# 1 Opening items

## 1.1 Module introduction

Section 2 of this module is concerned with Demoivre’s theorem and its applications. We start in Subsection 2.1 by proving the theorem which states that

(cosθ + isinθ)n = cos() + isin()

(where i2 = −1), and then use it to derive trigonometric identities, in Subsection 2.2, and to find all solutions to the equation zn − 1 = 0 (the roots of unity) in Subsection 2.3.

The remainder of this module is concerned with complex algebra; that is the manipulation of expressions involving complex variables. In Subsection 3.1Subsections 3.1 and Subsection 3.23.2 we solve some algebraic equations and consider the related problem of factorization. In Subsection 3.3 we point out techniques for simplifying complex algebraic expressions. Subsection 3.4 is concerned with the complex binomial expansion; that is, expanding (a + b)n in terms of powers of the variables a and b. Proofs of this theorem do not usually distinguish between real and complex variables, but there are applications which are specific to the complex case. Finally in Subsection 3.5 we mention the complex form of the geometric series and use it to obtain more trigonometric identities. Don’t worry if you are unfamiliar with the physics used in the examples in this module.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Ready to study? in Subsection 1.3.

## 1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 4.1Module summary and the Subsection 4.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 4.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

Question F1

Use Demoivre’s theorem to find all the roots of zn − 1 = 0, where n is a positive integer. For n = 3, plot your results on an Argand diagram.

We can write the number, 1, as

1 = e2πk = cos(2πk) + isin(2πk)

where k is any integer (positive, negative or zero). So z is given by

z = 12 = [cos(2πk) + isin(2π k)]2

and Demoivre’s theorem then allows us to rewrite the right–hand side as

$z = \cos\left(\dfrac{2\pi k}{n}\right) + i\sin\left(\dfrac{2\pi k}{n}\right)$

The sine and cosine functions are periodic with period 2π, so it follows that all the different integer values of k only give n distinct values of z and it is convenient to use k = 0, 1, 2, ..., (n − 1) to generate these values. For n = 3 the roots are given by

$z = \cos\left(\dfrac{2\pi k}{3}\right) + i\sin\left(\dfrac{2\pi k}{3}\right)$

where k = 0, 1, 2. These roots are plotted on an Argand diagram in Figure 2.

Question F2

Use Demoivre’s theorem to find z5 in its simplest form, where z = 2[cos(π/10) + isin(π/10)]

Using Demoivre’s theorem we have

z5 = 25[cos(π/10) + isin(π/10)]5 = 25[cos(5π/10) + isin(5π/10)] = 25[cos(π/2) + isin(π/2)] = 25i = 32i

Question F3

Use Demoivre’s theorem, together with the complex binomial theorem, to show that

(a) cos(4θ) = cos4θ − 6cos2θsin2θ + sin4θ

(b) sin(4θ) = 4cos3θsinθ − 4cosθsin3θ

From Demoivre’s theorem we have

cos(4θ) + isin(4θ) = [cosθ + isinθ]4

The right–hand side of this expression can be expanded using the binomial theorem to give

$\displaystyle (a+b)^4 = \sum_{r=0}^n({}^4C_{4-r}a^{4-r}b^r) = {}^4C_4a^4 + {}^4C_3a^3b + {}^4C_2a^2b^2 + {}^4C_1ab^3 + {}^4C_0b^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4$

where ${}^nC_r = \dfrac{n!}{r!(n-r)!}$ and n! = n(n − 1)(n − 2)(n − 3) ... 2 × 1 for n ≥ 1 with 0! = 1

So, putting together the results obtained from Demoivre’s theorem and the binomial theorem, we obtain

cos(4θ) + isin(4θ) = cos2θ + i4cos2θsinθ − 6cos2θsin2θi4cosθsin2θ + sin2θ
cos(4θ) + isin(4θ) = [cos2θ − 6cos2θsin2θ + sin2θ] + i[4cos2θsinθ − 4cosθsin2θ]

Equating real and imaginary parts we find

cos(4θ) = cos2θ − 6cos2θsin2θ + sin2θ

sin(4θ) = 4cos2θsinθ − 4cosθsin2θ

Question F4

Z is given by

$Z = R + i\left(\omega L-\dfrac{1}{\omega C}\right)$i

where R, ω, L and C are all real. Find the real and imaginary parts of Z−1.

Z can be written as

$Z = \dfrac{\omega RC+i\left(\omega^2LC-1\right)}{\omega C}$

and therefore

$Z^{-1} = \dfrac{\omega C}{\omega RC+i\left(\omega^2LC-1\right)}$

$\phantom{Z^{-1} }= \dfrac{\omega C}{\omega RC+i\left(\omega^2LC-1\right)} \times \dfrac{\omega RC-i\left(\omega^2LC-1\right)}{\omega RC-i\left(\omega^2LC-1\right)}$

$\phantom{Z^{-1} }= \dfrac{\omega C[\omega RC-i\left(\omega^2 LC-1\right)]}{\omega^2 R^2C^2+\left(\omega^2RC-1\right)^2}$

${\rm{Re}}\left(z\right) = \dfrac{\omega^2 C^2}{\omega^2 R^2C^2+\left(\omega^2RC-1\right)^2}~~~~~~~~{\rm{Im}}\left(z\right) = \dfrac{\omega C\left(1-\omega^2 LC\right)}{\omega^2 R^2C^2+\left(\omega^2RC-1\right)^2}$

Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to the following Ready to study? Subsection.

Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Section 4Closing items.

i

Study comment To begin the study of this module you need to be familiar with the following: the representation of a complex number on an Argand diagram; the modulus_of_a_complex_numbermodulus, complex conjugate and argument (including the principal value) of a complex number, Euler’s formula; the rationalizing_a_complex_quotientrationalization of complex quotients; trigonometric identities, such as cos(2θ) = cos2θ − sin2θ and results such as sin(π/3) = $\sqrt{3\os}$/2 and cos(π/3) = 1/2. You will also need to be familiar with the binomial expansion (for real numbers), the sum of a geometric series of real numbers, the formula for the solution of a quadratic equation and the fundamental theorem of algebra. If you are uncertain about any of these terms, you can review them now by reference to the Glossary, which will indicate where in FLAP they are developed. The following Ready to study questions will help you to establish whether you need to review some of the above topics before embarking on this module.

Throughout this module $\sqrt{x\os}$ means the positive square root so that $\sqrt{4\os} = 2$, and i2 = −1.

Question R1

A complex number, z, is such that its real part has the value, 1, and its imaginary part is $\sqrt{3\os}$.

(a) Express z in cartesian_form_of_a_complex_numberCartesian, polar_form_of_a_complex_numberpolar and exponential_form_of_a_complex_numberexponential forms.

(b) Express the complex conjugate of z, z*, and z−1 in exponential form.

(a) The cartesian_form_of_a_complex_numberCartesian form of a complex number is z = x + iy where x and y are real. In this case we have

$z=1+i\sqrt{3\os}$

The polar_form_of_a_complex_numberpolar form of a complex number is

z = r(cosθ + isinθ)

where r and θ are real. In this case we have

$r=\sqrt{x^2+y^2} = \sqrt{1+3\os} = 2\cos\theta = \dfrac12\quad\text{and}\quad\sin\theta = \dfrac{\sqrt{3\os}}{2}$

From the equilateral triangle shown in Figure 3 we find θ = π/3 and therefore the polar form of z is

z = 2[cos(π/3) + isin(π/3)]

The exponential_form_of_a_complex_numberexponential form of a complex number is z = re, where r and θ are again real. Using Euler’s formula

e = cosθ + isinθ

and comparing the result just found for the polar form, we have

z = 2e/3

(b) For an arbitrary complex number, z = x + iy, the complex conjugate is given by z* = xiy. If z is in exponential form, z = re then this corresponds to z* = re.

For the complex number given in this question, we have

z* = 2e/3

For an arbitrary complex number given in exponential form z = re, the inverse is given by z−1 = r−1e.

If z = 2e/3 we have

$z^{-1} = \dfrac12{\rm e}^{-i\pi/3}$

Consult the relevant terms in the Glossary for further details.

Question R2

A complex number,w, is defined by w = 1 + i   and   z = 1 + i$\sqrt{3\os}$.

(a) Express w in polar form. Use this result, together with your answer to Question R1, to find the polar form of zw.

(b) Find Re(zw) and Im(zw) (that is, the real and imaginary parts of zw).

