoiqa ajroÐsmata.

Sqetik me thn sunrthsh f2(z) = sin2 z = 12− cos 2z

2parathroÔme ìti èqei pargousa thn

F (z) = z2− sin 2z

4opìte∫

γ1

f2(z)dz = F (−1)− F (1),

∫γ1+∗γ2

f2(z)dz = F (ı)− F (1)

kai ∫γ1+∗γ2+∗γ3

f2(z)dz = F (1)− F (1) = 0.

45. Na upologist to olokl rwma In =∫γnLogzdz ìpou γn eÐnai h polugwnik gramm

[−1− ın,−1− ı, 1, 1− ı,−1 + ı

n], n = 1, 2, . . .. Sth sunèqeia na upologisteÐ to lim In.

LÔsh.

An jewr soume ton tìpo C−π pou prokÔptei an apì to epÐpedo afairèsoume thn arnhtik

hmieujeÐa kai to mhdèn tìte h sunrthsh Logz èqei pargousa se autìn thn sunrthshF (z) = zLogz − z kai epeid h polugwnik gramm perièqetai ston tìpo èqoume

In = F (−1+ı

n)−F (−1− ı

n) = (−1+

ı

n)Log(−1+

ı

n)−(−1+

ı

n)−(−1− ı

n)Log(−1− ı

n)+(−1− ı

n).

JewroÔme thn akoloujÐa Arg(−1 − ın) parathroÔme ìti (−1 − ı

n) → −1 apì to ktw

hmiepÐpedo opìte

Arg(−1− ı

n)→ −πkai Log(−1− ı

n) = Log|(−1− ı

n)|+ ıArg(−1− ı

n)→ −π.

OmoÐwc paÐrnoume ìti Log(−1 + ın)→ π opìte lim In = 2π.

46. Na upologisteÐ to epikampÔlio olokl rwma thc sunrthshc f(z) = 1z2+1

+ 11−z2 +cos 1

z−ı +ez

z+ıupernw thc kampÔlhc tou paraktw sq matoc qrhsimopoi¸ntac epoptikì upologismì

twn deikt¸n strof c n(γ,±1), n(γ,±ı) qwrÐc qr sh jewrhmtwn Cauchy.

Sq ma 1:

LÔsh.

Katarq n 1z2+1

= ı2

1z+ı− ı

21z+ı

opìte lambnontac upìyh ton orismì tou deÐkth strof c

n(γ, a) = 12πı

∫γ

dzz−a paÐrnoume

26

∫γ

1

z2 + 1dz = −πn(γ,−ı) + πn(γ, ı)

kai omoÐwc

∫γ

1

z2 − 1dz = πın(γ, 1)− πın(γ,−1).

An apì to anptugma

sin1

z − ı=

1

z − ı− 1

3!(z − ı)3+ . . .

exairèsoume ton arqikì ìro to upìloipo mèroc èqei pargousa pou shmaÐnei ìti mhdenÐzetaito olokl rwm tou opìte

∫γ

sin1

z − ı=

∫γ

1

z − ıdz = 2πın(γ, ı).

Epeid

ez

z + ı= e−ı

ez+ı

z + ı=

e−ı

z + ı(1 + (z + ı) + . . .) =

e−ı

z + ı+ . . .

èqoume ìti

∫γ

ez

z + ı= 2πın(γ,−ı).

