Math 185 – Solutions to Midterm I
1.
z
6

1
√
2
+
i
1
√
2
= 0 if and only if
z
6
=
1
√
2

i
1
√
2
. So we want to find all sixth roots of the
number
1
√
2

i
1
√
2
=
e

π
4
.
If
z
=
re
iθ
, the equation
z
6
=
e

π
4
becomes
r
6
e
i
6
θ
=
e

π
4
so
r
= 1 and 6
θ
=

π
4
+2
kπ
for some
k
∈
Z
. Thus the solutions to the equation are the complex numbers of the form
e
i
(

π
24
+
2
kπ
6
)
for
k
= 0
,
1
, . . . ,
5.
2.
(a) The function
z
→
(1+
i
)
z
+1 is a composition of the map
w
= (1+
i
)
z
(tilting by
π
4
+scaling
by a factor of
√
2) and the translation
w
→
w
+1. Hence it maps rectangles to rectangles.
The image of
S
is thus the rectangle with vertices 1
,
2+
i,
2

2
π
+
i
(2
π
+1)
,
1

2
π
+
i
2
π
which are the images of the vertices 0, 1,1 + 2
πi
and 2
πi
respectively.
(b) Since
e
z
=
e
x
cos
(
y
) +
ie
x
sin
(
y
) The horizontal line segments
y
= 0 and
y
= 2
π
are both
mapped to the ray 1
≤
r
≤
e
,
θ
= 0 and the vertical lines segments
x
= 0 and
x
= 1
are mapped to circles centered at the origin of radii 1 and
e
respectively. The region is
mapped to the annulus 1
≤
r
≤
e
and 0
≤
θ
≤
2
π
.
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 Fall '07
 Lim
 Math, Derivative, Partial differential equation, Complex number, Cauchy's integral theorem, Cauchy–Riemann equations

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