(a) We can represent w in polar form by

w = ρ(cosφ + isinφ)

In the present case the value of ρ is given by

$\rho = \sqrt{1^2+1^2} = \sqrt{2\os}$

andφ by $\cos\phi=\dfrac{1}{\sqrt{2\os}}$   and   $\sin\phi=\dfrac{1}{\sqrt{2\os}}$

From the right–angled triangle with two equal sides shown in Figure 4 we see that φ is π/4 and therefore the polar form of w is

w = 2[cos(π/4) + isin(π/4)]

In general, if complex numbers, z and w, are written in polar form as

z = r(cosθ + isinθ)   and   w = ρ(cosφ + isinφ)

then their product is given by zw = [cos(θ + φ) + isin(θ + φ)]

(b) This means that the modulus_of_a_complex_numbermoduli are multiplied and the arguments summed. For the numbers given in this and the previous questions we have

$zw = 2\sqrt{2\os}\left[\cos\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right) + i\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)\right]$

$\phantom{zw }= 2\sqrt{2\os}\left[\cos\left(\dfrac{7\pi}{12}\right) + i\sin\left(\dfrac{7\pi}{12}\right)\right]$

and the real_partreal and imaginary parts are given by

${\rm{Re}}\left(zw\right) = 2\sqrt{2\os}\cos\left(\dfrac{7\pi}{12}\right) = -0.732$

${\rm{Im}}\left(zw\right) = 2\sqrt{2\os}\sin\left(\dfrac{7\pi}{12}\right) = 2.732$

Consult the relevant terms in the Glossary for further details.

Question R3

Given that z = 3 + 4i, plot z and z* on an Argand diagram. What is the value of |z|?

The complex numbers z = 3 + 4i and z* = 3 − 4i are plotted on an Argand diagram in Figure 5. For an arbitrary complex number, z = x + iy, the modulus_of_a_complex_numbermodulus is given by $\lvert\,z\,\rvert = \sqrt{x^2+y^2}$ and for the present case we have

$\lvert\,z\,\rvert = \sqrt{3^2+4^2} = 5$

Consult the relevant terms in the Glossary for further details.

Question R4

Rationalize the expression   $\dfrac{2+3i}{(1+i)(1-2i)}$

Rationalizing a complex quotient means bringing it to the form x + iy, where x and y are real. This can be achieved by multiplying the numerator and the denominator by the complex conjugate of the denominator. In this case we have

$\dfrac{2 + 3i}{(1+ i)(1 - 2i)} = \dfrac{(2 + 3i)(1 - i)(1 + 2i)}{(1+ i)(1 - i)(1 - 2i)(1+ 2i)}$

$\phantom{\dfrac{2 + 3i}{(1+ i)(1 - 2i)} }= \dfrac{1}{2\times 5}[(2+3i)(1+i+2)]$

$\phantom{\dfrac{2 + 3i}{(1+ i)(1 - 2i)} }= \dfrac{(2+3i)(3+i)}{10} = \dfrac{6-3+9i+2i}{10}$

$\phantom{\dfrac{2 + 3i}{(1+ i)(1 - 2i)} }= \dfrac{3}{10} +\dfrac{11}{10}i$

Consult complex roots in the Glossary for further information.

Question R5

(a) State the fundamental theorem of algebra.

(b) How many roots would you expect the following equation to have

z5 + z4 + z3 + z2 + z + 1 = 0

(c) If z is restricted to real values, what does the fundamental theorem of algebra tell us about the number of real

roots of this equation?

(a) The fundamental theorem of algebra states that each polynomial with complex number coefficients and of degree n has, counting multiple root_of_an_equationroots an appropriate number of times, exactly n complex roots.

(b) The polynomial given in the question is of order 5 and therefore it has 5 roots.

(c) If z is restricted to take only real values then the theorem tells us nothing about the number of real roots except for the fact that (in this case) there can be no more than five.

Consult the relevant terms in the Glossary for further details.

Question R6

(a) Sum the series 1 + x + x2 + ... + x10   (for x ≠ 1)

(b) Expand (1 + x)7 in powers of x.

(a) Using the formula for the sum of a geometric series we find

$1+x+x^2+\dots +x^{10} = \dfrac{1-x^{11}}{1-x}$

(b) The binomial expansion gives

(1 + x)7 = 1 + 7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7

Consult geometric series for complex numbers in the Glossary for further information.

# 2 Demoivre’s theorem

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## 2.1 Introduction

If we have an arbitrary complex number, z, then we can choose to write it in polar form as

z = r(cosθ + isinθ)

where r and θ are real (and i2 = −1). Furthermore, if we have another complex number

w = ρ(cosφ + isinφ)

then the product of z and w can be written as

zw = [cos(θ + φ) + isin(θ + φ)](1)

In other words, the moduli of the numbers are multiplied together and the arguments are summed. It is easy to generalize this result to n complex numbers in the following way

z1 = r1[cos(θ1) + isin(θ1)]

z2 = r2[cos(θ2) + isin(θ2)]

z2 = r2$~~\vdots~~~~~~~~~~\vdots~~~~~~~~~~\vdots$

zn = rn[cos(θn) + isin(θn)]

to obtain the product

z1z2 × ... × zn = (r1r2 × ... × rn)[cos(θ1 + θ2 + ... + θn) + isin(θ1 + θ2 + ... + θn)]

Setting   r1 = r2 = ... = rn = 1   and   θ1 = θ2 = ... = θn = θ   we obtain Demoivre’s theorem

(cosθ + isinθ)n = cos() + isin()   Demoivre’s theorem(2)

Question T1

Evaluate both (cosθ + isinθ)2 and [cos(2θ) + isin(2θ)] for θ = 0, (π/4) and (π/2) rad. Show that your results are consistent with Demoivre’s theorem for n = 2.

For convenience, define L = (cosθ + isinθ)2   and   R = cos(2θ) + isin(2θ).

For θ = 0 we have cos(0) = 1 and sin(0) = 0 and therefore

L = (1 + i)2 = 1  and  R = (1 + i) = 1.

For θ = (π/4) rad we have sin(π/4) = cos(π/4) = $\dfrac{1}{\sqrt{2\os}}$, cos(π/2) = 0 and sin(π/2) = 1, so that

$L = \left(\dfrac{1}{\sqrt{2\os}}+i\dfrac{1}{\sqrt{2\os}}\right)^2 = 2i\dfrac{1}{\sqrt{2\os}}\dfrac{1}{\sqrt{2\os}} = i$

andR = (0 + i) = i.

For θ = (π/2) rad we have sin(π/2) = 1, cos(π/2) = 0, sin(π) = 0, cos(π) = −1 and hence

L = (0 + i)2 = −1 and R = (−1 + i) = −1.

In all cases we have L = R, which is consistent with Demoivre’s theorem.

The proof we have given for Demoivre’s theorem is only valid if n is a positive integer, but it is possible to show that the theorem is true for any real n and we will make this assumption for the remainder of this module.

✦ Use Demoivre’s theorem to show that one of the square roots of i − 1 is   21/4[cos(3π/8) + isin(3π/8).

i − 1 = 21/2[cos(3π/4) + isin(3π/4)] i and therefore, using Demoivre’s theorem with n = 1/2,

(i − 1)1/2 = 21/4[cos(3π/8) + isin(3π/8)].

Question T2

Use Demoivre’s theorem to show that one of the square roots of 1 + i is

21/4[cos(π/8) + isin(π/8)]

[Hint: First write 1 + i in polar form.]

Using the fact that cos(π/4) = sin(π/4) = $\dfrac{1}{\sqrt{2\os}}$ (see Figure 4) we can write (1 + i) as 2[cos(π /4) + isin(π /4)].

