Exercises 5.1
1. (a) 6 + 4i.
(b) 5 + 14i.
(c) 28 + 96i.
(d)
(e)
(f)
11
+ 10
i.
17
17
3
− 25 i.
2
−31
+ 367
i.
200
200
2. (a) ± √12 (1 − i).
√
√
(b) ±( 2 + 3i).
(c) ±(6 − 7i).
√
3. (a) 21 (−1 ± 3i).
√
(b) 14 (3 ± 7i).
(c) The roots are 1 + 2i and 1 + i.
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Exercises 5.2
1. (a) The quotient is x2 + 2x − 3 and the remainder is −7.
(b) The quotient is x2 − 3x + 8 and the remainder is −27x + 7.
(c) The quotient is 2x2 − 3x and the remainder is 1.
2. (a) We are given that 4 is a root and so we know that x − 4 is a factor.
Dividing out we get 3x2 − 8x + 4. This is a quadratic and so we
can find its roots by means of completing the square. We get 2
and 32 . Thus the roots are 4, 2, 32 .
(b) We are given that −1 and −2 are roots and so (x + 1)(x + 2) is a
factor. Dividing
out we get x2 − x + 1. The roots of√this quadratic
√
are 12 (1 ± 3i). Thus the roots are −1, −2, 21 (1 ± 3i).
3. The required cubic is (x − 2)(x + 3)(x − 4) = x3 − 3x2 − 10x + 24.
4. The required quartic is (x − i)(x + i)(x − 1 − i)(x − 1 + i) = x4 − 2x3 +
3x2 − 2x + 2.
5. By assumption x3 + ax2 + bx + c = (x − α)(x − β)(x − γ). Multiplying
out the RHS we get
x3 − (α + β + γ)x2 + (αβ + αγ + βγ)x − αβγ.
Now equate with the coefficients of the LHS to get
a = −(α + β + γ), b = αβ + αγ + βγ, c = −αβγ.
6. The polynomial in question has real coefficients and so the complex
√
roots come in complex conjugate
follows therefore that 3−i 2
√ pairs. It√
is also a root. Thus (x−3−i 2)(x−3+i 2) = x2 −6x+11 is a factor.
Dividing out we get x2 + 7x + 6. This factorises as (x + 1)(x
√+ 6) and√so
its roots are −1 and −6. Thus the roots are −1, −6, 3 + i 2, 3 − i 2.
√
7. The polynomial in question
has real√roots and so 1 + i 5 is another
√
root. Thus (x − 1 − i 5)(x − 1 + i 5) is a factor.
√ Dividing
√ out by
x2 − 2x + 6 we get√x2 − 2. √
This√factorises
as
(x
−
2)(x
+
2). Thus
√
the roots are 1 + i 5, 1 − i 5, 2, − 2.
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8. (a) −1 is a root and so x + 1 is a factor. We can write the polynomial
as the product (x + 1)(x2 + 1). The roots are therefore −1, i, −i.
(b) −2 is a root and so x + 2 is a factor. We can therefore write
the polynomial
as (x +√2)(x2 − 3x + 3). The roots are therefore
√
−2, 21 (3 + i 3), 12 (3 − i 3).
(c) 1 is a root and so we get a first factorisation of our polynomial
as (x − 1)(x3 + 5x + 6). −1 is a root of x3 + 5x + 6. We may
2
therefore factorise x3 +
√5x + 6 = (x + 1)(x − x + 6). The quadratic
1
has the roots 2 (1 ± i 23). The roots are therefore 1, −1, 12 (1 +
√
√
i 23), 21 (1 − i 23).
9. (a) Show that 1 is a root and then divide by x − 1 to get the required
factorisation (x − 1)(x2 + x + 1). Observe that x2 + x + 1 has
complex roots and so cannot be factorised further in terms of real
polynomials.
(b) This is a difference of two squares and so a first factorisation is
(x2 + 1)(x2 − 1) and thus the required factorisation is (x − 1)(x +
1)(x2 + 1). Observe that x2 + 1 has complex roots and so cannot
be factorised further in terms of real polynomials.
(c) Put y = x2 and Solve y 2 + 1 = 0. The solutions are ±i. Thus
x2 = i or x2 = −i. Taking square roots, yields x = √12 (1 +
−1
i), √
(1+i), √12 (−1+i), √12 (1−i). Now we collect together complex
2
√
conjugate pairs, to get (x− √12 (1+i))(x− √12 (1−i)) = x2 − 2x+1,
√
√
√
and x2 + 2x + 1. Thus x4 + 1 = (x2 − 2x + 1)(x2 + 2x + 1).
A student made a very nice observation that leads to a much
quicker solution to this question. Observe that x4 + 1 = (x2 +
1)2 − 2x2 . How does this help?
19
Exercises 5.3
1. By De Moivre’s theorem
(cos x + i sin x)5 = cos 5x + i sin 5x.
Expand the LHS to get
(cos x)5 + 5(cos x)4 (i sin x) + 10(cos x)3 (i sin x)2 + 10(cos x)2 (i sin x)3 +
5(cos x)(i sin x)4 +(i sin x)5 . Simplifying and equating real and complex
parts we get
cos 5x = cos5 x − 10 cos3 sin x + 5 cos x sin4 x
and
sin 5x = 5 cos4 x sin x − 10 cos2 x sin3 x + sin5 x.
2. (a) We have that eix = cos x+i sin x and e−ix = cos(−x)+i sin(−x) =
cos x − i sin x. Thus eix − e−ix = 2i sin x. It follows that sin x =
1
(eix − e−ix )
2i
(b) Using the calculations of (a) above, we have that eix + e−ix =
2 cos x and so we get the result.
We have that cos x = 21 (eix + e−ix ). Taking fourth powers of both sides
we get that
cos4 x =
This simplifies to
1 4ix
e + 4e2ix + 6 + 4e−2ix + e−4ix .
16
1
(cos 4x + 4 cos 2x + 3) ,
8
as required.
3. 1, i, −1, −i.
√ 1 + i 3 . Then the roots are 1, ω, ω 2 , ω 3 , ω 4 , ω 5 .
4. Let ω =
1
2
5. Let ω =
√1 (1
2
+ i). Then the roots are 1, ω, ω 2 , ω 3 , ω 4 , ω 5 , ω 6 , ω 7 .
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6. We have that
−8i = 8 (cos 270◦ + i sin 270◦ ) .
Thus the first cube root is
1
2i = 8 3 (cos 90◦ + i sin 90◦ ) .
It follows that the cube roots are
2i, 2ωi, 2ω 2 i
where we have used the cube roots of unity described in the text. It
follows that the required cube roots are
√
√
2i, − 3 − i, 3 − i.
7. Put z = ii . Then ln(z) = i ln(i). Now exp(i( π2 + 2πk)) = i where k is
any integer. It follows that ln(i) = i( π2 +2πk). Thus ln(z) = −( π2 +2πk).
Hence ii = exp(−( π2 + 2πk)). All of these values are real.
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