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ME5332 MIDTERM EXAMINATION
4:00-5:20pm, March 23, 2022
Closed book/note/no calculator
2
1.
Simplify
⎛
⎝
1 + √3
i
√3 −
i
⎞
⎠
101
.
2.
Compute
⌠
⌡
∞
−∞
1
x
2
−2
x
+2
dx
.
3.
Cite an example of complex analytic functions which are bounded on the real axis (
x
) but unbounded in the entire plane.
4.
Obtain the first two (2) terms of Laurent series of
f
(
z
) =
1
e
z
− 1 −
z
in the immediate neighborhood of
z
= 0.
5.
Find all the roots for
z
3
= 1 −
i
.
6.
Evaluate
⌠
(⎜)
⌡
|
z
| = 1
sin
z
z
6
d
z
where the integral path is the ccw unit circle.
7.
Express the imaginary part of [1/(
z
2
)] in
x
and
y
where
z
=
x
+
i
y
.
8.
Evaluate
⌠
⌡
∞
−∞
sin
x
1 −
x
4
dx
.
9.
The real part of an analytic function,
f
(
z
), is given as
u
=
y
x
2
+
y
2
.
Find
f
(
z
).
10.
Map |
z
− 1 | = 3 by
w
=
1 −
z
2 +
z
and draw the result on the (
u
,
v
) plane.
On my honor, I have neither given nor received aid on this test.
―
Your signature
Solution
1.
1 + √3
i
√3 −
i
=
1 + √3
i
√3 −
i
√3 +
i
√3 +
i
=
√3 +
i
+ 3
i
− √3
3 + 1
=
i
,
so
⎛
⎝
1 + √3
i
√3 −
i
⎞
⎠
101
=
i
.
2.
I
= 2 π
i
Res
⎛
⎝
1
z
2
−2
z
+2
; 1+
i
⎞
⎠
=π.
(1)
3.
f
(
z
) =
e
−
z
2
, sin
z
,
1
1 +
z
2
, cos
z
…
4.
e
z
− 1 −
z
=
z
2
2!
+
z
3
3!
+ …
so
1
e
z
− 1 −
z
=
1
z
2
2!
+
z
3
3!
+ …
= … =
2
z
2
−
2
3
z
+ …
5.
Note that the given equation is written as
z
3
=
1 −
i
(2)
=
√2
e
(− π/4 + 2
n
π)
i
,
n
= 0, 1, 2.
(3)
Let
z
=
r
e
i
θ
then,
r
3
e
3
i
θ
= √2
e
(− π/4 + 2
n
π)
i
.
Therefore,
r
= 2
1/6
, θ = −
π
12
+
2
n
π
3
,
n
= 0, 1, 2.
6.
I
=
⌠
(⎜)
⌡
z
−
z
3
/3! +
z
5
/5!− …
z
6
d
z
(4)
=
…+
1
120
⌠
(⎜)
⌡
1
z
dz
+ …
(5)
=
2 π
i
×
1
120
(6)
=
π
i
60
.
(7)
7.
1
z
2
=
-
z
2
z
2
-
z
2
=
(
x
−
i
y
)
2
(
x
2
+
y
2
)
2
=
x
2
−
y
2
(
x
2
+
y
2
)
2
−
i
2
x
y
(
x
2
+
y
2
)
2
,
(8)
Therefore, the imaginary part is
−
2
x
y
(
x
2
+
y
2
)
2
.
8.
This is an odd function so its integral is 0.
9.
By guessing, try
v
=
x
x
2
+
y
2
.
We have
u
+
i
v
=
y
+
x
i
x
2
+
y
2
=
i
(
x
−
i
y
)
x
2
+
y
2
=
i
-
z
z
-
z
=
i
z
.
10.
w
= [(1−
z
)/(2+
z
)] can be solved for
z
as
z
=
1−2
w
1+
w
.
So
z
−1 =
1−2
w
1+
w
− 1 =
−3
w
1+
w
,
and |
z
−1|=3 is transformed into
3 |
w
w
+1
| = 3,
or
|
w
| = |
w
+ 1|,
which represents the middle line between
w
= 0 and
w
= −1, i.e.
u
= −1/2.
File translated from T
E
X by
T
T
H
, version 4.03.
On 05 Mar 2023, 22:24.