A First Course in Linear Algebra, 2017a

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6.1.6 If p(z)=0, then you have
p(z)=0 = a n z n + a n−1 z n−1 + ···+ a 1 z + a 0
6.1.7 The problem is that there is no single √ −1.
= a n z n + a n−1 z n−1 + ···+ a 1 z + a 0
= a n z n + a n−1 z n−1 + ···+ a 1 z + a 0
= a n z n + a n−1 z n−1 + ···+ a 1 z + a 0
= p(z)
6.2.5 You have z = |z|(cosθ + isinθ) and w = |w|(cosφ + isinφ). Then when you multiply these, you
get
|z||w|(cosθ + isinθ)(cosφ + isinφ)
= |z||w|(cosθ cosφ − sinθ sinφ + i(cosθ sinφ + cosφ sinθ))
= |z||w|(cos(θ + φ)+isin(θ + φ))
6.3.1 Solution is:
(1 − i) √ 2,−(1 + i) √ 2,−(1 − i) √ 2,(1 + i) √ 2
6.3.2 The cube roots are the solutions to z 3 + 8 = 0, Solution is: i √ 3 + 1,1 − i √ 3,−2
6.3.3 The fourth roots of −16 are the solutions of x 4 + 16 = 0 and thus this is the same problem as 6.3.1
above.
6.3.4 Yes, it holds for all integers. First of all, it clearly holds if n = 0. Suppose now that n is a negative
integer. Then −n > 0andso
[r (cost + isint)] n =
1
[r (cost + isint)] −n = 1
r −n (cos(−nt)+isin(−nt))
r n
=
(cos(nt)− isin(nt)) = r n (cos(nt)+isin(nt))
(cos(nt)− isin(nt))(cos(nt)+isin(nt))
= r n (cos(nt)+isin(nt))
because (cos(nt) − isin(nt))(cos(nt)+isin(nt)) = 1.
6.3.5 Solution is: i √ 3 + 1,1 − i √ 3,−2 and so this polynomial equals
( (
(x + 2) x − i √ ))( (
3 + 1 x − 1 − i √ ))
3
6.3.6 x 3 + 27 =(x + 3) ( x 2 − 3x + 9 )