A First Course in Linear Algebra, 2017a

336 Complex Numbers
Therefore, x 3 − 27 =
( ( √
−1 3
(x − 3) x − 3
2 + i 2
))
( (
x − 3 −1
2
+ i
√ ))( (
3
2
x − 3 −1
2
− i
√
3
2
))(
x − 3
( √ ))
−1 3
2 − i 2
Note also
= x 2 + 3x + 9andso
x 3 − 27 =(x − 3) ( x 2 + 3x + 9 )
where the quadratic polynomial x 2 + 3x + 9 cannot be factored without using complex numbers.
( Note that even though the polynomial x 3 − 27 has all real coefficients, it has some complex zeros,
√ ) ( √ )
−1 3
3
2 + i −1 3
,and3
2
2 − i . These zeros are complex conjugates of each other. It is always
2
the case that if a polynomial has real coefficients and a complex root, it will also have a root equal to the
complex conjugate.
Exercises
Exercise 6.3.1 Give the complete solution to x 4 + 16 = 0.
Exercise 6.3.2 Find the complex cube roots of −8.
Exercise 6.3.3 Find the four fourth roots of −16.
Exercise 6.3.4 De Moivre’s theorem says [r (cost + isint)] n = r n (cosnt + isinnt) for n a positive integer.
Does this formula continue to hold for all integers n, even negative integers? Explain.
Exercise 6.3.5 Factor x 3 + 8 as a product of linear factors. Hint: Use the result of 6.3.2.
Exercise 6.3.6 Write x 3 + 27 in the form (x + 3) ( x 2 + ax + b ) where x 2 + ax + b cannot be factored any
more using only real numbers.
Exercise 6.3.7 Completely factor x 4 + 16 as a product of linear factors. Hint: Use the result of 6.3.3.
Exercise 6.3.8 Factor x 4 + 16 as the product of two quadratic polynomials each of which cannot be
factored further without using complex numbers.
Exercise 6.3.9 If n is an integer, is it always true that (cosθ − isinθ) n = cos(nθ) − isin(nθ)? Explain.
Exercise 6.3.10 Suppose p(x)=a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 is a polynomial and it has n zeros,
z 1 ,z 2 ,···,z n
listed according to multiplicity. (z is a root of multiplicity m if the polynomial f (x)=(x − z) m divides p(x)
but (x − z) f (x) does not.) Show that
p(x)=a n (x − z 1 )(x − z 2 )···(x − z n )
♠