Setting n = 1/2 in Demoivre’s theorem we have

(cosθ + isinθ)1/2 = cos(θ/2) + isin(θ/2)

and therefore

$(1+i)^{1/2} = (\sqrt{2\os})^{1/2}[\cos(\pi/4)+i\sin(\pi/4)]^{1/2}$

$\phantom{(1+i)^{1/2} }= (\sqrt{2\os})^{1/4}[\cos(\pi/8)+i\sin(\pi/8)]$

The significance of Demoivre’s theorem is that instead of calculating expressions such as (cosθ + isinθ)n by writing out the n + 1 individual terms in its binomial expansion, we know that the answer must be cos() + isin(). To emphasize the advantage of Demoivre’s theorem, consider the evaluation of z8 where

z = 2[cos(π/8) + isin(π/8)]

Without using Demoivre’s theorem, we could write z = x + iy and use a calculator to discover that x ≈ 1.847 759 and y ≈ 0.765 367, so that

z8 = (x + iy)8 = x8 + 8ix7y − 28x6y2 − 56ix5y3 + 70x4y4 + 56ix3y5 − 28x2y6 − 8ixy7 + y8 i

and, after a considerable amount of arithmetic, we would obtain the approximate answer

z8 ≈ −256.0 − 1.53 × 10−4i

Compare this brute force approach with the elegance of Demoivre’s theorem which gives the exact answer

z8 = {2[cos(π/8) + isin(π/8)]}8 = 28[cos(π) + isin(π)] = −28 = −256

## 2.2 Trigonometric identities

Demoivre’s theorem can be used to obtain a variety of useful identities involving cosn1θ, sinn1θ, cos() and sin(). The trick is to let z = e, from which we obtain z−1 = e and therefore from Euler’s formula

$z = \cos\theta + i\sin\theta$(3)

$\dfrac1z = \cos\theta - i\sin\theta$(4)

Adding and subtracting these two equations gives the useful relations

$z+\dfrac1z = 2\cos\theta$(5)

$z-\dfrac1z = 2i\sin\theta$(6)

More generally, we have zn = einθ and zn = einθ and in this case Demoivre’s theorem gives

$z^n = \cos(n\theta) + i\sin(n\theta)$(7)

$\dfrac{1}{z^n} = \cos(n\theta) - i\sin(n\theta)$(8)

Again, we can either add or subtract these two equations to obtain

$z^n+\dfrac{1}{z^n} = 2\cos(n\theta)$(9)

$z^n-\dfrac{1}{z^n} = 2i\sin(n\theta)$(10)

These results can be used in two ways; that is, to write

cosnθ or sinnθ in terms of cos() or sin() for various values of m

or

cos() or sin() in terms of cosmθ or sinmθ for various values of m

Although it is possible to obtain general identities, they are quite complicated and the technique is better illustrated by considering specific examples.

### Example 1

Express cos2θ in terms of the cosines of multiples of θ.

#### Solution

From Equation 5,

$2^2 \cos^2\theta = \left(z+\dfrac1z\right)^2 = z^2 + 2 + \dfrac{1}{z^2} = \left(z^2+\dfrac{1}{z^2}\right)+2$

The term in parentheses has precisely the correct form for Equation 9 (with n = 2), we have

$\left(z^2+\dfrac{1}{z^2}\right)+2 = 2\cos(2\theta)+2$

and therefore   cos2θ = 1 + cos(2θ)

which is the desired result.

Question T3

Use Demoivre’s theorem to show that

$\cos^3\theta = \dfrac{\cos(3\theta)+3\cos\theta}{4}$(11)

[Hint: Let z = e and then expand (z + 1/z)3.] i

If we let z = e, we can write

$2^3\cos^3\theta = \left(z+\dfrac1z\right)^3 = \left(z^2+2+\dfrac{1}{z^2}\right)\left(z+\dfrac1z\right)$

$\phantom{2^3\cos^3\theta }= \left(z^3+\dfrac{1}{z^3}\right) + 3\left(z+\dfrac1z\right)$

and therefore (using Equation 9)   $\cos^3\theta = \dfrac{\cos(3\theta)+3\cos\theta}{4}$

### To express an odd power of sinθ in terms of sin(mθ)

Similar techniques can be used to express an odd power of sinθ in terms of the sines of multiples of θ.

### Example 2

Express sin7θ in terms of the sines of multiples of θ.

#### Solution

Starting from the identity for sinθ in terms of z and 1/z we proceed as follows

$(2i)^7\sin^7\theta = \left(z-\dfrac1z\right)^7$ i

$\phantom{(2i)^7\sin^7\theta }= \left(z^7-\dfrac{1}{z^7}\right)-7\left(z^5-\dfrac{1}{z^5}\right)+2i\left(z^3-\dfrac{1}{z^3}\right)-35\left(z-\dfrac1z\right)$

$\phantom{(2i)^7\sin^7\theta }= 2i\sin(7\theta)-14i\sin(5\theta)+42i\sin(3\theta)-70i\sin\theta$

So the required result is

$\sin^7\theta = \dfrac{35\sin\theta-21\sin(3\theta)+7\sin(5\theta)-\sin(7\theta)}{2^6}$(12)

✦ Express sin5θ in terms of the sines of multiples of θ. i

✧ Let z = e then

$\sin^5\theta = \dfrac{1}{(2i)^5}\left(\dfrac1z\right)^5 = -\dfrac{i}{32}\left[\left(z^5-\dfrac{1}{z^5}\right)-5\left(z^3-\dfrac{1}{z^3}\right)+10\left(z-\dfrac1z\right)\right]$

$\phantom{\sin^5\theta }= -\dfrac{i}{32}\left[2i\sin(5\theta)-5\times 2i\sin(3\theta)+10\times 2i\sin\theta\right]$ i

$\phantom{\sin^5\theta }= \dfrac{1}{16}\left[\sin(5\theta)-5\sin(3\theta)+10\sin\theta\right]$

Question T4

Use Demoivre’s theorem to show that

$\sin^3\theta = \dfrac{3\sin\theta-\sin(3\theta)}{4}$(13)

If we let z = e we can write

$(2i)^3\sin^3\theta = \left(z-\dfrac1z\right)^3 = \left(z-\dfrac1z\right)\left(z^2-2+\dfrac{1}{z^2}\right)$

$\phantom{(2i)^3\sin^3\theta }= \left(z^3-2z+\dfrac1z\right) + \left(-z+\dfrac2z-\dfrac{1}{z^3}\right)$

$\phantom{(2i)^3\sin^3\theta }= \left(z-\dfrac1z\right) - 3\left(z-\dfrac1z\right)$

$\phantom{(2i)^3\sin^3\theta }= 2i \sin(3\theta) - 3 \times 2i\sin\theta$

which can be rearranged to give the required result   $\sin^3\theta = \dfrac{3\sin\theta-\sin(3\theta)}{4}$

### To express cos(nθ) and sin(nθ) in terms of cosθ and sinθ

Again it is best to consider an example rather than the general case, so let us suppose that we want to express cos(2θ) or sin(2θ) in terms of cosθ and sinθ. (In fact, we can derive two identities at the same time.) We start by using Demoivre’s theorem with n = 2

cos(2θ) + isin(2θ) = (cosθ + isinθ)2

and then expand the right–hand side to give

cos(2θ) + isin(2θ) = cos2θ + 2isinθcosθ − sin2θ

We can now equate the real and imaginary parts to obtain two identities

cos(2θ) = cos2θ − sin2θ(14)

sin(2θ) = 2sinθcosθ(15)

which are the required results.

✦ Use Demoivre’s theorem to obtain identities for cos(5θ) and sin(5θ) in terms of cosθ and sinθ.

✧ From Demoivre’s theorem, with n = 5, we have

cos(5θ) + isin(5θ) = (cosθ + isinθ)5

cos(5θ) + isin(5θ) = cos5θ + 5icos4θsinθ − 10cos3θsin2θ − 10icos2θsin3θ + 5cosθ sin4θ + isin5θ i

and equating real and imaginary parts gives us

cos(5θ) = cos5θ − 10cos3θsin2θ + 5cosθsin4θ and sin(5θ) = 5cos4θsinθ − 10cos2θsin3θ + sin5θ

Question T5

Use Demoivre’s theorem to obtain identities for cos(3θ) and sin(3θ) in terms of cosθ and sinθ.

We start by using Demoivre’s theorem with n = 3

cos(3θ) + isin(3θ) = (cosθ + isinθ)2

and then expand the right–hand side to obtain

cos(3θ) + isin(3θ) = cos2θ + 3cosθ(isinθ)2 + 3icos2θsinθ + (isinθ)2

cos(3θ) + isin(3θ) = cos2θ − 3cosθsin2θ + i(3cos2θsinθ − sin2θ)

We can now equate the real and imaginary parts to obtain two identities

cos(3θ) = cos2θ − 3cosθsin2θ

sin(3θ) = 3sinθcos2θ − sin2θ

## 2.3 Roots of unity

When we multiply two complex numbers we multiply their moduli and add their arguments; so to square a complex number we square the modulus and double the argument. i

We know how to find the square root of a positive real number, but how can we find the square root of a complex number? Obviously we reverse the process of squaring, and find the square root of the modulus and halve the argument. However, a complex number has many different arguments, for example

1 = e0i    or   e2πi   or   e4πi   or   e6πi   and so on

so it follows that

11/2 = ei    or    eπi    or    e2πi    or    e3πi  and so on

From this it would at first sight appear that we have found an embarrassingly large number of square roots of 1, but in fact 1 = e0i = e2πi = e4πi ..., whereas −1 = eπi = e3πi = e5πi ..., so that we have actually found just the two square roots that we expect.

