Complex Analysis 1988

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UPSC Civil Services Main 1988 - Mathematics



Complex AnalysisSunder LalRetired Professor of MathematicsPanjab University

ChandigarhJuly 19, 2010Question 1(a) By evaluating∫z dz+ 2over a suitable contour C prove that∫π0

1 5 + + 2cosθ4cosθdθ = 0Solution. By using the unit circle |z| = 1 as contour, and integrating∫|z|=1

z dz+ 2, we haveproved

∫2π0

1 ∫θ = 2π − φ so 5 + + that2cosθ4cosθdθ = 0 — see 1997, question 1(b). Now in2ππ

1 5 + + 2cosθ4cosθdθ put∫2ππ

1 5 + + 2cosθ4cosθ



dθ =∫π0

1 5 + ∫

+ 2cos(2π 4cos(2π − − φ)φ)(−dφ) =π

1 0

5 + + 2cosφ4cosφdφThus∫

2π0

1 ∫5 + + 2cosθ4cosθdθ = 2π

1 0

5 + + 2cosθ4cosθ

dθ showing

that∫

π0

1 5 + + 2cosθ4cosθdθ = 0.Note: If the contour was not prescribed, we could have put z = eiθ to get∫2π

01 5 + + 2cosθ4cosθdθ =1i∫



|z|=1

z(5z z2 + + 2z2 z + + 12)dz1



The integrand has residue at z = 0 is two poles at z = 0,z 1



2

and the residue at z = = −1

−1

22

inside is |z| = 1, which −1

2

, so we getare simple poles. The∫|z|=1

z2 + z + 1∫2π

z(5z + 2z2 + 2)0

1 5 + + 2cosθ4cosθdθ = 0Question 1(b) If f(z) is analytic in |z| ≤ R and x,y lie inside the disc, evaluate theintegral

dz = 0 ⇒

∫f(z)dzz is a constant.|z|=R

(z − x)(z − y)and deduce that a function analytic and bounded for all finiteSolution. Cauchy’s integral formula states that if f(z) is analytic on and within the disc|z| ≤ R, then for any ζ which lies within the discf(ζ) =∫|z|=R

f(z)dzζ − zThus∫|z|=R

12πif(z)dz(z − x)(z − y)=



1[∫x − y|z|=R

]

=∫|z|=R

]We now prove the remaining part, which is Liouville’s theorem.Let |f(z)| ≤ M for every z. Clearly |z − x| ≥ |z| − |x| = R − |x| and similarly |z − y| ≥R − |y| on |z| = R, and therefore

∣

∣

∣∣

f(z)dzz − x−f(z)dzz − y2πix − y[

f(x) − f(y)∫

∣

|z|=R

∣

∣

∣

M · 2πR(R − |x|)(R − |y|)

Thus |f(x)−f(y)| ≤(z − f(z)dzx)(z − y)≤

∣

∣



∣

∣

1

∣

2πi∣

∣

∣

(R |x − − y| · M · 2πR|x|)(R − |y|)R(R − |x|)(R − |y|)→ 0 as R → ∞,

it follows that |f(x) − f(y)| = 0 or f(x) = f(y), so f is constant.Question 1(c) If f(z) =. Since∑that∞n=0

zn has radius of convergence R and 0 < r < R, prove12πan

∫2π0

|f(reiθ)|2 dθ =∞∑|2r 2n

n=0

Solution.|f(z)|2 = f(z) · f(z) =

|an

∞∑an

zn

∞∑



am

zm =∞∑∑

ap

aq

zpzq

n=0m=0n=0p+q=n

2



We know that if a power series has a radius of convergence R, then it is uniformly and



absolutely convergent in |z| ≤ r where 0 < r < R, therefore12π∫2π

0|f(z)|2 dθ =∞∑n=0

2π1∫2π0

∑r pr qei(p−q)θ dθp+q=n

Sinceap

aq

∫02π

ei(p−q)θ dθ = 0 when p = q, we get

12π∫2π0

|f(z)|2 dθ =∞∑|an

|2r 2n

n=0

(Note: This shows that if |f(z)| ≤ M on |z| = r, then∑∞n=0

|2r 2n ≤ M2.)Question 2(a) Evaluate|an



∫C

zez (z − dza)3if a lies inside the closed contour C.

isSolution. Clearly the only pole of (z zez

− a)3is of order 3 at z = a. The residue at this pole12!d2

((z − a)3zez

)(dz2

(z − a)3z=a

)(zez + ez

z=a

)

zez + ez + ez

z=a

()1 +Thus by Cauchy’s residue theorem,∫C

12

=ddz=12= ea



a2(1 +)

= πiea(2 + a)Question 2(b) Prove∫∞0

zez dz(z − a)3= 2πi · ea

a2

e−x2

cos(2bx)dx =√2πe−b2

(b > 0)by integrating e−z2 along the boundary of the rectangle |x| ≤ R,0 ≤ y ≤ b.Solution.

