Math 185 – Solutions to Midterm II
1.
(a) The polynomial
z
4
+ 1 factors as a product (
z

a
1
)
· · ·
(
z

a
4
) where
a
1
, . . . , a
4
are its
four roots. These are easy to find in polar coordinates since, if
z
=
re
iθ
,we have
z
4
=

1
iff
r
4
e
4
iθ
=
e
iπ
so
r
= 1 and 4
θ
=
π
+ 2
kπ
for
k
∈
Z
. As a result the four roots are:
e
i
π
4
, e
i
3
π
4
, e
i
5
π
4
, e
i
7
π
4
and the polynomial can be factored as
z
4
+ 1 = (
z

e
i
π
4
)(
z

e
i
3
π
4
)(
z

e
i
5
π
4
)(
z

e
i
7
π
4
)
(b) The distance between any two of the above roots is at least
√
2 so the circles
C
1
and
C
2
contain only the roots
e
i
π
4
and
e
i
3
π
4
respectively. Now we will study the integrals of
1
z
4
+1
around
C
1
and around
C
2
independently.
•
Define
g
(
z
) =
1
(
z

e
i
3
π
4
)(
z

e
i
5
π
4
)(
z

e
i
7
π
4
)
. This is an analytic function at all points where
the denominator does not vanish so in particular it is analytic in
C
1
and its interior.
As a result, we can apply the Cauchy integral formula and obtain
C
1
1
z
4
+ 1
dz
=
C
1
g
(
z
)
(
z

e
i
π
4
)
dz
= 2
πig
(
e
i
π
4
)
•
Similarly we can define
h
(
z
) =
1
(
z

e
i
π
4
)(
z

e
i
5
π
4
)(
z

e
i
7
π
4
)
and using the same reasoning
as in the last bullet conclude that
C
2
1
z
4
+ 1
dz
=
C
2
h
(
z
)
(
z

e
i
3
π
4
)
dz
= 2
πih
(
e
i
3
π
4
)
2.
(a) The only points where the function
1
z
4
+1
is not analytic lie in the unit circle. As a result,
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 Fall '07
 Lim
 Calculus, Factors, Polar Coordinates, Trigraph, dz

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