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Quiz 9 (Notes, books, and calculators are not authorized) Show all your work in the blank space
you are given on the exam sheet. Answers with no justification will not be graded.
Question 1: Consider the function f (z) = z1 . Is f holomorphic? Specify where f is holomorphic.
1
1
Let P (x, y) = <( x+iy
) and Q(x, y) = =( x+iy
). The function P and Q are differentiable in R2 \{0}.
x
Observe that P (x, y) = x2 +y2 and Q(x, y) = x2−y
+y 2 . Moreover, the following computation
∂x P =
∂y Q =
x2 + y 2 − 2x2
−x2 + y 2
=
(x2 + y 2 )2
(x2 + y 2 )2
−x2 + y 2
−x2 − y 2 + 2y 2
= 2
2
2
2
(x + y )
(x + y 2 )2
∂y P =
∂x Q =
−2xy
+ y 2 )2
(x2
(x2
2xy
+ y 2 )2
shows that the Cauchy-Riemann identities hold; hence f is holomorphic in C \ {0}.
Question 2: Consider the paths γ : [0, 2π] −→ C such that γ(t) = 2eit and ψ : [0, 4π] −→ C
such that ψ(t) = 2eit (a) Describe in geometrical terms the paths C := γ([0, 2π]) and C 0 :=
ψ([0, 2π]).
C is the circle of radius 2 centered at 0 and traversed only once. C 0 is the circle of radius 2 centered
at 0 and traversed two times.
R
R
(b) Compute the integral I := C z1 dz and I 0 := C 0 z1 dz.
By definition
Z
C
1
dz =
z
Z
0
2π
1 0
γ (t)dt =
γ(t)
Z
0
2π
2−1 e−it (2ieit )dt =
Z
2π
2−1 (2i)dt = 2iπ.
0
i.e., I = 2iπ. Similarly
Z
Z 4π
Z 4π
Z 4π
1
1 0
dz =
ψ (t)dt =
2−1 e−it (2ieit )dt =
2−1 (2i)dt = 4iπ.
ψ(t)
C0 z
0
0
0
i.e., I 0 = 4iπ.
2
Quiz 10, December 4, 2014
Question 3: Consider the paths γ1 : [0, π] −→ C such that γ1 (t) = 2eit and γ2 : [0, π] −→ C
such that γ2 (t) = 2e−it . (a) Describe in geometrical terms the paths C1 := γ1 ([0, π]) and
C2 := γ2 ([0, π]).
C1 is the half circle of radius 2 centered at 0 starting from 2 passing through 2i and ending at −2,
i.e., C1 is contained in the upper half plane. C2 is the half circle of radius 2 centered at 0 starting
from 2 passing through −2i and ending at −2, i.e., C2 is contained in the lower half plane.
R
R
(b) Compute I1 := C1 z1 dz and I2 := C2 z1 dz.
By definition
Z
C1
1
dz =
z
Z
π
2−1 e−it (2ieit )dt =
π
Z
0
2−1 (2i)dt = iπ.
0
and
Z
C2
1
dz =
z
Z
π
−1 it
2
0
e (−2ie
−it
Z
)dt =
π
2−1 (−2i)dt = −iπ,
0
i.e., I1 = iπ and I2 = −iπ.
(c) Consider the paths γ3 : [0, π] −→ C, such that γ3 (t) = 2 cos(t) + 3i sin(t), γ4 [0, π] :−→ C,
such that γ4 (t) = 2 cos(t)−5i sin(t). (a) Describe in geometrical terms the paths C3 := γ1 ([0, π])
and C4 := γ2 ([0, π]).
C3 is the half ellipse centered at 0, of minor radius 2, major radius 3, starting from 2 passing through
3i and ending at −2, i.e., C3 is contained in the upper half plane. C4 is the half ellipse centered
at 0, of minor radius 2, major radius 5, starting from 2 passing through −5i and ending at −2, i.e.,
C4 is contained in the lower half plane.
R
R
(d) Compute I3 := C3 z1 dz and I4 := C4 z1 dz in terms on I1 and I2 .
From Question 1 we know that the function z −→ z1 is holomorphic in the domain enclosed by the
paths C1 and C3 . The Cauchy Theorem implies that I3 = I1 . The same reasoning applies to the
paths C2 and C4 , i.e., I4 = I2 .