Download Solutions for HW # 1

MATH 331 — Complex Analysis / Spring 2015
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Solutions to Homework Assignment # 1
Required problems
Problem 1
z = x + iy
⇒
1/z = (x − iy)/(x2 + y 2 )
⇒
Re(1/z) = x/(x2 + y 2 ).
Then
Re(1/z) ≥ a
⇒
x2 − (x/a) + y 2 ≤ 0
⇒
(x − a/2)2 + y 2 ≤ (a/2)2 .
Answer: inside of the circle |z − a/2| = a/2.
Problem 2
By the meaning of | . . . |, these are points equidistant from z = 0 and z = a, and hence they form the
perpendicular bisector of segment [0, a].
Algebraic specification, given z = x + iy, a = u + iv:
x2 + y 2 = (x − u)2 + (y − v)2
⇒
2xu + 2yv = u2 + v 2 .
Problem 3
√
By the Triangle inequality, |z + a| ≤ |z| + |a| = 1 + 13.
Problem 4
Let z = r eiθ . Then z = r e−iθ , so we need to solve
{
√
1 4iθ
i arctan(4/3)
e
=
3
+
4i
=
5
e
⇒
r
=
1/5,
r2
}
2πn
1
, n = 0, 1, 2, 3.
θ = arctan(4/3) +
4
4
Problem 5
(a) Using arctan(−x) = −arctan(x), we get:
(√
) (√
)
5ei arctan(2) /
2e−i arctan(1) .
(b)
(1 + 2i)(1 + i)/(12 + 12 ) = (−1 + 3i)/2.
(c)
arg of part (a) = arctan(2) + arctan(1).
arg of part (b) = arctan(−3) + π.
Hence arctan(1)+arctan(2)+arctan(3) = π.
1
Problem 6
(a) z = ei arctan(2) t;
(b) z = 1 − it. In both cases, t ≥ 0.
Problem 7
(a) arg(2z)=arg(2)+arg(z) = 0 + θ; then similarly:
(b) arg(iz) = π/2 + θ;
(c) arg(−z) = π + θ;
(d) arg(z) = −θ.
Problem 8
It is any sector
of angular width 2π/3 in C.
Example 1: {z arg(z) ∈ [0, 2π/3)}; Example 2: {z arg(z) ∈ (π/3, π]}.
Problem 9
1/eiθ = e−iθ
⇒
w( upper half of unit circle ) = lower half of unit circle.
w( [0, 1] ) = [1, +∞),
w( [−1, 0] ) = (∞, −1];
w(i/2) = −2i.
Answer: Region below the union of lines: {z ∈ (∞, −1] } ∪ {|z| = 1 & arg(z) ∈ [−π, π] } ∪ {z ∈ [1, +∞) }.
Bonus problems
B-1
(a) Ellipse and (b) Hyperbola with foci at z = a and z = b.
B-2
Let z = x + iy, a = p + iq; then:
3
3 2
x − 2xa + a2 + y 2 − 2yb + b2 = 0.
4
4
Let us work on the x-part; the y-part will transform similarly.
(
( )2 )
(
)
3
8
4
4 2
3
4 2 1 2
3 2
2
2
2
x − 2xa + a =
x − xa +
a
− a +a =
x− a − a .
4
4
3
3
3
4
3
3
x2 + y 2 = 4(x − a)2 + 4(y − b)2
⇒
Thus, the original equation becomes
(
)
)
(
4 2
4 2 4 2
x − a + y − b = (a + b2 ),
3
3
9
which describes a circle with the center outside of segment [0, a]. This cirle is called Apollonius circle of points
0 and a and plays a fundamental role in non-Euclidean (hyperbolic) geometry.
√
1 + t2 ei arctan(t)
z=√
= e2i arctan(t) ,
1 + t2 e−i arctan(t)
This is a unit circle with point z = −1 excluded.
B-3
arctan(t) ∈ (−π/2, π/2).
B-4
{
}
Let z ∈ eiθ , θ ∈ [0, π] . Then w = 2 cos θ, so w ∈ [−1, 1].
Let z ∈ [−1, 1]; then w ∈ (−∞, −2] ∪ [2, ∞).
Finally, w(i/2) = −3i/2. Thus, the required image is the lower half of C.
B-5
{
}
Let z ∈ 2 eiθ , θ ∈ [0, π] . Then w = (5/2) cos θ + i(3/2) sin θ, so w is on the upper half of the ellipse
(x/2.5)2 + (y/1.5)2 = 1.
Let z ∈ [−2, 2]; then w ∈ (−∞, −2.5] ∪ [2/5, ∞).
Finally, w(i/2) = −3i/2. Thus, the required image is the region below the union of rays (−∞, −2.5],
[2/5, ∞) and the aforementioned upper arc of the ellipse.
2