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Math 412 Group Work #3 Spring 2009 SOLUTIONS
Problem 1. Compute
(1 − i
3 (
3 + i) 2 .
SOLUTION: The computation is greatly simplified if we write 1 − i
3 and
3 + i
in polar form:
1 − i
3 = 2e −iπ/ 3 and
3 + i = 2e iπ/ 6 .
Thus,
(1 − i
3 (
3 + i) 2 =
2 e −iπ/ 3
2 e iπ/ 6
= 32e −iπ e iπ/ 3 = 32e −i 23 π
[
cos(
2 π
3
) − i sin(
2 π
3
]
[
− i
]
3)i.
Problem 2. Find all cube roots of the complex number −2 + 2i.
SOLUTION: We begin by expressing z = −2 + 2i in polar form:
−2 + 2i = (
8)e i( 34 π +2πk) .
Thus,
(−2 + 2i) 1 / 3 =
2 e i( π 4 + 2 πk 3 ) .
To get the three cube roots of −2 + 2i, we substitute k = 0, k = 1, and k = 2:
k = 0 gives z 0 =
2 e iπ/ 4 = 1 + i.
k = 1 gives z 1 =
2 e i 1112 π =
[
cos
11 π
12
11 π
12
)]
k = 2 givesz 2 =
2 ei^
19 π (^12) =
[
cos
19 π
12
19 π
12
)]
Problem 3. Find all roots of the equation
z 4 − 4 z 3
You may freely use the fact that z = i is one of the roots.
SOLUTION: Since the polynomial in question contains only real coefficients, any
complex roots must arise in conjugate pairs by Problem 5(a) in Section 1.2. Therefore,
z = i = −i is also a root. Hence, p(z) = z 4 − 4 z 3
- 6z 2 − 4 z + 5 is divisible by
(z − i)(z + i) = z 2
- In fact, long division reveals that
p(z) = (z 2
so the two remaining roots of p(z) are precisely the roots of the quadratic equation
z^2 − 4 z + 5 = 0. From the quadratic equation, these roots are z = 4 ±
√ − 4 2 = 2 ± i.
Therefore, the four roots of the given equation are
z = i, z = −i, z = 2 + i, z = 2 − i.
