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Solutions to the first midterm exam
Complex variables, 802346.
December 31, 2009
Exercises and solutions:
Exercise 1:Sketch the graph of the given equation in the complex
plane
a. |2z 1| = 4 b. Im{z 1} = Re{z + 4 3i}
Solution of exercise 1:
1. Note that the expression |2z 1| = 4 can be written as 2|z 12|
= 4
and |z 12| = 2 or |2(x + iy) 1| = 4 and |(2x 1) + i2y| = 4,
then
(2x 1)2 + (2y)2 = 42 or 4(x 12
)2 + 4y2 = 42 and (x 12
)2 + y2 = 4,this is the equation of a circle centered at (1
2, 0) with radius 2, see
figure 1.
2. Using the fact z = x + iy, x = Re{z}, y = Im{z} then Im{z 1}
=Re{z + 4 3i} gives Im{(x 1) + iy} = Re{(x + 4) + i(y 3)} ory = x +
4, see figure 2.
Exercise 2:Use the equation
wk = r1/n
cos
+ 2k
n
+ isin
+ 2k
n
,
for k = 0, 1, 2, . . . , n 1, to compute all roots of the
following expres-sions. Give the principal nth root in each case.
Sketch w0, w1, w2, . . . , wn1,on an appropriate circle centered at
the origin.
a. (1 + i)1/5
b. (1)1/4
y
x
2
-2
2.5-1.5
z-plane
Figure 1: Circle of exercise 1
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y
x
4
-4
y=x+4
Figure 2: Line of exercise 1
w0
w1
w2
w3w4
Figure 3: The five roots of 1 + i of exercise 2
Solution to exercise 2:Using the equation
wk = r1/n
cos
+ 2k
n
+ isin
+ 2k
n
,
for k = 0, 1, 2, . . . , n 1, we will compute the roots of1. (1
+ i)
1/5, where z = 1+ i and n = 5, thus r =
2, (
2)1/5 = 1.0718,
= tan1(1) = 4 and k = 0, 1, 2, 3, 4. Let k = 1n( + 2k),
seefigure 3. For
(a) k = 0, 0 =1
5(4
) = 20
, then w0 = (
2)1/5
cos20
+ isin
20
=
1.07189 = 1.0586 + i0.1677.
(b) k = 1, 1 =1
5(4
+2) = 920
, then w1 = (
2)1/5
cos920
+ isin
920
=
1.071881 = 0.1677 + i1.0586.
(c) k = 2, 2 =1
5(4
+4) = 1720
, then w2 = (
2)1/5
cos1720
+ isin
1720
=
1.0718153 = 0.9549 + i0.4868.(d) k = 3, 3 =
1
5(4
+6) = 54
= 34
, then w3 = (
2)1/5
cos34
isin 34
=
1.0718 135 = 0.7579 i0.7579.(e) k = 3, 3 =
1
5(4
+8) = 3320
= 720
, then w4 = (
2)1/5
cos720
isin 720
=
1.0718
63 = 0.4866
i0.9549.
2. (1)1/4, where z = 1 and n = 4, thus r = 1, = and k =0, 1, 2,
3. Let k =
1
n ( + 2k), see figure 4. For
(a) k = 0, 0 =1
4() =
4, then w0 = cos
4
+ isin
4
= 145 =
0.7071 + i0.7071.
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w0
x
y
w1
w2 w3
Figure 4: The four roots of 1 of exercise 2
x
S
y=x
y
Figure 5: The set S of y < x in exercise 3
(b) k = 1, 1 =1
4( + 2) = 3
4, then w1 = cos
3pi4
+ isin
34
=
1135 = 0.7071 + i0.7071.(c) k = 2, 2 =
1
4( + 4) = 5
4= 3
4, then w2 = cos
3pi4
isin
34
= 1 135 = 0.7071 i0.7071.
(d) k = 3, 3 =1
4( + 6) = 7
4=
4, then w3 = cos
pi4
isin
4
= 1 45 = 0.7071 i0.7071.
Exercise 3:
Sketch the set S of points in the complex plane satisfying the
given in-equality
1. Im {z} < Re {z} ,2. 2 |z 3 + 4i| 5.
Determine whether the set is
a. open b. closed c. a domain
d. bounded e. connected
Solution to exercise 3:
1. Note that Im {z} < Re {z} , gives y < x. That is, S is
every pointbelow the line y = x, this is all the region below the
line y = x, seefigure 5. The set S is open,because every point in S
has a neighbor-hood that is also in S.Since S is open , it cannot
be closed.Because it is open, it is a domain, but not bounded. It
is connected.
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x
y
S
3
-4
Figure 6: The set S of 2 |z 3 + 4i| 5 in exercise 3
2. The inequality 2 |z 3 + 4i| 5 gives 2 |(x 3) + i(y + 4)|
5,see figure 6. S is not open, it is closed. Because, it is not
open, itis not a domain, it is bounded. But it is not connected,
because itincludes its boundary points.
Exercise 4:
Usez
=x
+iy
,x
=Re
{z},
y=
Im{
z} and
Re{
z} |
Re{
z}| |
z|, where|z| =
x2 + y2. To prove the inequality
|Re{z}| + |Im{z}|
2|z|.
Solution to exercise 4:Using z = x + iy, x = Re{z}, y = Im{z},
Re{z} |Re{z}| |z|, andIm{z} |Im{z}| |z|, where |z| =
x2 + y2, then x |x|
x2 + y2
and y |y|
x2 + y2. The addition of the last two expressions gives
(x + y) (|x| + |y|) 2
x2 + y2, or
|Re{z}| + |Im{z}| 2|z|.
To get a lower bound to (|x| + |y|) , we need to use the
inequality (|x| |y|)2 0. The latter implies that |x|2 + |y|2
2|x||y| 0 or |x|2 + |y|2 2|x||y| and 2(|x|2 + |y|2) = |x|2 + |y|2 +
|x|2 + |y|2 |x|2 + |y|2 + 2|x||y| =(|x| + |y|)2 and 2(|x|2 + |y|2)
(|x| + |y|)2. But x |x| and y |y|, sox2 |x|2, y2 |y|2, and x2 + y2
|x|2 + |y|2. Hence, 2(|x|2 + |y|2) 2(x2 + y2) (|x| + |y|)2 and
2
x2 + y2 |x| + |y|, or
2|z| |Re{z}| + |Im{z}|.
Exercise 5:Suppose that z = r(cos + isin). Ifn is an integer,
evaluate zn + zn andzn zn.
Solution to exercise 5:Since z = r(cos + isin) and z = r(cos
isin), then DeMoivre formulagives zn = rn [cos(n) + isin(n)] and zn
= rn [cos(n) isin(n)], for
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an integer n, and
zn + zn = rn [cos(n) + isin(n)] + rn [cos(n) isin(n)]= rn
[cos(n) + isin(n) + cos(n) isin(n)]= 2rncos(n).
In a similar manner,
zn zn = rn [cos(n) + isin(n)] rn [cos(n) isin(n)]= rn [cos(n) +
isin(n) cos(n) + isin(n)]= i2rnsin(n).
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