Download Homework # 1 Solutions Problem 11, p. 4 Solve z 2 + z +1=0. Strictly

Homework # 1 Solutions
Problem 11, p. 4 Solve z 2 + z + 1 = 0.
Strictly speaking, we don’t have the quadratic formula at our disposal for
complex numbers, but it does work. The method suggested in the book is
essentially to try it out on real x, see there are no real solutions, and then
solve some equations in x and y. But, let’s just use the quadratic formula:
p
−1 ± 12 − 4(1)(1)
z =
√2
1
3
= − ±i
2
2
Problem 1b, p. 7 Reduce to a real number:
5i
(1 − i)(2 − i)(3 − i)
5i
(1 − 3i)(3 − i)
5i
=
−10i
1
= −
2
=
Problem 5b, p. 13 Sketch the set of points determined by the condition
|z + i| ≤ 3.
This is the set of points inside or on the circle centered at −i of radius 3:
Problem 2a, p. 16 Sketch the set of points determined by the condition
Re(z − i) = 2.
Writing z = x − iy, we have Re(z − i) = Re(x − i(y + 1)) = x. So the set of
points is simply the line x = 2.
2
Problem 5b, p. 23 Show that
5i
2+i
5i
= 1 + 2i.
2+i
5eiπ/2
√
5ei arctan(1/2)
√ i(π/2−arctan(1/2))
5e
=
√ i arctan(2)
=
5e
= 1 + 2i
=
Here, we have used the identity π/2 − arctan(x) = arctan(1/x) for x > 0.
Problem 4b, √
p. 30 √
Find the√sixth roots of 8.
√
6
Here we have 6 8 = 23 = 2. So the modulus of any sixth root is 2.
The arguments are
2nπ
π 2π
4π 5π
= 0, , , π, , .
6
3 3
3 3
The sixth roots are then
√
√
√
1 + 3i 1 − 3i
± 2, ± √
,± √
.
2
2
Problem 5, p. 34 Let S be the open set consisting of all points z such that
|z| < 1 or |z − 2| < 1. Explain why S is not connected.
The set S consists of the points interior to two circles, one centered at 0
and one centered at 2, both of radius 1. These circles do not overlap and
therefore no point in one may be joined to a point in the other by a path
that does not leave the set.