To jroisma twn epimèrouc oloklhrwmtwn dÐnei to zhtoÔmeno. Mac apomènoun oi deÐktecstrof c. Enac sÐgouroc trìpoc epoptikoÔ upologismoÔ ìso sÔnjeto sq ma kai an èqounoi kleistèc kampÔlec eÐnai o ex c. Fèroume opoiad pote hmieujeÐa me arq to shmeÐo poujèloume na upologÐsoume ton deÐkth strof c tou. Gia thn dik mac dieukìlunsh protimmekapoia na tèmnei ìso ligìterec forèc mporoÔme to Ðqnoc thc kampÔlhc. Sth sunèqeia sekje shmeÐo pou h hmieujeÐa tèmnei to Ðqnoc thc kampÔlhc antistoiqoÔme ton arijmì +1 −1 anloga an thn tèmnei kat thn jetik arnhtik for. To algebrikì jroismadÐnei ton deÐkth strof c. An h hmieujeÐa den tèmnei o deÐkthc strof c eÐnai mhdèn. SthnperÐptws mac n(γ, 1) = 1−1−1 = −1, n(ı, γ) = +1−1−1+1 = 0, n(γ, 0) = −1−1 = −2,n(γ,−ı) = +1 + 1− 1 = 1,n(γ,−1) = −1− 1 = −2

47. An mia analutik sunrthsh f parousizei memonwmènh anwmalÐa sto mhdèn breÐte ticdunatèc anwmalÐec stic paraktw peript¸seic.

1.f( 1n) → 0. 2. f( 1

n) → +∞ 3. f( 1

2n) → 1 kai f( 1

2n+1) → −1. 4. f( 1

n) → 0 kai 5.

f( ın) → +∞. Sthn perÐptwsh 3 exetste an uprqei zn → 0 me f(zn) → +∞ kai sthn

perÐptwsh 4 an uprqei zn → 0 me f(zn)→ 8.


LÔsh. Sthn pr¸th perÐptwsh apokleÐetai h perÐptwsh tou pìlou opìte oi endeqìmenecanwmalÐec eÐnai ousi¸dhc apaleÐyimh. Sthn deÔterh perÐptwsh apokleÐetai h apaleÐyimh.Sthn trÐth kai tètarth perÐptwsh efìson den èqoume eniaÐo ìrio apokleÐontai o pìloc kaito apaleÐyimo opìte ja èqoume ousi¸dh anwmalÐa kai apì to je¸rhma C-W opoiod poteshmeÐo tou epipèdou apoteleÐ ìrio katllhlhc akoloujÐac f(zn) me zn → 0. MporoÔme naapodeÐxoume ìti ìrio tètoiac akoloujÐac eÐnai kai to +∞. Pragmati efìson to f(D(0, 1

n))

eÐnai puknì sto migadikì epÐpedo sÔmfwna me to je¸rhma C-W uprqei zn ∈ D(0, 1n) me

f(zn) ∈ D(n, 1) opìte zn → 0 kai f(zn)→ +∞.

Ta parapnw den apoteloÔn oloklhrwmènh apnthsh diìti prèpei na apodeÐxoume ìti oiparapnw peript¸seic eÐnai kai dunatèc , dhlad kje perÐptwsh prèpei na sunodeuteÐ kaime antÐstoiqo pardeigma.

Sqetik me thn perÐptwsh 1 . Oi sunart seic z2

z, ze

2πz ikanopoioÔn tic proôpojèseic all

to mhdèn sthn pr¸th eÐnai apaleÐyimo sthn deÔterh ousi¸dec an¸malo shmeÐo.

Sqetik me thn perÐptwsh 2 . Oi sunart seic 1z, e

2πz

zikanopoioÔn tic proôpojèseic all

to mhdèn sthn pr¸th eÐnai pìloc kai sthn deÔterh ousi¸dec an¸malo shmeÐo.

H sunrthsh e2πız apoteleÐ katllhlo pardeigma gia thn perÐptwsh 3. kai h sunrthsh

ze2πız apoteleÐ katllhlo pardeigma gia thn perÐptwsh 4.

48. 1. QarakthrÐste thn anwmalÐa sto mhdèn twn sunart sewn fgkai g

fme thn bo jeia thc

txhc thc tou wc rÐza twn sunart sewn

f(z) = z2(e2z2 − 2z2 − 1) kai g(z) = 3 sin(z9) + z39 − z9.

