MATH 3.3: Demoivre’s theorem and complex algebra 
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PPLATO / FLAP (Flexible Learning Approach To Physics) 


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
(cos_{ }θ + i_{ }sin_{ }θ)^{ n} = cos(nθ) + i_{ }sin(nθ)
(where i^{2} = −1), and then use it to derive trigonometric identities, in Subsection 2.2, and to find all solutions to the equation z^{ n} − 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.
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 z^{ n} − 1 = 0, where n is a positive integer. For n = 3, plot your results on an Argand diagram.
Figure 2 An Argand diagram showing the three roots of z^{3} − 1 = 0.
Answer F1
We can write the number, 1, as
1 = e^{2πk} = cos(2πk) + i_{ }sin(2πk)
where k is any integer (positive, negative or zero). So z is given by
z = 1^{2} = [cos(2πk) + i_{ }sin(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 z^{5} in its simplest form, where z = 2_{ }[cos(π/10) + i_{ }sin(π/10)]
Answer F2
Using Demoivre’s theorem we have
z^{5} = 2^{5}[cos(π/10) + i_{ }sin(π/10)]^{5} = 2^{5}[cos(5π/10) + i_{ }sin(5π/10)] = 2^{5}[cos(π/2) + i_{ }sin(π/2)] = 2^{5}i = 32_{ }i
Question F3
Use Demoivre’s theorem, together with the complex binomial theorem, to show that
(a) cos(4θ) = cos^{4}_{ }θ − 6_{ }cos^{2} θ_{ }sin^{2}_{ }θ + sin^{4}_{ }θ
(b) sin(4θ) = 4_{ }cos^{3}_{ }θ_{ }sin_{ }θ − 4_{ }cos_{ }θ_{ }sin^{3}_{ }θ
Answer F3
From Demoivre’s theorem we have
cos(4θ) + i_{ }sin(4θ) = [cos_{ }θ + i_{ }sin_{ }θ]^{4}
The right–hand side of this expression can be expanded using the binomial theorem to give
$\displaystyle (a+b)^4 = \sum_{r=0}^n({}^4C_{4r}a^{4r}b^r) = {}^4C_4a^4 + {}^4C_3a^3b + {}^4C_2a^2b^2 + {}^4C_1ab^3 + {}^4C_0b^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4$
where ${}^nC_r = \dfrac{n!}{r!(nr)!}$ 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) + i_{ }sin(4θ) = cos^{2}_{ }θ + i_{ }4_{ }cos^{2} θsin_{ }θ − 6_{ }cos^{2}_{ }θ_{ }sin_{ }^{2}_{ }θ − i_{ }4_{ }cosθ_{ }sin^{2}_{ }θ + sin^{2}_{ }θ
cos(4θ) + i_{ }sin(4θ) = [cos^{2}_{ }θ − 6_{ }cos^{2}_{ }θ_{ }sin^{2}_{ }θ + sin^{2}_{ }θ] + i_{ }[4_{ }cos^{2}_{ }θ_{ }sin_{ }θ − 4_{ }cosθ_{ }sin^{2}_{ }θ]
Equating real and imaginary parts we find
cos(4θ) = cos^{2}_{ }θ − 6_{ }cos^{2}_{ }θ_{ }sin^{2}_{ }θ + sin^{2}_{ }θ
sin(4θ) = 4_{ }cos^{2}_{ }θ_{ }sin_{ }θ − 4_{ }cosθ_{ }sin^{2}_{ }θ
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}.
Answer F4
Z can be written as
$Z = \dfrac{\omega RC+i\left(\omega^2LC1\right)}{\omega C}$
and therefore
$Z^{1} = \dfrac{\omega C}{\omega RC+i\left(\omega^2LC1\right)}$
$\phantom{Z^{1} }= \dfrac{\omega C}{\omega RC+i\left(\omega^2LC1\right)} \times \dfrac{\omega RCi\left(\omega^2LC1\right)}{\omega RCi\left(\omega^2LC1\right)}$
$\phantom{Z^{1} }= \dfrac{\omega C[\omega RCi\left(\omega^2 LC1\right)]}{\omega^2 R^2C^2+\left(\omega^2RC1\right)^2}$
${\rm{Re}}\left(z\right) = \dfrac{\omega^2 C^2}{\omega^2 R^2C^2+\left(\omega^2RC1\right)^2}~~~~~~~~{\rm{Im}}\left(z\right) = \dfrac{\omega C\left(1\omega^2 LC\right)}{\omega^2 R^2C^2+\left(\omega^2RC1\right)^2}$
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) = cos^{2}_{ }θ − sin^{2}_{ }θ 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 i^{2} = −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.
Figure 3 See Answer R1.
Answer R1
(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_{ }(cos_{ }θ + i_{ }sin_{ }θ)
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) + i_{ }sin(π/3)]
The exponential_form_of_a_complex_numberexponential form of a complex number is z = r_{ }e^{iθ}, where r and θ are again real. Using Euler’s formula
e^{iθ} = cos_{ }θ + i_{ }sin_{ }θ
and comparing the result just found for the polar form, we have
z = 2e^{iπ/3}
(b) For an arbitrary complex number, z = x + iy, the complex conjugate is given by z* = x − iy. If z is in exponential form, z = r_{ }e^{iθ} then this corresponds to z* = r_{ }e^{−iθ}.
