Download COMPLEX ANALYSIS - HOMEWORK 1 Problems: §1.1 # 5 a

COMPLEX ANALYSIS - HOMEWORK 1
RICKY NG
Problems:
§1.1 # 5 a-c, 6 a-c, 7 a-c, 9, 14, 15, 17, 19
§1.2 # 3, 4, 7 c-d
Section 1.1
5. (3 pts) Write the number in the form a + bi.
Solution.
(a) −3
i
2
= 0 + − 23 i.
(b) (8 + i) − (5 + i) = (8 − 5) + (i − i) = 3 + 0i.
(c)
2
i
=
2
i
· ii =
2i
i2
= −2i = 0 + (−2)i.
6. (3 pts) Write the number in the form a + bi.
Solution.
(a) (−1 + i)2 = 1 − 2i + i2 = 0 + (−2)i.
(b)
2−i
1
3
= (2 − i) · 3 = 6 − 3i = 6 + (−3)i.
(c) i(π − 4i) = πi − 4i2 = πi + 4 = 4 + πi.
7. (3 pts) Write the number in the form a + bi.
Solution.
(a)
8i − 1
8i − 1 i
8i2 − i
=
· =
= −(−8 − i) = 8 + i.
i
i
i
i2
(b)
−1 + 5i
−1 + 5i 2 − 3i
13 + 13i
=
·
=
= 1 + i.
2 + 3i
2 + 3i 2 − 3i
13
(c)
3 · 3 i2
8
8 i
8
8
3 i
+ =
+
=
=
· =− i=0+ −
i.
i 3
3i
3i
3i
3i i
3
3
1
9. (2 pts) Write the number in the form a + bi.
Solution.
(2 + 3i)(6 − i) − (8 + i)(1 + 2i)
2 + 3i 8 + i
−
=
1 + 2i 6 − i
(1 + 2i)(6 − i)
9 − i 8 − 11i
9−i
=
·
=
8 + 11i
8 + 11i 8 − 11i
61
107
61 − 107i
=
+ −
i.
=
185
185
185
14. (2 pts) Show that Re(iz) = − Im(z) for every complex number z.
Proof. Write z = a + bi, where a and b are real numbers. Then iz = i(a + ib) = −b + ia,
so Re(iz) = −b. On the other hand, Im(z) = b and we have Re(iz) = − Im(z).
15. (2pts) Let k be an integer. Show that i4k = 1, i4k+1 = i, i4k+2 = −1, and
i4k+3 = −1.
Proof. Since i2 = −1, i4k = i2·2k = (i2 )2k = (−1)2k . 2k is an even number, so (−1)2k = 1
and i4k = 1. From this we deduce
i4k+1 = i4k · i = 1 · i = i;
i4k+2 = i4k · i2 = 1 · i2 = −1;
i4k+3 = i4k · i3 = 1 · i3 = −i.
17. (2pts) Use # 15 to evaluate the 3i11 + 6i3 +
8
i20
+ i−1 .
Solution. Note that i11 = i3 = −i, i20 = 1, and i−1 = i3 = −i. So the above expression is
equivalent to
−3i − 6i + 8 − i = 8 − 10i.
19. (3pts) Write the complex equation z 3 + 5z 2 = z + 3i as two real equations.
Solution. Write z = a + bi, where a and b are real numbers. Rewrite the expressions on
the LHS in terms of their real and imaginary parts:
z 3 = (a + bi)3 = a3 + 3a2 bi − 3ab2 − b3 i
= (a3 − 3ab2 ) + (3a2 b − b3 )i
5z 2 = 5(a + bi)2 = 5(a2 + 2abi − b2 )
= (5a2 − 5b2 ) + (10ab)i.
Thus
Re(z 3 + 5z 2 ) = a3 − 3ab2 + 5a2 − 5b2
Im(z 3 + 5z 2 ) = 3a2 b − b3 + 10ab,
which should agree with
Re(z + 3i) = a
Im(z + 3i) = b + 3,
2
respectively. Therefore, the two real equations are
a3 − 3ab2 + 5a2 − 5b2 = a
(1)
3a2 b − b3 + 10ab = b + 3.
Section 1.2
3. (2pts) Which of the points i, 2 − i, and −3 is farthest from the origin?
Solution. We will compute the modulus of each point.
√
|i| = 02 + 12 = 1;
p
√
|2 − i| = 22 + (−1)2 = 5;
| − 3| = 3;
so −3 is the farthest from the origin.
4. (6 pts) Let z = 3 − 2i. Plot the points z, −z, z, −z, and
the same for z = 2 + 3i and z = −2i.
1
z
in the complex plane. Do
Solution.
A = 3 − 2i = (3, −2);
B = −(3 − 2i) = −3 + 2i = (−3, 2);
C = 3 − 2i = 3 + 2i = (3, 2);
D = −3 − 2i = −3 − 2i = (−3, −2).
For E = z1 ,
1
3 + 2i
3 + 2i
·
=
=
3 − 2i 3 + 2i
13
3 2
,
13 13
.
Plot:
Similarly, here is the plot for z = 2 + 3i:
A = (2, −3); B = (−2, 3); C = (2, 3); D = (−2, −3); E =
3
2
3
,−
13 13
Plot for z = −2i:
A = D(0, −2); B = C = (0, 2); E =
1
0,
2
.
7. (6 pts) Describe the set of points z in the complex plane that satisfies the following.
Solution. (c)
|2z − i| = 4.
Recall that the equation |z − z0 | = r, for a fixed complex number z0 and positive number
r, represents the circle centered at z0 with radius r. Hence, it suffices to rewrite the given
equation into such form. Indeed, dividing the above equation by 2, we get
i
z − = 2.
2
Since the complex number 2i is equivalent to the point (0, 21 ) in the complex plane, the
given equation is a circle centered at the point (0, 12 ) with radius 2.
4
Solution. (d)
|z − 1| = |z + i|.
An algebraic approach: Write z = x + iy and simplify
|x + iy − 1| = |x + iy + i|
(x − 1)2 + y 2 = x2 + (y + 1)2
x2 − 2x + 1 + y 2 = x2 + y 2 + 2y + 1
y = −x.
This is the straight line y = −x.
A geometric approach: Plot the complex numbers 1 and −i in the complex plane, i.e. the
points (1, 0) and (0, −1). Then the above equation describes z as having equal distance
from both points, hence representing the line y = −x.
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