2. Me ton Ðdio trìpo qarakthrÐste ìla ta an¸mala shmeÐa sthn perÐptwsh f(z) = z sin z,g(z) = cos z.

LÔsh. Epeid g′(z) = z8h(z), h(z) = 27 cos(z9) + 39z30 − 9 kai h(0) 6= 0, to mhdèn eÐnairÐza txhc 8 sthn g′ kai epomènwc txhc 9 sthn g.

An to mhdèn eÐnai rÐza txhc k sthn f tìte profan¸c èqei txh k − 2 sthn sunrthsh(e2z2 − 2z2− 1), k− 3 sthn pargwgì thc 4z((e2z2 − 1)), k− 4 sthn ((e2z2 − 1)) kai k− 5sthn pargwgì thc 4z(e2z2). T¸ra eÐnai profanèc ìti k − 5 = 1.

Me bsh ta parapnw to mhdèn eÐnai pìloc txhc 9−6 = 5 sthn fgkai rÐza txhc 9−6 = 5

sthn gf.

Sqetik me to deÔtero er¸thma eÔkola diapist¸noume ìti to mhdèn eÐnai rÐza txhc 2 kaioi arijmoÐ kπ, k = ±1,±2 . . . aplèc rÐzec sthn f . EpÐshc oi arijmoÐ kπ+ π

2, k = ±1,±2 . . .

eÐnai aplèc rÐzec sthn g. Epeid den uprqei koin rÐza twn sunart sewn prokÔptei ìtioi rÐzec thc g apoteloÔn aploÔc pìlouc sthn f

g, to mhdèn pìlo deutèrac txhc kai oi

upìloipec rÐzec thc f aploÐ pìloi sthn gf.

49. An z1, z2, a, b ∈ C me a 6= b d¸ste paradeÐgmata analutik¸n sunart sewn sto C−z1, z2me Re(f, z1) = a,Re(f, z2) = b stic paraktw peript¸seic

28

1. z1 pìloc txhc k, z2 pìloc txhc m.

2 . z1 pìloc txhc k, z2 ousi¸dec an¸malo.

3. z1 ousi¸dec an¸malo, z2 ousi¸dec an¸malo.

LÔsh.

An

f1(z) = 1(z−z1)k

+ az−z1 , f2(z) = 1

(z−z2)m+ b

z−z2 kai f3(z) = ea

z−z1 , f4(z) = eb

z−z2

tìte oi sunart seic g1 = f1 + f2, g2 = f1 + f4 kai g3 = f3 + f4 ikanopoioÔn thn pr¸th, thndeÔterh kai thn trÐth perÐptwsh antÐstoiqa.

50. An z0 apaleÐyimo, pìloc txhc k, ousi¸dec an¸malo shmeÐo sunrthshc f tìte eÐnai a-paleÐyimo pìloc txhc k + 1 ousi¸dec an¸malo antÐstoiqa sthn sunrthsh f ′. Me thnbo jeia tou parapnw apodeÐxte ìti an z0 eÐnai pìloc thc sunrthshc f tìte eÐnai ousi¸decan¸malo sthn sunrthsh ef .

LÔsh.

PaÐrnontac ta anaptÔgmata Laurent∑+∞

n=−∞ an(z − z0)n kai∑+∞

n=−∞ nan(z − z0)n−1 twnsunart sewn f kai f ′ antÐstoiqa kai susqetÐzontc ta me touc orismoÔc twn an¸malwnshmeÐwn sthn kje perÐptwsh prokÔptei mesa to pr¸to mèroc. Apì to pr¸to mèroc epÐshcprokÔptei ìti apokleÐetai h pargwgoc na èqei na èqei to z0 pìlo txhc 1.