The method can clearly be extended to cube roots. To find the cube root of a given complex number, we first write it in exponential form, then find the cube root of the modulus (a positive real number), and divide the modulus by three.

✦ Find a cube root of the complex number 1 + i.

✧ First we write the complex number in exponential form $1+i=\sqrt{2\os}{\rm e}^{i\pi/4}$, then we take the cube root of (the real number) $\sqrt{2\os}$ and divide π/4 by three to give

$(1+i)^{1/3} = 2^{1/6}{\rm e}^{i\pi/12}$. i

Although we have been able to find a cube root of a given complex number, there is one question we have not addressed. Is there more than one cube root; if so, what are the others? Demoivre’s theorem provides a complete answer to such questions.

According to the fundamental theorem of algebra, each polynomial with complex number coefficients and of degree n has, counting multiple roots an appropriate number of times, exactly n complex roots. More specifically, the theorem tells us that the equation

zn − 1 = 0(16)

where n is a positive integer, has precisely n roots. These roots are known as the roots_of_unitynth roots of unity (because we can rewrite the equation as z = 11/n). To find these roots we use the fact that we can write the number 1 as i

1 = e2πki = cos(2πk) + isin(2πk)(17)

where k is any integer (positive, negative or zero).

So z is given by

z = 11/n = [cos(2πk) + isin(2πk)]1/n

and Demoivre’s theorem then allows us to rewrite the right–hand side, obtaining

$z = \cos\left(\dfrac{2\pi k}{n}\right) + i\sin\left(\dfrac{2\pi k}{n}\right)$

Don’t forget that, whereas k can take any integer value, n is fixed (even if we don’t specify what it is at the moment). However, the sine and cosine functions are periodic with period π, i.e.

$\cos\left(\dfrac{2\pi k}{n}+2\pi\right) = \cos\left(\dfrac{2\pi k}{n}\right)$(18)

$\sin\left(\dfrac{2\pi k}{n}+2\pi\right) = \sin\left(\dfrac{2\pi k}{n}\right)$(19)

So it follows that all the different integer values of k only give n distinct values of z and it is convenient to use k = 0, 1, 2, ..., (n −1) to generate these values.

In the case of n = 2, the two roots are given by

z = cos(πk) + isin(πk)

where k = 0, 1 or, more explicitly

z0 = cos(0) + isin(0) = 1

z1 = cos(π) + isin(π) = −1

And once again we have the familiar result that 11/2 = ±1 which is plotted on an Argand diagram in Figure 1.

### Example 3

Find the three cube roots of 1.

#### Solution

Arguing as before, we could express 1 in exponential form (using different values of the argument) then find the cube root of the modulus and divide the argument by three. More formally, using Demoivre’s theorem, the three roots are given by

z = cos(2πk/3) + isin(2πk/3)

where k = 0, 1, 2 or, more explicitly

$z_0 = \cos(0) + i\sin(0) = 1$

$z_1 = \cos(2\pi/3) + i\sin(2\pi/3) = -\dfrac12+\dfrac{\sqrt{3\os}}{2}i$

$z_2 = \cos(4\pi/3) + i\sin(4\pi/3) = -\dfrac12-\dfrac{\sqrt{3\os}}{2}i$

and these are plotted on an Argand diagram in Figure 2.

Question T6

If z1 and z2 are as given above, find: (a) |z1| and |z2|, (b) |z1|3 and |z23|.

(a) For any complex number, z = x + iy, the modulus, |z|, is defined by $\lvert\,z_1\,\rvert = \sqrt{x^2+y^2}$ where both x and y are real. The modulus of each root is therefore given by

$\lvert\,z_1\,\rvert = \sqrt{(-1/2)^2+(\sqrt{3}/2)^2}$

$\lvert\,z_2\,\rvert = \sqrt{(-1/2)^2+(-\sqrt{3}/2)^2}$

(b) From the previous part |z1| = 1, so that |z2|3 = 1. The hard way to do this part (and not recommended) is to calculate z3 explicitly, and then find its modulus.

Since

$\left(-\dfrac12-i\dfrac{\sqrt{3\os}}{2}\right)^2 = \left(-\dfrac12-i\dfrac{\sqrt{3\os}}{2}\right)\left(-\dfrac12-i\dfrac{\sqrt{3\os}}{2}\right) = -\dfrac12+i\dfrac{\sqrt{3\os}}{2}$

it follows that

$z_2^3 = \left(-\dfrac12-i\dfrac{\sqrt{3\os}}{2}\right)^3 = \left(-\dfrac12-i\dfrac{\sqrt{3\os}}{2}\right)\left(-\dfrac12+i\dfrac{\sqrt{3\os}}{2}\right) = \dfrac14+\dfrac34 = 1$

and therefore |z23| = 1.

A slightly easier method would be to use Demoivre’s theorem and write

z23 = [cos(4π/3) + isin(4π/3)]2 = cos(4π) + isin(4π)

so that|z23| = cos2(4π) + sin2(4π) = 1

The easiest way is to notice that |z23| = |z2|3; then the answer is obviously 1 from part (a).

Aside There is a slight problem here with notation (which can only be resolved properly by a discussion which is beyond the scope of FLAP). For a positive real variable x we know that x1/2 and $\sqrt{x\os}$ are often used to denote the positive root of x, usually because we require the expression to define a function and so it must have just one value.

In this module we are following the convention that square roots of real numbers, such as $\sqrt{x\os}$ and 31/2, are positive, because they usually arise as moduli of complex numbers, which must be positive.

For complex numbers we are often interested in all the n roots of unity, and we certainly do not wish to restrict the discussion to just one of them. So we follow the convention that for a complex number z, z1/n means all n values. This means that we will have to distinguish carefully between the square root of the real number 2, which takes only the positive value, and the square root of the complex number 2, which takes both positive and negative values. In practice the context would usually make the meaning clear, and this minor problem will cause us no great difficulty.

✦ Find the three values of (1 + i)1/3 in exponential form.

✧ Writing 1 + i in the alternative exponential form we have $1+i=\sqrt{2\os}{\rm e}^{i\pi/4}{\rm e}^{2\pi ki}$ where k = 0, 1, 2. The three values are

z0 = 21/6e/12e0i = 21/6e/12,

z1 = 21/6e/12e2/3 = 21/6e3/4,

and

z2 = 21/6e/12e4/3 = 21/6e17/12. i

Question T7

Find all roots of the equation z6 − 1 = 0 and plot your results on an Argand diagram.

The six roots are given by

$z_k = \cos(\pi k/3) + i\sin(\pi k/3)$   where k = 0, 1, 2, 3, 4, 5; so that

$z_0 = \cos(0) + i\sin(0) = 1$

$z_1 = \cos(\pi/3) + i\sin(\pi/3) = \dfrac12+i\dfrac{\sqrt{3\os}}{2}$

$z_2 = \cos(2\pi/3) + i\sin(2\pi/3) = -\dfrac12+i\dfrac{\sqrt{3\os}}{2}$

$z_3 = \cos(\pi) + i\sin(\pi) = -1$

$z_4 = \cos(4\pi/3) + i\sin(4\pi/3) = -\dfrac12+i\dfrac{\sqrt{3\os}}{2}$

$z_5 = \cos(5\pi/3) + i\sin(5\pi/3) = \dfrac12-i\dfrac{\sqrt{3\os}}{2}$

The values obtained for the six roots of z6 − 1 = 0 are plotted on an Argand diagram in Figure 6.

Question T8

Show that all roots of the equation zn − 1 = 0, where n is a positive integer, satisfy |z| = 1. Describe the geometric figure on which all such points lie, when plotted on an Argand diagram.

The n roots of zn − 1 = 0 are given by

$z_k = \cos\left(\dfrac{2\pi k}{n}\right) + i\sin\left(\dfrac{2\pi k}{n}\right)$

with k = 0, 1, 2, ..., (n − 1).

