4. [3 marks]
(a) Show that if

z

= 2, then

3
z
2

4
z
+ 1
 ≥
3.
(b) Use the Estimation Lemma and the result in the previous part to show that
I

z

=2
iz
2
+ 5
3
z
2

4
z
+ 1
dz
≤
12
π.
Solution:
If

z

= 2, then by the circle inequality, we have

3
z
2

4
z
+ 1
 ≥ 
3
z
2
   
4
z
+ 1

= 12
  
4
z
+ 1
 ≥
12

9 = 3
,
since by the triangle inequality
 
4
z
+ 1
 ≤  
4
z

+ 1 = 9. Then by the Estimation
Lemma, we have
I

z

=2
iz
2
+ 5
3
z
2

4
z
+ 1
dz
≤
M
·
L,
where
M
is any bound of
iz
2
+ 5
3
z
2

4
z
+ 1
on the contour and
L
= 4
π
is the length of the
contour. By the triangle inequality, we have

iz
2
+ 5
 ≤ 
iz
2

+ 5 = 9 on the coutour, and
by the previous part

3
z
2

4
z
+ 1
 ≥
3. Therefore we can take
M
=
9
3
= 3, and then
ML
= 12
π
.
5. [3 marks]
(a) Evaluate
H
γ
e
i
(
z

2)
dz
, where
γ
is the circle

z

= 3, traversed anticlockwise.
(b) Evaluate
R
γ
Log(
z
)
z
dz
, where
γ
is the semicircle

z

= 2
,
Im(
z
)
≥
0, traversed anti
clockwise.
Justify your answers.
Solution:
Since
e
iz
is entire, the first integral is zero by the CauchyGoursat Theorem.
For the second integral, we use the parametrization
z
=
γ
(
t
) = 2
e
it
,
0
≤
t
≤
π
, so that
γ
0
(
t
) =
iz
. Then the integrand is
ln 2 +
it
z
·
iz
=

t
+
i
ln 2
,
so the integral is
π
Z
0
(

t
+
i
ln 2)
dt
=

π
2
2
+
iπ
ln 2
.