complex analysis 1996

8/3/2019 Complex Analysis 1996

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UPSC Civil Services Main 1996 - Mathematics


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Complex AnalysisSunder LalRetired Professor of MathematicsPanjab UniversityChandigarhFebruary 20, 2010Question 1(a) Sketch the ellipse C described in the complex plane byz = Acost + iB sint,A > Bwhere t is a real variable and A,B, are positive constants.If C is the trajectory of a particle with z(t) as the position vector of the particle at timet, identify with justication1. the two positions where the velocity is minimum.2. the two positions where the acceleration is maximum.Solution. We are given that x = Acost,y = B sint which implies thatA2x2y2Since A > B, it follows that it is the standard ellipse with 2A as the major axis +and B22B = as1.the minor axis.B(0,B)

CA(A,0)O A(A,0)B (0,B)1


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=

A22 sin2 t + B22 cos2 t= (A2 B2)sin2 t + B2Since A2B2 > 0, the speed is minimum when sin2 t = 0 i.e. when x(t) = A,y(t) =0 i.e. when the particle is at the two ends of the major axis, the points A and A inthe gure.2. Acceleration =d2zdt2

= A2 Magnitude of acceleration = cost iB2 sint.2A2 cos2 t + B2 sin2 t = 2(A2 B2)cos2 t + B2.Since A2 B2 > 0, acceleration is maximum when cos2 t = 1 cost = 1 i.e. theparticle is at either end of the major axis, A or A . (Note that acceleration is minimumwhen cos2 t = 0 i.e. the particle is at either end of the minor axis).

Question 1(b) Evaluate z0

lim1 sin(z2) cosz.Solution.z0lim1 cosz2sin2 z2sin2 z(z2

2

sin(z2)sin(z2)2


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412Note that sinz has a simple zero at z = 0 and sinz = z(z) where (z) is analytic and(0) = 1, so z0lim= z0lim= z0lim)2sin(z2)z2=sinzz= 1.Question 1(c) Show that z = 0 is not a branch point for the function f(z) =sinzz. Is it

a removable singularity?we Solution. x a branch We know of w =that z, w sin=z z is is a multiple valued function and has two branches. Onceanalytic, andsinz =z (3!


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z)3+(5!z)5+ ...or sinzz= 1

3!z+z25!z37!+ ...2 1. The velocity v =

dzdt

= Asint + iBcost.Speed = magnitude of velocity =

dzdt


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Thus z0limsinzz= 1, so z = 0 is not a branch point of the function f(z) =sinzz. In factz = 0 is a removable singularity of f(z). In fact

F(z) ={sinzz, z = 01, z = 0is analytic everywhere once a branch of

z is specied.Question 2(a) Prove that every polynomial equation a0z2 +...+anzn = 0,an=0,n 1 has exactly n roots.Solution. Let P(z) = a0

+a1z +a2

for any z C . Let f(z) =P(z)+ 1


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a1z , + a2z2 + ... + a

nzn. Suppose, if possible, that P(z) = 0then f(z) is an entire function i.e. f(z) is analytic in thewhole complex plane. We shall now show that f(z) is bounded.P(z) = zn)Since(an

+an1z+az2n2+ ... +zn

a0a

jznj 0 as z , for 0 j < n, is follows that given =|an|2nthere exists R > 0such that |z| > R

a j


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RSince |z| R is a compact set and f(z) is analytic on it, f(z) is bounded on |z| R.Consequently f(z) is bounded on the whole complex plane. Now we use Liouvilles theorem If an entire function is bounded on the whole complex plane, then it is a constant. Thusf(z) and therefore P(z) is a constant, which is not true, hence our assumption that P(z) = 0for all z C is false. So there is at least one z1

C where P(z1) = 0. (This result is calledthe fundamental theorem of algebra.)We now prove by induction one zero namely z = aa0

1

on n that P(z) has n zeros. If n = 1, P(z) = a0+ a1z has.Assume as induction hypothesis that any polynomial of degree n1 has n1 zeros. By


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Euclids algorithm, , we get P1(z) and R(z) such that P(z) = (z z1)P