An z0 eÐnai pìloc txhcm thc sunrthshc ef opìte eÐnai pìloc txhcm+1 thc sunrthshc(ef )′ = f ′ef pou shmaÐnei ìti eÐnai pìloc txhc èna sthn f ′ kti pou èqei apokleisteÐ. An z0

eÐnai apaleÐyimo kai rÐza txhc m thc sunrthshc epèktashc thc ef tìte eÐnai pìloc txhcm thc sunrthshc 1

ef= e−f opìte epanerqìmaste sthn prohgoÔmenh perÐptwsh. An z0

eÐnai apaleÐyimo sth sunrthsh ef qwrÐc na eÐnai rÐza thc epèktashc tìte eÐnai apaleÐyimo

kai sthn sunrthsh (ef )′

ef= f ′ diìti me bsh to pr¸to skèloc kai thn upìjesh ja eÐnai

pìloc sthn f ′. H mình perÐptwsh pou apomènei eÐnai tou ousi¸douc an¸malou shmeÐou.

51. An f akeraÐa sunrthsh apodeÐxte ìti

n!

∫γ

f(z)

(z − a)n+1dz = (n− k)!

∫γ

f (k)(z)

(z − a)n+1−k dz, k = 0, 1, 2 . . . .

LÔsh. PaÐrnoume dÔo forèc thn tim f (n)(a) me ton tÔpo tou Cauchy. Thn pr¸th forton efarmìzoume sthn sunrthsh f gia thn n-txhc pargwgo kai thn deÔterh for sthnsunrthsh f (k) gia thn (n− k)- txhc pargwgo.

52. Na genikeÔsete to sumpèrasma thc prohgoÔmenhc skhshc sthn perÐptwsh pou h sunrthshf eÐnai analutik sto C− 0.


LÔsh.

An jewr soume ta anaptÔgmata Laurent

f(z) =+∞∑

n=−∞

an(z − a)n, f ′(z) =+∞∑

n=−∞

nan(z − a)n−1

tìte

f(z)

(z − a)n+1= . . .+

anz − a

+ . . . ,f ′(z)

(z − a)n= . . .+

nan

z − a+ . . . ,

∫γ

f(z)

(z − a)n+1dz = 2πın(γ, a)an,

∫γ

f ′(z)

(z − a)ndz = 2πın(γ, a)nan = n

∫γ

f(z)

(z − a)n+1dz.

Pollaplasizontac me (n−1)! paÐrnoume to zhtoÔmeno gia thn pr¸th pargwgo kai suneq'-izontac diadoqik paÐrnoume to sumpèrasma gia opoiod pote k.

53. JewroÔme tic sunart seic

z2n sin2k 1z, z2n+1 sin2k+1 1

z,(z − 1)2n cosk 1

(z−1)n, cos 1

2z−1e(z− 1

2)2 .

Na apodeiqjeÐ ìti ìpou èqoun nìhma ta epikampÔlia oloklhr¸mata twn parapnw sunart -sewn se kleistèc kampÔlec eÐnai mhdèn.

LÔsh. H apìdeixh sthrÐzetai sthn ex c apl all idiaÐtera qr simh parat rhsh. Anmia analutik sunrthsh eÐnai rtia to anptugma Laurent ekfrzetai san jroisma ìr-wn rtiac txhc me thn ènnoia ìti oi peritt c txhc mhdenÐzontai opìte to oloklhrwtikìupìloipo sto mhdèn eÐnai mhdèn An eÐnai peritt mhdenÐzontai oi ìroi rtiac txhc. To pr¸-to prokÔptei mesa apì thn sqèsh

∑+∞n=−∞ anz

n = f(z) = f(−z) =∑+∞

n=−∞(−1)nanzn,

opìte a2k+1 = 0. Genik¸tera an h sunrthsh eÐnai << rtia wc proc z − a >> dhlad f(a + Z) = f(a − Z), Z = z − a tìte to anptugma Laurent sto a ekfrzetai san -jroisma ìrwn rtiac txhc kai to antÐstoiqo oloklhrwtikì upìloipo eÐnai mhdèn. 'Olec oisunart seic tou probl matoc èqoun akrib¸c èna an¸malo shmeÐo kai èqoun thn prohgoÔ-menh idiìthta na eÐnai summetrikèc wc proc autì opìte to oloklhrwtikì upìloipo kai wcek toÔtou to olokl rwma eÐnai mhdèn.