Since

$\lvert\,z_k\,\rvert = \cos^2\left(\dfrac{2\pi k}{n}\right) + \sin^2\left(\dfrac{2\pi k}{n}\right) = 1$

we have |zk| = 1, and the roots therefore lie on a circle of radius 1 centred at the origin.

# 3 Complex algebra

In this section we consider complex algebra; that is the manipulation of expressions involving complex variables.

## 3.1 Solving equations

Finding the nth root of unity has already provided us with experience of using complex algebra to solve equations. In this subsection we consider other examples of solving equations.

### Example 4

Solve the following equation

z2 + izγ + ω2 = 0(20) i

where γ and ω are real constants, which occurs in the theory of damped oscillations.

#### Solution

We can solve for z by using the well–known formula for the roots of the quadratic equation

az2 + bz + c = 0

namely

$z = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In our particular example we have a = −1, b = and c = ω2 and therefore

$z = \dfrac{-i\gamma\pm\sqrt{\smash[b]{(i\gamma)^2-4(-1)\omega^2}}}{-2}$

$z = \dfrac{i\gamma\pm\sqrt{\smash[b]{4\omega^2-\gamma^2}\us}}{2} = \dfrac{i\gamma}{2}\pm\sqrt{\smash[b]{\omega^2-\gamma^2/4}\us}$

✦ Find the roots of the equation z2 + z + i = 0 in the form x + iy.

✧ In this case (and using the usual notation for a quadratic equation) a = 1, b = 1 and c = i.

The roots of the equation are therefore,

$z = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{-1\pm\sqrt{1-4i\os}}{2}$

However, we can find a square root of 1 − 4i by writing   1 − 4i = 4.123e−1.3258i   so that

$\sqrt{1-4i\os} \approx (4.1231)^{1/2}{\rm e}^{-0.6629} \approx 2.0305[\cos(0.6629) - i\sin(0.6629)] \approx 1.6005 - 1.2496i$

and therefore the roots of the equation are   $\dfrac{-1\pm(1.6005-1.2496i)}{2}$

which gives   0.300 − 0.625i   and  −1.300 + 0.625i   as the roots of the equation (which we can easily check by substituting these values into the original equation).

Question T9

Find the roots of the equation 2z2 − 11zi − 5 = 0. Many of the algebraic operations for complex variables are almost identical to those for real variables; for example, the solution of simultaneous equations.

For the equation az2 + bz + c = 0 the solutions are given by

$z = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case we have a = 2, b = −11i and c = −5 and therefore

$z=\dfrac{11i\pm\sqrt{\smash[b]{-(11)^2+4\times 2\times 5}}}{4} = \dfrac{11i\pm9i}{4}$ = 5i or i/2

✦ Solve the pair of simultaneous equations

2z + iw = 5 + i(21)

iz − 3w = i(22)

✧ Multiplying the first equation by i and the second by 2 we obtain

2izw = 5i − 1(23)

2iz − 6w = 2i(24)

Subtracting Equation 24 from Equation 23 we obtain

5w = 3i − 1   so that   $w = \dfrac{3i-1}{5}$

and substituting this value for w into Equation 22 we have

$z = -3iw + 1 = 1 - 3i\left(\dfrac{3i-1}{5}\right) = \dfrac{3i+14}{5}$

### Example 5

Suppose we want to solve the pair of equations

I1(Z1 + Z2) + I2Z2 = ε1 i

I1Z2 + I2Z2 = ε2

for I1 and I2 where all the variables are complex.

#### Solution

Equations such as these (but sometimes having many more variables) often arise in the mesh analysis of a.c. circuits. We can subtract the second equation from the first to find

I1Z1 = ε1ε2

so that$I_1 = \dfrac{\varepsilon_1 - \varepsilon_2}{Z_1}$

Substituting this result back in the second equation gives us

$\dfrac{\varepsilon_1 - \varepsilon_2}{Z_1}Z_2 + I_2Z_2 = \varepsilon_2$

and hence   I2Z1Z2 = ε2Z1 − (ε1ε2)Z2 = ε2(Z1 + Z2) − ε1Z2

so that$I_2 = \dfrac{\varepsilon_2(Z_1+Z_2)-\varepsilon_1 Z_2}{Z_1 Z_2}$

So far the algebra would have been no different for real variables, but now suppose we want to find the real and imaginary parts of I1 and we are told that

Z1 = Z2 = R +iX,   ε1 = ε   and  ε2 = εe i

where R, X, ε and φ are all real. I1 is given by

$I_1 = \dfrac{\varepsilon_1-\varepsilon_2}{R+iX} = \dfrac{ [\overbrace{\vphantom{0} \varepsilon}^{\large\color{purple}{\varepsilon_1}}-\overbrace{\varepsilon(\cos\phi+i\sin\phi)}^{\large\color{purple}{\varepsilon_2}}]}{R+iX} \times \dfrac{R-iX}{R+iX}$

$\phantom{I_1 }=\dfrac{\varepsilon(1-\cos\phi-i\sin\phi)}{R^2+X^2}(R-iX)$

$\phantom{I_1 }=\dfrac{\varepsilon}{R^2+X^2}\left\{[R(1-\cos\phi)-X\sin\phi]+i[-R\sin\phi-X(1-\cos\phi)]\right\}$

and therefore

${\rm{Re}}(I_1) = \dfrac{\varepsilon}{R^2+X^2}[R(1-\cos\phi) - X\sin\phi]$

${\rm{Im}}(I_1) = \dfrac{\varepsilon}{R^2+X^2}[-R\sin\phi) - X(1-\cos\phi)]$

Question T10

Find Re(I2) and Im(I2) in Example 5, i.e. $I_2 = \dfrac{\varepsilon_2(Z_1+Z_2)-\varepsilon_1 Z_2}{Z_1 Z_2}$   and   Z1 = Z2 = R + iX

From the text we have Z1 = Z2 = R + iX so that

$I_2 = \dfrac{\varepsilon_2(Z_1+Z_2)-\varepsilon_1 Z_2}{Z_1 Z_2} = \dfrac{\varepsilon_2(Z_1+Z_1)-\varepsilon_1 Z_1}{Z_1^2} = \dfrac{2\varepsilon_2-\varepsilon_1}{Z_1} = \dfrac{\varepsilon}{R+iX}(\cos\phi +2i\sin\phi-1)$

$\phantom{I_2 }= \dfrac{\varepsilon}{R^2+X^2}(\cos\phi +2i\sin\phi-1)(r-iX)$

$\phantom{I_2 }= \dfrac{\varepsilon}{R^2+X^2}\left\{[R(\cos\phi-1) +2X\sin\phi]) + i[2R\sin\phi + X(1-2\cos\phi)]\right\}$

which gives

${\rm{Re}}(I_2) = \dfrac{\varepsilon}{R^2+X^2}[R(\cos\phi-1)+2X\sin\phi]$

and${\rm{Re}}(I_2) = \dfrac{\varepsilon}{R^2+X^2}[R\sin\phi+X(1-2\cos\phi)]$

## 3.2 Factorizing

Complex numbers enable us to factorize expressions with real coefficients that are impossible to factorize in terms of real numbers. For example, expanding the right–hand side of the equation will verify the equation

z2 + 1 = (zi)(z + i)

and soz2 + 1 (which is a polynomial with real coefficients) has complex factors. Complex numbers are also involved in the factorization of expressions, which have complex coefficients, as in

z2 + (1 + i)z + i = (z + i)(z + 1)

In simple cases, factorization can be achieved by spotting values which make the expression zero (in this case noticing that z = −i and z = −1 are values for which z2 + (1 + i)z + i = 0). More complicated cases may involve finding the roots by numerical means, and there are actually computer programs designed to do precisely this.

If we need to factorize an expression of the form

zn + an−1zn−1 + an−2zn−2 + ... + a0

for some (possibly quite large) positive integer, n, then the n roots, z1, z2, ... , zn, of the equation

zn + an−1zn−1 + an−2zn−2 + ... + a0 = 0

correspond to the factorization

zn + an−1zn−1 + an−2zn−2 + ... + a0 = (zz1)(zz2) ... (zzn)

In practice, it is unlikely that you will need to perform factorizations for large values of n. However, it is important that you should know how many roots (and therefore factors) to look for.

### Example 6

Factorize 3z2 + 27.