1(z) + R(z), whereR(z) 0 or degR(z) < 1 i.e. R(z) is a constant. Putting z = z1we get R(z) 0, soP(z) = (z z1)P1(z). Since P1

(z) is a polynomial of degree n 1, by induction hypothesisit has n 1 roots in C , and therefore P(z) has n roots in C .3


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,...,z


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We now prove that P(z) has exactly n roots. Let z1,z2n

be the (not necessarilydistinct) roots of P(z). Let g(z) =(z z1P(z). Clearly g(z) is analytic inthe whole complex plane. Sincezlim)(z z2

)...(z zn)g(z) = zlimP(z)(z z1)(z z2

)...(z zn)=an+an1z+an2z2+ ... + zna0

(1 zz


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1

)(1 zz2

)...(1 z

zn)= anit follows that given > 0 there exists R such that |g(z) an| < for |z| > R, so g(z) isbounded in the region |z| > R. The function g(z) being analytic is bounded in the compactregion |z| R. Thus by Liouvilles theorem g(z) is a constant, in fact g(z) = an

, andthereforeP(z) = an(z z1)(z z2)...(z zn)Thus if is a zero of P(z), then = z

jfor some j, 1 j n. Thus P(z) has exactly nzeroes.Alternate Proof: We shall use Rouches theorem Let be a simple closed rectiablecurve. Let f(z),g(z) be analytic on and within . Suppose |g(z)| < |f(z)| on , then f(z)and f(z) g(z) have the same number of zeroes inside .Let f(z) = anzn and g(z) = an1zn1 + ... + a0. Let R be so large that |g(z)| < |f(z)|on |z| = R. Then f(z) and f(z) + g(z) = P(z) have the same number of zeroes within|z| = R. But whatever R > 0 we take, f(z) has exactly n zeroes in |z| = R, therefore P(z)has exactly n zeroes in C .


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Note: Rouches theorem follows from the Argument Principle Note that (arg(f(z)+g(z))) =change in argument of f(z)+g(z) as z moves along =

argf(z)+arg(1+f(z)g(z))as f(z) = 0 along . But arg(1+f(z)g(z)

) = 0 because |f(z)g(z)| < 1 and thereforeg(z)f(z)continuesto lie in the disc |w 1| < 1 as z moves on i.e. does not go around the origin.Question 2(b) By using the residue theorem, evaluate

0loge(x2 x2 + + 11)dxSolution.Let f(z) =log(z log(z + i) in C + 1 {z + z2| i)

and z = we consider iy,y 1},where it is single-valued. Let be the con-tour consisting of the line joining (R,0) and(R,0) and , which is the arc of the circle of radius R and center (0,0) lying in the upper half plane. is oriented counter-clockwise.


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(R,0) (0,0)(R,0) 4


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Clearly f(z) has a simple pole at z = i in the upper half plane. The residue at z = i is


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limzi(z + i)log(z 1 + z2+ i)=log2i2i=2i1log2ei2

=2i1[log2 + i2]=412ilog 2Thus by Cauchys residue theoremRlimlog(z 1 + + z2i)= Rlimlog(z 1 +


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[]4as z = x on the real axis.We shall now show that Rlim+ z2i)+log(x 1 + + x2i)dx = 2i1

2ilog2

log(z

1 + + z2i)= 0. On , z = Rei, so

log(z + i)1 + z2=


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( + R2 log R)R1log(z 0. 0 andR|log(1+ R21Rei)| 0 as R , it follows that Rlim1 + + z2i)=

Thus i22Equating real and imaginary parts, we get0log(x 1 + + x2

i)dx = log2 + ilog(1 1 + + x2x2)dx =12

log(1 1 + + x2x2)dx =12


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(z + 12)= (z + 2)(1z + 2)2k1 5(1z + 2)2k1=k=1(1)k1(2k 1)!k=1(1)k1(2k 1)!k=0(z + ak2)k, a2k2=(2k (1)k1 1)!

, a2k1=5(1)k1(2k 1)!5


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The region of convergence of the series is 0 < |z + 2| < . The Laurent expansion shows


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that the function has an essential singularity at z = 2 this also follows from the factthat limz0sin 1z

does not exist.6

complex analysis 1996

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