MATH2016 Analysis
TUTORIAL ANSWERS VI
1. (i) The numerator is non-zero, so the pole has order of the zero of z 3 − 1
iz
eiz
ei
at z = 1, and is simple. By Lemma 14.5, res( z3e−1 , 1) = limz→1 3z
2 = 3.
(ii) The poles are at z = 0, −2i. Since the numerator is non-zero in each
case, the orders of the poles are 1 and 2 respectively. Again using
Lemmas 14.3
πz )
−1
=
= 14 and
and 14.5, the residues are equal to res(f, a) = limz→0 cos(πe
(z+2i)2
4i2
res(f, −2i) = limz→−2i
d cos(πeπz )
( z )
dz
= limz→−2i
−2π 2 eπz sin(πeπz )−cos(πeπz )
z2
sin z
(z−π)6
1
− 120
.
(iii) We write w = z − π to ease calculations. Then we find that
sin(w+π)
w6
=
− sin w
w6
2
2
=
− w16 (w
−
w3
3!
2 2
+
w5
5!
− . . .) so the residue is
2 4
− 5!1
)
(iv) cos z(2z
= z15 (1 − (2z2! ) + (2z4! ) − . . .)2 = z15 (1 − 2z 4 +
5
1
(1 − 4z 4 + . . .) = z15 − z4 + . . . so the residue is −4.
z5
=
2z 8
3
= − 14 .
=
− . . .)2 =
2. This has simple poles at z = ±i.
iπ
ezπ
res(f, i) = limz→i z+i
= e2i = −1
= 2i . Similarly, res(f, −i) = − 2i .
2i
3. (i) Let z = γ(t) = eit , −π ≤ t ≤ Rπ, and we use Rcos t = 21 z + z1 .
π
2z
1
dt
2z
= 2+ 1 1z+ 1 = z2 +4z+1
. Hence −π 2+cos
= −i γ z2 +4z+1
z −1 dz =
Then 2+cos
t
t
(
)
2
z
√
√
R
R
1
1
−2i γ z2 +4z+1
dz = −2i γ (z−a)(z−b)
dz where a = −2 + 3 and b = −2 − 3.
There are simple poles at a and b. Only a lies within the unit circle and so by
Cauchy’s
Residue Formula,
Z π
1
1
1
dt
= −2i×2πi lim (z−a)
= 4π lim
= 4π
=
z→a
z→a z − b
(z − a)(z − b)
a−b
−π 2 + cos t
2π
√ .
3
(ii) First we find the poles and residues in the upper half plane for the
function f (z) = z 2 /(z 4√+ z 2 + 1). We treat z 4 + z 2 + 1 as a quadratic in z 2 ,
3
which gives z 2 = −1±i
= e2πi/3 and e4πi/3 . [We put in polar form to make
2
it easier to take square roots.] Taking square roots we find that the poles in
the upper √half plane are
at eπi/3 and e2πi/3 , which (using Lemma 14.3) have
√
i 3−1
√ and −i 3−1
√ .
residues −6+2
i 3
6+2 i 3
R
R
Let γ be the usual semi-circular contour. Then γ f (z)dz = γ1 f (z)dz +
R
f (z)dz. From the above, by Cauchy’s Residue Theorem, for R > 1,
Rγ2
z2
f (z)dz = 2πi res(f, eπi/3 ) + res(f, e2πi/3 ) = √π3 . Since on γ1 , | z4 +z
2 +1 | ≤
γ
R
R2
πR3
, by the Estimation Lemma, | γ1 f (z)dz| ≤ R4 −R2 −1 → 0 as R → ∞.
R4 −R2 −1
R
R
R
Thus √π3 = limR→∞ γ f (z)dz = limR→∞ γ1 f (z)dz + limR→∞ γ2 f (z)dz =
R∞
RR
2
2
limR→∞ −R x4x+1 dx + 0 = −∞ x4x+1 dx.
iz
4. (i) The function z3e−1 has poles at the the cube roots of unity. Of these,
only z = 1 lies within the stated
contour, so by Cauchy’s Residue
R semi-circular
iz
i
i
.
Theorem and question 1(i), γ1 z3e−1 dz = 2πi × 1 × e3 = 2πie
3
(ii) The contour has only the pole z = 0 inside, so by question 1(ii), and
R
πz )
as the winding number is 1, γ2 cos(πe
dz = 2πi × 14 = πi2 .
z(z+2i)2
(iii) This contour contains both poles, again winding number 1, so from
R
πz )
question 1(ii), γ3 cos(πe
dz = 2πi × ( 41 − 14 ) = 0.
z(z+2i)2
(iv) There is a pole only at z = 0, and the winding number is 2. Thus
R 3z4 +2z2 +cos2 (2z2 )
R
R
R
2
2)
dz = γ4 z3 dz + γ4 z23 dz + γ4 cos z(2z
dz
5
z5
γ4
2
2
)
, 0)) = 4πi(3 + 0 − 4) = −4πi.
= 2πi × 2 × (res( z3 , 0) + res( z23 , 0) + res( cos z(2z
5
P
n
5. (i) Let ez = ∞
n=0 bn z . Multiplying the two power series,
z
e f (z) =
∞
X
n
cn z , with cn = a0 bn + a1 bn−1 + . . . + an b0 =
n=0
n
X
ak bn−k .
k=0
But bn−k = 1/(n − k)!, hence the result. Note that both power series have
infinite radius of convergence, so the product is valid everywhere.
(ii) We have
d ez
z
(e ) = ez ee = ez f (z),
f 0 (z) =
dz
and saw earlier that
to differentiate power series term-by-term,
P
P we are allowed
∞
n
n−1
na
z
=
so that f 0 (z) = ∞
n
j=0 cn z above. Equating coefficients of
n=0
z n−1 gives
n−1
X
ak
nan =
.
(n
−
1
−
k)!
k=0
6.(i) We note that f ◦ γ is a continuous function on a closed bounded
interval [a, b] and by Lemma 2.3 is therefore bounded, i.e., |f (γ(t))| ≤ M for
a ≤ t ≤ b.
(ii) This follows directly from the Maximum Modulus Principle, Theorem
12.3.
(iii) Let z ∈ C. We can find m, n ∈ Z such that z − m − ni lies in the
square with vertices 0, 1, i, 1 + i. Then |f (z)| = |f (z − m − ni)| by (∗) and so
|f (z)| ≤ M by (ii).
(iv) The function f is a bounded function differentiable on all of C. Hence,
by Liouville’s theorem, Theorem 12.2, it is constant.
2