#### Solution

Notice that the expression is zero for z = ±3i and therefore (z − 3i) and (z + 3i) are factors. Since the highest power of z in the polynomial 3z2 + 27 is z2, the fundamental theorem of algebra tells us that these are the only roots, and therefore

3z2 + 27 = k(z − 3i)(z + 3i)

for some constant k. Comparing the coefficients on each side of this expression for any particular power of z (z2 is the most convenient in this case) we obtain k = 3, so that

3z2 + 27 = 3(z − 3i)(z + 3i)

Question T11

Factorize the expression 2z2 + 32.

First solve the equation 2z2 + 32 = 0 as follows: z2 = −16 and therefore z = ±4i. So we have

2z2 + 32 = k(z + 4i)(z − 4i)

for some constant k, and comparison of the coefficient of any particular power, say z2, gives the factorization

2z2 + 32 = 2(z + 4i)(z − 4i)

### Example 7

Factorize the expression −z2 + izγ + w2

where z is regarded as the ‘unknown variable’.

#### Solution

We need to find the two roots of Equation 20. From Example 4 we know that

$z = \dfrac{i\gamma}{2}\pm\sqrt{\smash[b]{\omega^2-\gamma^2\us}}$

and therefore, for some constant k,

$-z^2 - iz\gamma + w^2 = k\left(z -\dfrac{i\gamma}{2} - \sqrt{\smash[b]{\omega^2-\gamma^2\us}}\right)\left(z -\dfrac{i\gamma}{2} + \sqrt{\smash[b]{\omega^2-\gamma^2\us}}\right)$

Comparing the coefficients of z2 (or any other convenient power of z) tells us that k = −1, and so we have

$-z^2 - iz\gamma + w^2 = -\left(z -\dfrac{i\gamma}{2} - \sqrt{\smash[b]{\omega^2-\gamma^2\us}}\right)\left(z -\dfrac{i\gamma}{2} + \sqrt{\smash[b]{\omega^2-\gamma^2\us}}\right)$

Question T12

Factorize the expression z2 + iz + 2.

Inspection shows that the expression is zero for z = i, −2i (or, alternatively, use the formula for the roots of a quadratic equation). Therefore we have

z2 + iz + 2 = k(zi)(z + 2i)

for some constant k, and, equating coefficients, the final result is

z2 + iz + 2 = (zi)(z + 2i)

## 3.3 Simplifying

We simplify expressions involving complex numbers in much the same way that we simplify expressions involving real numbers, except that every occurrence of i2 may be replaced by −1. It may also be necessary to rationalize any complex quotients, in other words to convert such quotients into the form x + iy where x and y are real, in order to arrive at the simplest form. Since there is not really any general prescription for simplifying complex expressions, the best approach is to consider some typical examples.

### Example 8

Simplify the following expression

Z = (3R + 2iX) − (R + iX) + (7R + 3iX)

#### Solution

To simplify such an expression we treat the real and imaginary parts separately to obtain i

Z = 9R + 4iX

### Example 9

Find the real and imaginary parts of a complex number Z defined by

$Z = R_1 + \dfrac{1}{\cfrac{1}{R_2 + i\omega L}+ \cfrac{1}{1/(i\omega C)}}$(25)

where R, ω, L and C are real.

#### Solution

To find the real and imaginary parts of Z we rationalize the complex quotients

$Z = R_1 + \dfrac{R_2+i\omega L}{1+i\omega C(R_2+i\omega L)}$

$\phantom{Z }= R_1 + \dfrac{R_2+i\omega L}{1+i\omega C(R_2+i\omega L)} \times \dfrac{1-i\omega C(R_2+i\omega L)}{1-i\omega C(R_2+i\omega L)}$ i

$\phantom{Z }= R_1 + \dfrac{(R_2+i\omega L)[(1-\omega^2 LC)-i\omega CR_2]}{(1-\omega^2 LC)+\omega^2 C^2 R_2^2}$

$\phantom{Z }= R_1 + \dfrac{R_2+i[\omega L(1-\omega^2 LC)-\omega CR_2^2]}{(1-\omega^2 LC)+\omega^2 C^2 R_2^2}$

which gives the results

${\rm{Re}}(Z) = R_1+\dfrac{R_2}{(1-\omega^2 LC)+\omega^2 C^2 R_2^2}~~~~{\rm{Im}}(Z) = R_1+\dfrac{\omega L(1-\omega^2 LC)-\omega CR_2^2}{(1-\omega^2 LC)+\omega^2 C^2 R_2^2}$

✦ Simplify the expression $\dfrac{3+2i}{1+i}-\dfrac{1}{1+2i}$

✧ $\dfrac{3+2i}{1+i}-\dfrac{1}{1+2i} = \dfrac{3+2i}{1+i}\times \dfrac{1-i}{1-i} - \dfrac{1}{1+2i}\times \dfrac{1-2i}{1-2i} = \dfrac{5-i}{2} -\dfrac{1-2i}{5} = \dfrac{23-i}{10}$

Question T13

Simplify the expression $\dfrac{1}{1-i}+\dfrac{1+i}{2}$

One way to simplify $\dfrac{1}{1-i} + \dfrac{1+i}{2}$ is to put everything over a common denominator and then to rationalize the resulting expression as in

$\dfrac{1}{1-i} + \dfrac{1+i}{2} = \dfrac{1}{2(1-i)}[2+(1+i)(1-i)]$

$\phantom{\dfrac{1}{1-i} + \dfrac{1+i}{2} }= \dfrac{1}{2(1-i)}[2+1+1] = \dfrac{2(1+i)}{(1-i)(1+i)} = 1+i$

Alternatively we can rationalize the first term and then simplify it, as in the following

$\dfrac{1}{1-i} + \dfrac{1+i}{2} = \dfrac{1+i}{(1-i)(1+i)} + \dfrac{1+i}{2}$

$\phantom{\dfrac{1}{1-i} + \dfrac{1+i}{2} }= \dfrac{1+i}{2} + \dfrac{1+i}{2} = 1+i$

which is probably slightly easier.

## 3.4 Complex binomial expansion

You probably recall that the binomial coefficient, ${}^nC_r$, is defined by

${}^nC_r = \dfrac{n!}{r!(n-r)!}$(26)

where n! is known as n factorial, and is defined by

n! = n(n − 1)(n − 2)(n − 3) ... 2 × 1 for n ≥ 1 and 0! = 1(27)

The expression (a + b)n, where a and b are real and n is a positive integer, can be written in terms of powers of a and b by means of the binomial expansion i

(a + b)n = nCnan + nCn−1an−1b + ... + nCnranrbr + ... + nC1abn−1 + nC0bn = $\displaystyle \sum_{r=0}^n{}^nC_{n-r}a^{n-r}b^r$(28)

Question T14

(a) Evaluate 3Cr for r = 0, 1, 2, 3. (b) Prove the general result nCr = nCnr.

(a) Using ${}^nC_r = \dfrac{n!}{(n-r)!r!}$ we have

${}^3C_0 = \dfrac{3!}{3!0!} = 1$   ${}^3C_1 = \dfrac{3!}{2!1!} = 3$

${}^3C_2 = \dfrac{3!}{1!2!} = 3$   ${}^3C_1 = \dfrac{3!}{0!3!} = 1$

(b) Using the definition of ${}^nC_r$ we have

${}^nC_{n-r} = \dfrac{n!}{[n-(n-r)]!(n-r)!} = \dfrac{n!}{r!(n-r)!} = \dfrac{n!}{(n-r)!r!} = {}^nC_r$

There is nothing in the proof of the binomial theorem i which restricts it to real quantities a and b, and for this module our interest in the theorem lies in its complex applications. The following examples are relevant to Subsection 2.2.

✦ Expand and simplify $\left(z - \dfrac1z\right)^3$

✧ $\displaystyle \left(z-\dfrac1z\right)^3 = \sum_{r=0}^3 \left[{}^3C_{3-r}z^{3-r}\times\left(-\dfrac1z\right)^r\right]$

$\phantom{\left(z-\dfrac1z\right)^3 }={}^3C_3z^3 - {}^3C_2\dfrac{z^2}{2}+{}^3C_1\dfrac{z}{z^2}-{}^3C_0\dfrac{1}{z^3} = z^3 - 3z + \dfrac3z-\dfrac{1}{z^3}$

$\phantom{\left(z-\dfrac1z\right)^3 }=\left(z^3-\dfrac{1}{z^3}\right) - 3\left(z-\dfrac1z\right)$

✦ Expand and simplify $\left(z - \dfrac1z\right)^7$

✧ $\displaystyle \left(z-\dfrac1z\right)^7 = \sum_{r=0}^7 \left[{}^7C_{7-r}z^{7-r}\times\left(-\dfrac1z\right)^r\right]$

$\phantom{\left(z-\dfrac1z\right)^7 }= z^7 - 7z^5 + 21z^3 - 35z + \dfrac{35}{z} - \dfrac{21}{z^3} + \dfrac{7}{z^5} - \dfrac{1}{z^7}$

$\phantom{\left(z-\dfrac1z\right)^7 }=\left(z^7-\dfrac{1}{z^7}\right) -7 \left(z^5-\dfrac{1}{z^5}\right) + 21\left(z^3-\dfrac{1}{z^3}\right)$

Question T15

Use the binomial theorem to show that

$\left(z - \dfrac1z\right)^4 = \left(z^4 - \dfrac{1}{z^4}\right) + 4\left(z^2 - \dfrac{1}{z^2}\right) + 6$

[Hint: What is the relationship between nCr = nCnr?]

