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**Unformatted text preview: **1.5 SINGULARITIES. ZEROS AND RESIDUES 55 I has the general solution 2 = 1 + NTE (N: 0, i1, i2, . . . ). Thus, apart from N: 0,
all of these are zeros of ﬂz). (d) For 1
if—fu+n+n3 f (2) :
factorizing as in (b), we have _—1—
oilfc+lfh+¢hl+DfB—thl+nf so —1, +1, ,5; (l + j) and Vi; (—1 —j) are still singularities, but this time they are
triply repeated. Hence they are all poles of order three. There are no zeros. f(Z) = Determine the location of, and classify, the Expand each of the following functions in a Laurent
singularities and zeros of the following functions. series about 2 = O, and give the type of singularity
Specify also any zeros that may exist. (if any) in each case: cos 2 z 1 — cos .7
(a) 2 (bl —,2¥, (C) 4 (a) Z z (2+J) (2—1) z—l
- e: m
(d) cothz (e) Em Z, (f) e‘“ "l (b) ?
z" + II:
—] 7| Zﬁl Z+- (clz coshz
(g) 2 (h) —,J— ,, 2+1 (z+2) (3—3) ((1) tan (22+2_7+2)
(i) 1 a Show that if f(z) is the ratio of two polynomials z2(z2 — 42 + 5) then it cannot have an essential singularity. 1.5.3 Residues If a complex function f(z) has a pole at the point z = 20 then the coefﬁcient (1,1 of the
term l/(z _ 20) in the Laurent series expansion of ﬂz) about 2 = 20 is called the residue
of f(z) at the point z : 20. The importance of residues will become apparent when
we discuss integration in Section 1.6. Here we shall concentrate on efﬁcient ways
of calculating them, usually without ﬁnding the Laurent series expansion explicitly.
However, experience and judgement are sometimes the only help in ﬁnding the easiest way of calculating residues. First let us consider the case whenﬂz) has a simple pole
at z = 20. This implies, from the deﬁnition of a simple pole, that f(z)=Z£:’ln+a0+al(z—zo)+... in an appropriate annulus S < |z — zul < R. Multiplying by z — 26 gives (z—zo)f(z)=a,1 + (10(2—20) + . . . m 60 FUNCTIONS OF A COMPLEX VARIABLE sinz z2 z4
_=1.a_+__ z 3! 5’
giving SiIlZ 1 1 13
—2:“—gZ+T§{jZ—...
z 2 Taking the cube of this series, we have
sinz 3 l 3 3 l 1 z 1
[—2]=[——éz+f;-nz-...]=—3—3—5— —
z Z z z 6 z 22
Hence the residue at z = 0 is —§. (0) The function z4/(z + 1)3 has a triple pole at z = _1, so, using (1.38), d2 4
z—>—l 3 (z ) ml.— 2 4
residue: lim (2+ 1)} Z 3 }= lim
“*1 dz (2+ 1) = lim §x4x3z2 = 6(—i)2 = 6 24-1 Residues are sometimes difﬁcult to calculate using (1.38), especially if circular func—
tions are involved and the pole is of order three or more. In such cases direct calculation
of the Laurent series expansion using the standard series for sinz and cos 2 together with
the binomial series, as in Example 1.260)), is the best procedure. "“ ‘ - '
\. xﬁﬁzx‘C‘ \\ .
.. \ 1 3. Determine the residues of the following rational Z4 _ 1 _ m z _
functions at each pole in the ﬁnite 3 plane: (0) Z4 + 1 (Z ‘ e ) (d) E: (Z _ n)
22 + l 1 1 .
' (a) — (b) — (e) ——— (2 =1)
22—2—2 22(1—2) (2+1)2
2 3 2
(c) % (d) Z_‘%+Z__1 .- . , The following functions have poles at the points
(Z - 1X3 + 9) Z + 42 indicated. Determine the order of the pole and the
03) 26+424 +23 + 1 (f) [2 + 1J2 reSidue there.
5 i 1
(2—1) Z (a) 0032 (2:0)
+1 3+4 2
(g) 2+ (h) 7—2—2 2 _ 2
(z—l) (2+3) 2 +32 +22 (b) Z z (z=#1)
(z +1)2(22 + 4)
‘ Calculate the residues at the simple poles indicated (C) e2 (Z = mt n an inte er)
of the following ﬁmctions: 511122 ’ g
(a) 0052 (z = 0) (b) 4 511122 (2 : Ema) (Hint: use limu_,0(sin u)/u_ = 1 (u : z — me), after
2 z + z + l differentiating, to replace 5111 u by u under the limit.) 7 1.6 CONTOUR INTEGRATION 71 } _ where yis a circle centred at z = 1. Writing 13(2) = 2“, then
' min
%f(z)dz=f ﬁ(z)3dz
c 7 (Z— 1)
and, since f1(z) is analytic within and on the circle 7/, it follows from (1.48) that
w d_2.1d2 _.122
31,1 CﬂZ) 2‘- TEJE Efdz) —T€J( le=1
3:] so that "‘ “ " i " .J ‘7 ' "‘ i Al§L§¥¥N Using the Cauchy integral theorem, evaluate the ' .‘ﬁbmplex 2 plane: contour integral ithesti'aight1inejoining2+j0tg£+j2; zzdz
'thestrajght lines from 2 +j0 to 2 +’j2’and then {22 _ 1M2 + 2) 94:12; 5 -thecitclelz|=2from2+j0to0+j2inan wheIeCis
anticlockwise direction. (a) the circle lzl = l
uate @1524 — z‘ + 2) dz around the following (b) the Ciro“: '2 l = 3
sed contours C in the 2 plane: I the circle 12' = I; Using the Cauchy integral theorem, evaluate the D _ I , contour miegral
Eb) the square With vernces at 0 + JO, 1 +10,
1 + j] and O +j1; jg
c 52 dz ' life) ﬂlecurve consisting ofthe parabolasy=x2 from (Z + 1)'(z _ 2X2 + 4i) 0+j0tol+jlandy2=xﬁoml+jt00+j0. ._ where C is the result of Example 1.30, and show that (a) the circle |z| = 3
I dz _{j2n (n = g) (b) the circle|2|=5
“ T‘ 'c(z—zn)"“ 0 (17¢ 1) I Using the Cauchy integral theorem, evaluate the
- Where C is a simple closed contour surrounding following contour integrals: f4 ‘ 2:20. a
zJ + z
y . . (a) 3 dz
t Evaluate the contour Integral C (22 + 1)
_ dz where Cis the unit circle {2] = l;
. c z—_'- 4
anysimple closedcurve andz=4is c (2*1)(Z+2) (a). outside C (b) inside C where C is the circle |z| = 3. ...

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- Fall '10
- ChinC.Lee
- Laurent series expansion, Laurent, cauchy integral theorem, sinz