Let the

rectangle

be

ABCD

where

A

=(−R,0),B = (R,0),C = (R,b),D = (−R,b)

oriented positively. Since e−z2 has no pole in-side ABCD, we get R→∞

limD(−R,b)C(R,b) y = b∫x = −RC

x = R e−z2dz = 0.ABCD

A(−R,0) (0,0)B(R,0) 3y = 0



(Note that e−z2 has no pole in the entire complex plane.)



1. On BC, z = R + iy and 0 ≤ y ≤ b, therefore

∣

∣

∣

∣∫e−z2

dz

∣

∣

∣

∣

∣

∣

∣

∣

∫b

e−R2

e−2Riye−i2y2

idy=

∣∣

∣

∣

≤ e−R2

∫b

ey2

dy = (constant)e−R2

BC

00

Clearly e−R2 → 0 as R → ∞, so R→∞

lim∫BC

e−z2



dz = 0.2. On DA, z = −R + iy and 0 ≤ y ≤ b, therefore

∣

∣

∣∣

∫DA

∣

∣

∣

∣

∣

∣

∣

∣

∫b0

∣

∣

∣

∣∫b0

ey2

dyThus R→∞

lime−z2

dz

=e−R2

e2Riye−i2y2

idy≤ e−R2

∫DA



e−z2

dz = 0.3. On AB, z = x so R→∞

lim∫

∫∞AB−∞

√π.4. On CD, z = x + ib, thereforeR→∞

lime−z2

dz =e−x2

dx =∫∫e−z2

dz =∫−∞

e−x2

e−i2b2e−2ibx dx = −eb2

∞

e−x2

[cos2bx − isin2bx]dxCD∞−∞

Using the above calculations, we get0 = R→∞

lim

∫e−z2

dz =√π − eb2

∫∞



e−x2

[cos2bx − isin2bx]dxC−∞

Equating real and imaginary parts,

∫∞−∞

e−x2

sin2bxdx = 0and ∫∞−∞

√πe−b2

Thus

∫∞0

e−x2

cos2bxdx =e−x2

cos2bxdx =12∫

∞e−x2

cos2bxdx =√−∞

πe−b2

24



of Question 2(c) Prove that the coefficients c



n

the expansion11 − z − z2

∞∑

n=0zn

satisfy cn

=cn

= cn−1

+ cn−2

,n ≥ 2. Determine cn

.

Solution. z2 + z − 1 = 0 ⇒ z = analytic in the disc |z| < λ as both −1±

the 2√5

. singularities Let λ = −1+

at 2√

z 5,µ = λ = and −1−

2√5

. z = Thus µ lie outside f(z) = 1−z−z2

it. 1isThusf(z) powers Let has we f(z) Taylor get=∑series ∞n=0

cexpansion n

zn, then with (1 − center z − z2)z ∑



= ∞n=0

0.cn

zn = 1. Equating coefficients of likec0

= 1c1

= 0c2

− c0

= 0...cn

− c1

− c0

= 0Thus c

n− cn−1

− cn−2

= cn−1

+ cn−2

,n ≥ 2. The cn

’s are Fibonacci numbers.Nowf(z) =−1(z − λ)(z − µ)=]



=−1λ − µ[z − 1

λ−z − 1µ√−15[−

λ1(1 −λz)−1

−−1

µ]If we confine z to the disc |z| < λ, then | λz

(1 −µz)−1

| < 1,| µz

| < 1 and we havef(z) =√15



[∞∑]n=0

zn

λn+1

∞∑zn

n=0

µn+1

=∞∑zn

n=0

where c

n−cn

are given as above. But the Taylor series of a function is unique, therefore we havecn

=√1

5]=[1λn+1

−µn+1

1]

=√15[(√5 √



2− 5 + 1)n+1

(

)n+1

−1)√5 −2+ 1√5 − 1)5 − 1

5 − 1]=√15[(2()n+1

−(−2()n+1

[(√5 2+ 1)n+1

+ (−1)n√15(√5 − 1)



n+1

]25

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Complex Analysis 1988