$\displaystyle \left(z+\dfrac1z\right)^4 = \sum_{r=0}^4\left[{}^4C_{4-r}z^{4-r}\times\dfrac{1}{z^r}\right] = {}^4C_4z^4 + {}^4C_3z^2 + {}^4C_2 + {}^4C_1\dfrac{1}{z^2} + {}^4C_0\dfrac{1}{z^4}$

$\phantom{\displaystyle \left(z+\dfrac1z\right)^4 }=z^4+4z^2+6+4\dfrac{1}{z^2}+\dfrac{1}{z^4} = \left(z^4+\dfrac{1}{z^4}\right)+4\left(z^2+\dfrac{1}{z^2}\right)+6$

In Subsection 2.2 we showed how to write powers of trigonometric functions in terms of functions of multiple angles by setting z = e and consequently:

$z^n+\dfrac{1}{z^n} = 2\cos(n\theta)$(Eqn 9)

$z^n-\dfrac{1}{z^n} = 2i\sin(n\theta)$(Eqn 10)

The binomial theorem is a considerable help in using these results to derive identities for powers of trigonometric functions. For example, the answer to Question T15 enables us to write

$2^4\cos^4(\theta) = \left(z - \dfrac1z\right)^4 = \left(z^4 - \dfrac{1}{z^4}\right) + 4\left(z^2 - \dfrac{1}{z^2}\right) + 6$

$\phantom{2^4\cos^4(\theta) }= 2\cos(4\theta) + 4 \times 2\cos(2\theta) + 6$

and therefore

$\cos^4(\theta) = \dfrac18[\cos(4\theta) + 4\cos(2\theta) + 3]$

Question T16

(a) Use the binomial theorem to show that

$\left(z - \dfrac1z\right)^5 = \left(z^5 - \dfrac{1}{z^5}\right) + 5\left(z^3 - \dfrac{1}{z^3}\right) + 10 \left(z - \dfrac1z\right)$

(b) Use the first part of this question, together with Demoivre’s theorem, to express cos5θ in terms of cosines of

multiples of θ.

(a) Using the binomial theorem we have

$\displaystyle \left(z+\dfrac1z\right)^5 = \sum_{r=0}^5\left[{}^5C_{5-r}z^{5-r}\times\dfrac{1}{z^r}\right] = {}^5C_5z^5 + {}^5C_4z^3 + {}^5C_3z + {}^5C_2\dfrac1z + {}^5C_1\dfrac{1}{z^3}+ {}^5C_0\dfrac{1}{z^5}$

$\phantom{\displaystyle \left(z+\dfrac1z\right)^5 }= {}^5C_5\left(z^5+\dfrac{1}{z^5}\right) + {}^5C_4\left(z^3 + \dfrac{1}{z^3}\right) + {}^5C_3\left(z +\dfrac1z\right)$

$\phantom{\displaystyle \left(z+\dfrac1z\right)^4 }= \left(z^5+\dfrac{1}{z^5}\right) + 5\left(z^3 + \dfrac{1}{z^3}\right)+ 10\left(z +\dfrac1z\right)$

(b) Defining z = e, we have (using the previous part of the question)

$2^5\cos^5\theta = \left(z+\dfrac1z\right)^5 = \left(z^5+\dfrac{1}{z^5}\right) + 5\left(z^3 + \dfrac{1}{z^3}\right) + 10\left(z +\dfrac1z\right)$

$\phantom{2^5\cos^5\theta }= 2\cos(5\theta) + 5\times 2\cos(3\theta) + 10\times 2\cos\theta$

and therefore   $\cos^5\theta = \dfrac{1}{16}[\cos(5\theta) + 5\cos(3\theta) + 10\cos\theta]$

## 3.5 Complex geometric series

For real numbers a and x, the series a + ax + ax2 + ... + axn is known as a geometricseries, and its sum is

$a\dfrac{1-x^{n+1}}{1-x}$

provided that x ≠ 1.

The geometric series for complex numbers a and z is identical in form, and i

$a+az+az^2+\dots+az^n = a\dfrac{1-z^{n+1}}{1-z}$    if z ≠ 1(29)

✦ Write down the sum of the series 1 + e + e2 + e3.

✧ $1+{\rm e}^{i\theta} + {\rm e}^{2i\theta} + {\rm e}^{3i\theta} = \dfrac{1-{\rm e}^{4i\theta} }{1-{\rm e}^{i\theta} }$

Sums of this kind can be used to simplify certain trigonometric expressions, as, for example, in the following case.

✦ Sum the series sin(θ) + sin(2θ) + sin(3θ) + ... + sin(9θ).

✧ Let z = e then

sin(θ) + sin(2θ) + sin(3θ) + ... + sin(9θ) = Im(z + z2 + ... + z9)

$={\rm {Im}}\left(\dfrac{z-z^{10}}{1-z}\right) = {\rm {Im}}\left(\dfrac{{\rm e}^{i\theta}-{\rm e}^{10i\theta}}{1 - {\rm e}^{i\theta}}\right) = {\rm {Im}}\left[\dfrac{({\rm e}^{i\theta}-{\rm e}^{10i\theta})(1-{\rm e}^{-i\theta})}{(1-{\rm e}^{i\theta})(1-{\rm e}^{-i\theta})}\right]$ i

$=\dfrac{1}{2(1-\cos\theta)}{\rm {Im}}\left({\rm e}^{i\theta} - 1 - {\rm e}^{10i\theta} + {\rm e}^{9i\theta}\right) = \dfrac{\sin\theta-\sin(10\theta)+\sin(9\theta)}{2(1-\cos\theta)}$

Question T17

Sum the series cos(θ) + cos(2θ) + cos(3θ) + ... + cos(9θ).

cosθ + cos(2θ) + cos(3θ) + ... + cos(9θ)

$={\rm{Re}}\left(\dfrac{z-z^{10}}{1-z}\right) = {\rm{Re}}\left(\dfrac{{\rm e}^{i\theta}-{\rm e}^{10i\theta}}{1-{\rm e}^{i\theta}}\right) = {\rm{Re}}\left[\dfrac{({\rm e}^{i\theta}-{\rm e}^{10i\theta})(1-{\rm e}^{-i\theta})}{(1-{\rm e}^{i\theta})(1-{\rm e}^{-i\theta})}\right]$

$=\dfrac{1}{2(1-\cos\theta)}{\rm{Re}}({\rm e}^{i\theta}-1-{\rm e}^{10i\theta}+{\rm e}^{9i\theta})$

$=\dfrac{\cos\theta-1-\cos(10\theta)+\cos(9\theta)}{2(1-\cos\theta)}$

# 4 Closing items

## 4.1 Module summary

1

The identity

(cosθ + isinθ)n = cos() + isin()(Eqn 2)

is known as Section 2Demoivre’s theorem. It is valid for any real value of n.

2

Defining z = e and using Demoivre’s theorem gives us

$z^n+\dfrac{1}{z^n} = 2\cos(n\theta)$(Eqn 9)

$z^n-\dfrac{1}{z^n} = 2i\sin(n\theta)$(Eqn 10)

These results are useful for deriving trigonometric identities such as those which give

(a) cosnθ or sinnθ in terms of cos() or sin();

(b) cos() or sin() in terms of cosmθ and sinmθ.

3

The n solutions of the equation zn − 1 = 0 are known as the nth Subsection 2.3roots of unity and are given by

$z = \cos\left(\dfrac{2\pi k}{n}\right) + i\sin\left(\dfrac{2\pi k}{n}\right)$

where k = 0, 1, 2, ..., (n − 1). If plotted on an Argand diagram, the roots correspond to n equally spaced

points lying on a circle of unit radius, centred at the origin

4

The fundamental theorem of algebra states that each polynomial with complex number coefficients and of degree n has, counting multiple roots an appropriate number of times, exactly n complex roots.