54. JewroÔme tic sunart seic

z2n+1 sin2k 1z, z2n sin2k+1 1

z,(z − 1)2n+1 cosk 1

(z−1)n, cos 1

2z−1e(z− 1

2)3 .

Na apodeiqjeÐ ìti ta oloklhr¸mata twn parapnw sunart sewn upernw thc kampÔlhcC(1

4, 1

8) eÐnai mhdèn.

LÔsh.

30

Epeid se ìlec tic peript¸seic ta memonwmèna an¸mala shmeÐa eÐnai sto exwterikì thc kam-pÔlhc oi deÐktec strof c mhdenÐzontai opìte ja mhdenÐzontai kai ta antÐstoiqa epikampÔliaoloklhr¸mata.

55. Na upologistoÔn ta epikampÔlia oloklhr¸mata upernw thc kampÔlhc C(0, 4) twn sunart -sewn.

z339k−1sin1

zk, z339k sin

1

zk, z2 cos

1

(3z − 1), z cos2 z

(3z − 1).

LÔsh.

Epeid

z339k−1sin1

zk= z339k−1

+∞∑n=0

(−1)n

(2n+ 1)!(

1

zk)2n+1 =

+∞∑n=0

(−1)n

(2n+ 1)!

1

zk(2n+1)−339k+1,

h sqèsh k(2n+ 1)− 339k+ 1 = 1 dÐnei n = 169, o antÐstoiqoc suntelest c pou tautÐzetaime to oloklhrwtikì upìloipo eÐnai −1

339!kai epeid o deÐkthc strof c eÐnai èna to antÐstoiqo

olokl rwma eÐnai −2πı339!

.

Sto epìmeno olokl rwma akolouj¸ntac thn Ðdia diadikasÐa katal goume sthn exÐswshk(2n+ 1)− 339k = 1 pou den èqei akèraia lÔsh. Autì shmaÐnei ìti o ìroc txhc −1 poumac endiafèrei èqei paralhfjeÐ apì to jroisma epeid èqei suntelest mhdèn. Epeid tooloklhrwtikì upìloipo eÐnai mhdèn ja mhdenÐzetai kai to antÐstoiqo olokl rwma.

Epeid h epìmenh sunrthsh èqei memonwmèno an¸malo shmeÐo to 13gia praktikoÔc lìgouc

jètoume

z − 13

= Z opìte

z2 cos1

(3z − 1)= (

1

9+

2

3Z + Z2)

+∞∑n=0

(−1)n

(2n+ 1)!32n+1z2n+1=

+∞∑n=0

1

9

(−1)n

(2n+ 1)!32n+1Z2n+1+

+∞∑n=0

(−1)n

(2n+ 1)!32n+1Z2n−1+

+∞∑n=0

2

3

(−1)n

(2n+ 1)!32n+1Z2n.

Epilègontac kai ajroÐzontac touc suntelestèc txhc −1 sta parapnw ajroÐsmata brÐsk-oume oloklhrwtikì upìloipo sto 1

3thn tim 1

9.3− 1

3!33 + 0 = 154

opìte to olokl rwma èqeithn tim 2πı

54.

56. Na upologistoÔn ta epikampÔlia oloklhr¸mata

In =

∫C(0,nπ)

ez − e−z

ez + e−zdz, n = 1, 2 . . . .


LÔsh. H exÐswsh ez + e−z = 0 isodÔnama e2z = −1 = eıπ dÐnei rÐzec zk = ıπ2

+ ıkπ,k = 0 ± 1 ± 2 . . .. H pargwgoc tou paranomast stic rÐzec tou den mhdenÐzetai opìteeÐnai aplèc. EpÐshc kamÐa rÐza den mhdenÐzei ton arijmht opìte apoteloÔn pìlouc pr¸thc

txhc sto phlÐko. Sthn perÐptwsh aut Res( ez−e−zez+e−z

, zk) = ( (ez−e−z)(ez−e−z)′

)z=zk = 1

H sqèsh zk ∈ D(0, πn) dhlad −nπ < π2

+ kπ < nπ isodunameÐ me −n ≤ k ≤ n opìteIn = 2πı

∑nk=−nRes(f, zk) = 2πı(2n+ 1).

