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HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2HIGHWAY AND TRANSPORT ENGINERING EXAM AND ANSWER-2
Determine if the following series or improper integrals are converge.pdffashiondestinationld
Determine if the following series or improper integrals are convergent or divergent:
Solution
Here is a complex-analytic proof.
For zD=C{0,1}, let
R(z)=1log2z
where the sum is taken over all branches of the logarithm. Each point in D has a neighbourhood
on which the branches of log(z) are analytic. Since the series converges uniformly away from
z=1, R(z)is analytic on D.
Now a few observations:
(i) Each term of the series tends to 0 as z0. Thanks to the uniform convergence this implies that
the singularity at z=0 is removable and we can set R(0)=0.
(ii) The only singularity of R is a double pole at z=1 due to the contribution of the principal
branch of logz. Moreover, limz1(z1)2R(z)=1.
(iii) R(1/z)=R(z).
By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii)
the denominator of R(z) is (z1)2. Since R(0)=R()=0, the numerator has the form az. Then (ii)
implies a=1, so thatR(z)=z(z1)2.
Now, setting z=e2iw yieldsn=1(wn)2=2sin2(w)
down vote
Here is a complex-analytic proof.
For zD=C{0,1}, let
R(z)=1log2z
where the sum is taken over all branches of the logarithm. Each point in D has a
neighbourhood on which the branches of log(z) are analytic. Since the series converges
uniformly away from z=1, R(z)is analytic on D.
Now a few observations:
(i) Each term of the series tends to 0 as z0. Thanks to the uniform convergence this implies
that the singularity at z=0 is removable and we can set R(0)=0.
(ii) The only singularity of R is a double pole at z=1 due to the contribution of the principal
branch of logz. Moreover, limz1(z1)2R(z)=1.
(iii) R(1/z)=R(z).
By (i) and (iii) R is meromorphic on the extended complex plane, therefore it is rational. By (ii)
the denominator of R(z) is (z1)2. Since R(0)=R()=0, the numerator has the form az. Then (ii)
implies a=1, so thatR(z)=z(z1)2.
Now, setting z=e2iw yieldsn=1(wn)2=2sin2(w)
which implies thatk=01(2k+1)2=28,
and the identity (2)=2/6 follows..
1. Ma 416: Complex Variables
Solutions to Homework Assignment 5
Prof. Wickerhauser
Due Thursday, October 6th, 2005
Read R. P. Boas, Invitation to Complex Analysis, Chapter 2, sections 8A–8H.
1. Classify the following singularities as removable, poles, or essential. If the singularity
is a pole, state its order.
2
(a) 1/(ez − 1) at z = 0 (b) e1/z at z = 0 (c) z/ sin z at z = 0
2
Solution: (a) Pole of order 2, since ez − 1 = z 2 + O(z 4 ) as z → 0.
(b) Essential singularity, since e1/zn = 1 for zn = 1/2nπi → 0 but e1/zn = −1 for
zn = 1/(2n + 1)πi → 0 as n → ∞. Hence the function can have no limit as z → 0.
(c) Removable singularity, since limz→0 [z/ sin z] = 1/ limz→0 [(sin z)/z] = 1. 2
2. Find the residues of the following functions at the indicated points.
(a) 1/(ez − 1) at z = 0 (b) z 4 /(z − 1 z 3 − sin z) at z = 0
6
(c) (z 2 + 1)/z 4 − 1) at z = 1 and z = i.
Solution: (a) Since this is a simple pole, find the residue as follows:
(z − 0) z 1
lim z −1
= lim = lim = 1.
z→0 e z→0 z + O(z 2 ) z→0 1 + O(z)
(b) Note that z − 1 z 3 − sin z = − 120 z 5 + O(z 7 ) at z = 0. Hence z 4 /(z − 1 z 3 − sin z)
6
1
6
has a simple pole at z = 0. Find the residue as follows:
(z − 0)z 4 z5 −120
lim 1 3 = lim 1 5 = lim = −120.
z→0 z − z − sin z z→0 − z + O(z 7) z→0 1 + O(z 2 )
6 120
(c) Factor the numerator and denominator polynomials into
(z − i)(z + i) 1
= .
(z − i)(z + i)(z − 1)(z + 1) (z − 1)(z + 1)
Hence z = i (also z = −i) is a removable singularity, so the residue there is 0. However,
(z−1)
z = 1 is a simple pole with residue limz→1 (z−1)(z+1) = 1 .
2
2
1
2. 3. Find the residue of the function f (z) = 1/ sinh2 z at z = 0.
Solution: Since sinh z = O(z) as z → 0, we see that z = 0 is a pole of order 2 for
1/ sinh2 z. Find the residue as follows, setting n = 2 in the formula on page 73 of our
text:
(n−1)
1 d d z2
lim [f (z)(z − z0 )n ] = lim
z→z0 (n − 1)! dz z→0 dz sinh2 z
2z sinh z − 2z 2 cosh z
= lim
z→0 sinh3 z
z sinh z − z cosh z
= 2 lim
z→0 sinh z sinh2 z
L’Hˆpital’s rule allows us to evaluate the factor limits:
o
z 1
lim = lim = 1,
z→0 sinh z z→0 cosh z
and
sinh z − z cosh z z sinh z
lim 2 = lim = 0.
z→0 sinh z z→0 2 sinh z cosh z
Hence the residue is 0. 2
2