5

If we determine the roots, z1, z2, ..., zn, of the equation

zn + an−1zn−1 + an−2zn−2 + ... + a0 = 0

then we have the following Subsection 3.2factorization

zn + an−1zn−1 + an−2zn−2 + ... + a0 = (zz1)(zz2) ... (zzn)

6

The complex binomial expansion is (a + b)n = nCnan + nCn−1an−1b + ... + nCnranrbr + ... + nC1abn−1 + nC0bn

(a + b)n = nCnan + nCn−1an−1b + ... + nCnranrbr + ... + nC1abn−1 + nC0bn = $\\ \displaystyle \sum_{r=0}^n{}^nC_{n-r}a^{n-r}b^r$(Eqn 28)

where

${}^nC_r = \dfrac{n!}{r!(n-r)!}$(Eqn 26)

and n! = n(n − 1)(n − 2)(n − 3) ... 2 × 1, for n ≥ 1, with 0! = 1

7

The complex form of the Subsection 3.5geometric series is

$a+az+az^2+\dots+az^n = a\dfrac{1-z^{n+1}}{1-z}$    if z ≠ 1(Eqn 29)

The complex binomial expansion and the complex form of the geometric series may be used to derive certain trigonometric identities.

## 4.2 Achievements

Having completed this module, you should be able to:

A1

A1 Define the terms that are emboldened and flagged in the margins of the module.

A2

State and derive Demoivre’s theorem.

A3

Use Demoivre’s theorem to derive trigonometric identities.

A4

Explain what is meant by the nth roots of unity and find all such roots.

A5

Describe the relevance of the fundamental theorem of algebra to the solution of equations and the factorization of polynomials.

A6

Solve quadratic equations and linear equations involving complex variables.

A7

Factorize simple polynomials.

A8

Simplify complex expressions.

A9

State and apply the complex binomial expansion.

A10

Apply the complex form of the geometric series.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

## 4.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions, each of which tests one or more of the Achievements.

Question E1 (A2)

State Demoivre’s theorem and use it to simplify zn where

z = 3[cos(π/n) + isin(π/n)]

for any integer, n.

Demoivre’s theorem states that

(cosθ + isinθ)n = cos() + isin()

for any real number, n. Using this theorem we have

zn = 3n[cos(π/n) + isin(π/n)]n = 3n[cos(/n) + isin(/n)] = −3n

(since cos(π) = −1 and sin(π) = 0).

Question E2 (A2 and A4)

Use Demoivre’s theorem to find an expression for the roots of the equation zn − 1 = 0. Hence find the simple form of the solutions of this equation when n = 4.

The roots of zn − 1 = 0 are given by

$z = \cos\left(\dfrac{2\pi k\right}{n}) + i\sin\left(\dfrac{2\pi k\right}{n})$

where k = 0, 1, 2, ..., (n − 1)

For n = 4 we have

z0 = cos(0) + isin(0) = 1

z1 = cos(2π/4) + isin(2π/4) = cos(π/2) + isin(π/2) = i

z2 = cos(4π/4) + isin(4π/4) = cos(π) + isin(π) = −1

z3 = cos(6π/4) + isin(6π/4) = cos(3π/2) + isin(3π/2) = −i

Question E3 (A2, A3 and A9)

Expand the expression (cosθ + isinθ)2 by removing the brackets, and then rewrite it using Demoivre’s theorem. Hence show that cos(2θ) = cos2θ − sin2θ and sin(2θ) = 2sinθcosθ. Use a similar method and the complex binomial theorem, together with Demoivre’s theorem, to express cos(9θ) and sin(9θ) in terms of powers of cosθ and sinθ.

(cosθ + isinθ) = cos2θ + 2isinθcosθ − sin2θ   and, from Demoivre’s theorem:

(cosθ + isinθ)2 = cos(2θ) + isin(2θ)

Equating real and imaginary parts gives

cos(2θ) = cos2θ − sin2θ

andsin(2θ) = 2sinθcosθ

From Demoivre’s theorem we have

cos(9θ) + isin(9θ) = (cosθ + isinθ)9   and from the binomial theorem

(a + b)9 = $\displaystyle \sum_{r=0}^9{}^9C_{9-r}a^{9-r}b^r$

(a + b)9 = 9C9a9 + 9C8a8b + 9C7a7b2 + ... + 9C1ab8 + 9C0b9

(a + b)9 = a9 + 9a8b + 36a7b2 + 84a6b3 +126a5b4 +126a4b5 + 84a3b6 + 36a2b7 + 9ab8 + b9

So, putting together the results obtained from Demoivre’s theorem and the binomial theorem (with a = cosθ and b = isinθ) we find:

cos(9θ) + isin(9θ) = cos9θ + 9icos8θsinθ − 36cos7θsin2θ − 84icos6θsin3θ + 126cos5θsin4θ

cos(9θ) + isin(9θ) = + 126icos4θsin5θ − 84cos3θsin6θ −36icos2θsin7θ + 9cosθsin8θ + isin9θ

Comparing the coefficients of the real and imaginary parts on each side, we find

cos(9θ) = cos9θ − 36cos7θsin2θ + 126cos5θsin4θ − 84cos3θsin6θ + 9cosθsin8θ

andsin(9θ) = 9cos8θsinθ − 84cos6θsin3θ + 126cos4θsin5θ − 36cos2θsin7θ + sin9θ

It is very easy to make an arithmetic error in such a question, and a good deal of patience is needed to get it right. You should reread Subsection 3.4 if you had difficulty with the principles involved in this question.

Question E4 (A8)

The following equation occurs when solving the damped driven harmonic oscillator:

(−ω2iωβ0 + ω0) z0eiωt = α0eiωt

where t, α0, ω0, β0, and ω are real and positive, and z0 is complex. Find expressions for Re(z0), Im(z0) and |z0|.

The factor, eiωt, is non–zero and may be cancelled to give

$z_0 = \dfrac{\alpha_0}{\omega_0^2-\omega^2-i\omega\beta_0} = \dfrac{\alpha_0(\omega_0^2-\omega^2+i\omega\beta_0)}{(\omega_0^2-\omega^2)^2 +\omega^2\beta_0^2}$

Taking the real and imaginary parts, we have

${\rm {Re}}(z_0) = \dfrac{\alpha_0(\omega_0^2-\omega^2)}{(\omega_0^2-\omega^2)^2 +\omega^2\beta_0^2}$   and   ${\rm {Im}}(z_0) = \dfrac{\omega\alpha_0\beta_0}{(\omega_0^2-\omega^2)^2 +\omega^2\beta_0^2}$

To find |z0| we can use

$\lvert\,z_0\,\rvert^2 = z_0(z_0\cc) = \dfrac{\alpha_0^2}{(\omega_0^2-\omega^2)^2 +\omega^2\beta_0^2}$

so that$\lvert\,z_0\,\rvert^2 = \sqrt{z_0(z_0\cc)} = \dfrac{\alpha_0}{\sqrt{(\omega_0^2-\omega^2)^2 +\omega^2\beta_0^2}}$

Question E5 (A6 and A7)

Factorize the following expression:

2z2 − 11izx − 5x2

You should realize that z is not equal to x + iy unless we specifically choose to make such a definition, and in this case we should not assume that the x mentioned in the question is the real part of z.

You may be able to spot the values of z for which the expression 2z2 − 11izx − 5x2 is zero; these are z = xi/2 and for z = 5xi. (Alternatively, use the formula for finding the roots of a quadratic equation.) Since the expression we want to factorize is a polynomial of order 2 in z, the fundamental theorem of algebra tells us that zxi/2 and z − 5xi are the only factors and therefore

2z2 − 11izx − 5x2 = k(zxi/2)(z − 5xi)   for some constant k

Comparing the coefficients of z2, we see that the coefficient of proportionality is 2 and therefore

20z2 − 11izx − 5x2 = (2zxi)(z − 5xi)

Question E6 (A4 and A10)

Write down the sum of the following geometric series

1 + z + z2 + ... + z8

and hence show that the roots of the equation 1 + z + z2 + ... + z8 = 0 lie on a circle with its centre at the origin.

$1+z+z^2+\dots+z^8 = \dfrac{1-z^9}{1-z}$   if z ≠ 1