57. Na upologistoÔn ta kÔria mèrh twn anaptugmtwn Laurent me kèntro to mhdèn kai taantÐstoiqa oloklhrwtik upìloipa twn sunart sewn

z91 sin 1z, e

2z2−1

z99, 1

sink zgia k = 1, 2, 3, 4, kai (ez−1)k

sink+2 zgia opoiod pote akeraio k.

LÔsh.

Katarq n

z91 sin1

z= z91

+∞∑n=0

(−1)n1

(2n+ 1)!z2n+1=

+∞∑n=0

(−1)n1

(2n+ 1)!z2n+1−90.

Gia to kÔrio mèroc paÐrnoume to jroisma twn ìrwn me 2n+ 1 > 90

dhlad to+∞∑n=45

(−1)n1

(2n+ 1)!z2n+1−90.

Gia to oloklhrwtikì upìloipo jètoume 2n + 1 − 90 = 1, dhlad n = 45 pou eÐnai osuntelest c − 1

(91)!.

H sunrthsh e2z2−1

z99grafetai

e−1

z99

+∞∑n=0

2nz2n

(2n)!=

+∞∑n=0

e−12nz2n−99

(2n)!.

Gia to kÔrio mèroc paÐrnoume 2n < 99 dhlad to∑49

n=0e−12nz2n−99

(2n)!.

Gia to oloklhrwtikì upìloipo jètoume 2n− 99 = −1,

dhlad n = 49 pou antistoiqeÐ ston suntelest e−1249

98!.

Epeid zsin z→ 1 ìtan z → 0 , To mhdèn eÐnai pìloc pr¸thc txhc sthn 1

sin zto oloklhrwtikì

upìloipo sto mhdèn eÐnai èna kai to kÔrio mèroc eÐnai 1z.

Epeid z2

sin2 z→ 1 ìtan z → 1 , to mhdèn eÐnai pìloc deutèrac thn 1

sin2zopìte,

32

1

sin2 z=a−2

z2+a−1

z+ . . . ,

z2

sin2 z= a−2 + a−1z + . . . .

Epeid h sunrthsh eÐnai rtia oi ìroi peritt c txhc sthn seir mhdenÐzontai opìte a1 = 0.

Apì to ìrio sto mhdèn tou teleutaÐou anaptÔgmatoc prokÔptei a−2 = 1 kai kÔrio mèroc1z2.

Epeid h sunrthsh 1sin3 z

eÐnai peritt ja mhdenÐzontai oi ìroi artÐac txhc sto anptugmaLaurent kai epeid to mhdèn eÐnai pìloc trÐthc txhc èqoume thn morf

1

sin3 z=a−3

z3+a−1

z+ . . . .

An pollaplasisoume thn sunrthsh me z3 paÐrnoume thn sunrthsh

F (z) =z3

sin3 z= a−3 + a−1z

2 + . . .

pou profan¸c epekteÐnetai analutik sto mhdèn. kai isqÔei F (0) = a−3,F ′′

2!(0) = a−1 . . .

An den parèmboume sthn parapnw morf ja qreiasteÐ na upostoÔme epÐponh diadikasÐakìpwshc me pollaplèc efarmogèc tou kanìna Hospital . Ja to apofÔgoume ìmwc knontacèxupnh qr sh gnwst¸n anaptugmtwn Taylor.

Sthn perÐptws mac grfoume

sin z = z − z3

3!+ . . . = zq(z), q(z) = 1− z2

3!+ . . . , q(0) = 1, 0 = q′(0),− 1

3!=q”(0)

2!.

Epeid F (z) = 1q3(z)

èqoume

a−3 + a−1z2 + . . . = F (z) = q−3(z) = q−3(0)z +

(q−3)′′(0)

2!+ . . . .

Epeid

q−3(0) = 1, (q−3)′(0) = (−3q−4q′)(0) = 0, (q−3)′′(0) = (12q−5q′ − 3q−4q′′)(0) = 2

èqoume a−3 = 1, a−1 = 1.

Apì ta parapnw prokÔptei ìti kÔrio mèroc isoÔtai me 1z3

+ 1z. kai ìti to antÐstoiqo

oloklhrwtikì upìloipo sto mhdèn me 1.

OmoÐwc

1

sin4 z=a−4

z4+a−2

z2+ . . .


opìte

a−4 + a−2z2 + . . . =

z4

sin4 z=

1

q4= (

1

q4)(0) + (

1

q4)′(0)z +

1

2!(

1

q4)′′(0)z2 + . . .

kai telik

a−4 = 1, a−2 =1

2!(

1

q4)′′(0) =

1

2(20q−6q′ − 4q−5q′′)(0) = −2

3.

Profan¸c to kÔrio mèroc eÐnai 1z4− 2

3z3kai to antÐstoiqo oloklhrwtikì upìloipo eÐnai

ìpwc anamenìtan mhdèn.

Anloga ergazìmenoi kai sthn teleutaÐa perÐptwsh jètoume

ez − 1 = zh, h(z) = 1 +1

2!z +

1

3!z2 + . . .

opìte

h(0) = 1, h′(0) =1

2, h′′(0) =

1

3,

(ez − 1)k

sink+2 z=

hk

z2qk+2=a−2

z2+a−1

z+ . . .

kai

a−2 = (hk

qk+2)(0) = 1, a−1 = (

hk

qk+2)′(0) =

k

3.

58. An mia sunrthsh f eÐnai analutik sto C−0 apodeÐxte ìti∫ 2π

0f(reit)dt = 2πRes(f(z)

z, 0).

sth sunèqeia upologÐste ta oloklhr¸mata∫ 2π

0eae

ıktdt,∫ 2π

0cos (aeıkt)dt,

∫ 2π

0sin (aeıkt)dt, kai

∫ 2π

0ea cos kt cos(a sin kt)

ìpou a pragmatikìc kai k fusikìc arijmìc.

LÔsh.

An jèsoume z = reıt, dz = ıreıtdt, dt = dzız

èqoume∫ 2π

0

f(reit)dt =1

ı

∫C(0,r)

f(z)

zdz = 2πRes(

f(z)

z, 0).

To pr¸to olokl rwma apoteleÐ efarmog tou parapnw gia f(z) = eazk.

Epeid z f(z)z→ f(0) = 1 ìtan z → 0 to mhdèn eÐnai pìloc pr¸thc txhc sthn f(z)

zisqÔei

Res(f(z)z, 0) = 1 opìte me bsh to pr¸to skèloc to olokl rwma ja èqei tim 2π.

Sthn deÔterh perÐptwsh qrhsimopoioÔme thn sunrthsh cos(azk) kai epanalambnoume thndiadikasÐa thc prohgoÔmenhc perÐptwshc.

34

Sthn trÐth perÐptwsh qrhsimopoioÔme thn sunrthsh sin(azk) kai epeid to mhdèn eÐnai

apaleÐyimo sthn sin(azk)z

to oloklhrwma mhdenÐzetai. Sthn teleutaÐa perÐptwsh an denxefÔgei apì thn prosoq mac ìti h upì olokl rwsh sunrthsh eÐnai to pragmatikì mèrocthc sunrthshc tou pr¸tou oloklhr¸matoc apantme mesa ìti h tim tou oloklhr¸matoceÐnai èna. An xefÔgei ja èljoume sthn katllhlh morf an efarmìsoume ton tÔpo cosw =eıw+e−ıw

2.

59. QarakthrÐste ta memonwmèna an¸mala shmeÐa kai upologÐste ta antÐstoiqa oloklhrwtikupìloipa twn sunart sewn

sinn z

(eız − 1)k+n,

sinn 1z

(eı1z − 1)k+n

, k = 0, 1, 2., e3

2z−1 cos(3

2z − 1),

1 + z + z2 + z3 + z4 + z5

1 + z + z2, e

2z − 23

3!z3− 24

4!z4− . . . .

LÔsh.

Sqetik me thn pr¸th sunrthsh parathroÔme ìti oi sunart seic sin z, (eız − 1) èqoun toÐdio sÔnolo riz¸n pou eÐnai ta akèraia pollaplsia tou π kai eÐnai ìlec aplèc. H diadikasÐaìmwc ja dieukolunjeÐ an knoume thn antikatstash

sin z =eız − e−ız

2ı=e−ız(1 + eız)(eız − 1)

2ı

opìte

sinn z

(eız − 1)k+n=e−ınz(1 + eız)n

2nın(eız − 1)k.

Gia k = 0 eÐnai profanèc ìti ìla ta an¸mala shmeÐa eÐnai apaleÐyima kai h sunrthsh èqeiakèraia epèktash.

Gia k = 1 ìlec oi rÐzecmπ tou paranomast eÐnai txhc èna kai den mhdenÐzoun ton arijmht epomènwc eÐnai aploÐ pìloi kai to oloklhrwikì upìloipo dÐnetai apì thn tim

(e−ınz(1 + eız)n)z=mπ(2nın(eız − 1))′z=mπ

pou mhdenÐzetai gia n perittì kai paÐrnei thn tim 1ın+1 gia n rtio.

Sqetik me thn deÔterh sunrthsh èqoume katarq n memonwmèno an¸mala shmeÐa ta zm =1mπ,m = ±1,±2, . . .. Ed¸ qreizetai prosoq sthn pagÐda tou mhdenìc pou eÐnai an¸ma-

lo all ìqi memonvmèno (h akoloujÐa twn prohgoumènwn memonomènwn an¸malwn shmeÐwnsugklÐnei sto mhdèn) kai den isqÔoun ta gnwst perÐ oloklhrwtik¸n upoloÐpwn. Gia taupìloipa an¸mala shmeÐa h lÔsh af netai ston anagn¸sth na suneqÐsei akolouj¸ntacthn diadikasÐa thc prohgoÔmenhc perÐptwshc.


Sqetik me thn trÐth sunrthsh an jèsoume Z = z − 12èqoume

e3

2z−1 cos(2

2z − 1) = e

32Z cos

1

Z= e

32Z (

eıZ + e

−ıZ )

2=

1

2e

3+2ı2Z +

1

2e

3−2ı2Z =

+∞∑n=0

1

2n!

(3 + 2ı)n

2nZn+

+∞∑n=0

1

2n!

(3− 2ı)n

2nZn=

+∞∑n=0

1

2n!

((3 + 2ı)n + (3− 2ı)n)

2nZn.

Apì thn parapnw morf eÐnai profanèc ìti to 12eÐnai ousi¸dec an¸malo shmeÐo kai to

antÐstoiqo oloklhrwtikì upìloipo eÐnai 32.

Sqetik me thn epìmenh perÐptwsh an den proume upìyh tic rÐzec kai pollaplasisoumearijmht kai paranomast me z − 1 h sunarthsh gÐnetai z

6−1z3−1

= z3 + 1 pou shmaÐnei ìti toìrio thc arqik c sunrthshc uprqei se kje pragmatikì arijmì opìte ìla ta an¸malashmeÐa thc eÐnai apaleÐyima.

H teleutaÐa sunrthsh tautÐzetai me thn 1+ 2z

+ 22

2!z2pou profan¸c èqei monadikì an¸malo

shmeÐo to mhdèn pìlo deutèrac txhc kai oloklhrwtikì upìloipo 2.

Λυμένες ασκήσεις Μιγαδικές- Σάμαρης 2012

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