For the complex number given in this question, we have
z* = 2e^{−iπ/3}
For an arbitrary complex number given in exponential form z = r_{ }e^{iθ}, the inverse is given by z^{−1} = r^{−1}_{ }e^{−iθ}.
If z = 2e^{iπ/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).
Answer R2
Figure 4 See Answer R2.
(a) We can represent w in polar form by
w = ρ_{ }(cos_{ }φ + i_{ }sin_{ }φ)
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) + i_{ }sin(π/4)]
In general, if complex numbers, z and w, are written in polar form as
z = r_{ }(cos_{ }θ + i_{ }sin_{ }θ) and w = ρ_{ }(cos_{ }φ + i_{ }sin_{ }φ)
then their product is given by zw = rρ_{ }[cos(θ + φ) + i_{ }sin(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 + 4_{ }i, plot z and z* on an Argand diagram. What is the value of _{ }z_{ }?
Figure 5 See Answer R3.
Answer R3
The complex numbers z = 3 + 4_{ }i and z* = 3 − 4_{ }i 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 $\dfrac{2+3i}{(1+i)(12i)}$
Answer R4
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
$\dfrac{2 + 3i}{(1+ i)(1  2i)} = \dfrac{(2 + 3i)(1  i)(1 + 2i)}{(1+ i)(1  i)(1  2i)(1+ 2i)}$
$\phantom{\dfrac{2 + 3i}{(1+ i)(1  2i)} }= \dfrac{1}{2\times 5}[(2+3i)(1+i+2)]$
$\phantom{\dfrac{2 + 3i}{(1+ i)(1  2i)} }= \dfrac{(2+3i)(3+i)}{10} = \dfrac{63+9i+2i}{10}$
$\phantom{\dfrac{2 + 3i}{(1+ i)(1  2i)} }= \dfrac{3}{10} +\dfrac{11}{10}i$
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
z^{5} + z^{4} + z^{3} + z^{2} + 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?
Answer R5
(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 + x^{2} + ... + x^{10} (for x ≠ 1)
(b) Expand (1 + x)^{7} in powers of x.
Answer R6
(a) Using the formula for the sum of a geometric series we find
$1+x+x^2+\dots +x^{10} = \dfrac{1x^{11}}{1x}$
(b) The binomial expansion gives
(1 + x)^{7} = 1 + 7x + 21x^{2} + 35x^{3} + 35x^{4} + 21x^{5} + 7x^{6} + x^{7}
Consult geometric series for complex numbers in the Glossary for further information.
i
If we have an arbitrary complex number, z, then we can choose to write it in polar form as
z = r_{ }(cos_{ }θ + i_{ }sin_{ }θ)
where r and θ are real (and i^{2} = −1). Furthermore, if we have another complex number
w = ρ_{ }(cos_{ }φ + i_{ }sin_{ }φ)
then the product of z and w can be written as
zw = rρ_{ }[cos(θ + φ) + i_{ }sin(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
z_{1} = r_{1}_{ }[cos(θ_{1}) + i_{ }sin(θ_{1})]
z_{2} = r_{2}_{ }[cos(θ_{2}) + i_{ }sin(θ_{2})]
z_{2} = r_{2} $~~\vdots~~~~~~~~~~\vdots~~~~~~~~~~\vdots$
z_{n} = r_{n}_{ }[cos(θ_{n}) + i_{ }sin(θ_{n})]
to obtain the product
z_{1}z_{2} × ... × z_{n} = (r_{1}r_{2} × ... × r_{n})[cos(θ_{1} + θ_{2} + ... + θ_{n}) + i_{ }sin(θ_{1} + θ_{2} + ... + θ_{n})]
Setting r_{1} = r_{2} = ... = r_{n} = 1 and θ_{1} = θ_{2} = ... = θ_{n} = θ we obtain Demoivre’s theorem
(cos_{ }θ + i_{ }sin_{ }θ)^{ n} = cos(nθ) + i_{ }sin(nθ) Demoivre’s theorem(2)
Question T1
Evaluate both (cos_{ }θ + i_{ }sin_{ }θ)^{2} and [cos(2θ) + i_{ }sin(2θ)] for θ = 0, (π/4) and (π/2) rad. Show that your results are consistent with Demoivre’s theorem for n = 2.
Answer T1
For convenience, define L = (cos_{ }θ + i_{ }sin_{ }θ)^{2} and R = cos(2θ) + i_{ }sin(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 2^{1/4}[cos(3π/8) + i_{ }sin(3π/8).
✧ i − 1 = 2^{1/2}[cos(3π/4) + i_{ }sin(3π/4)] i and therefore, using Demoivre’s theorem with n = 1/2,
(i − 1)^{1/2} = 2^{1/4}[cos(3π/8) + i_{ }sin(3π/8)].
Question T2
Use Demoivre’s theorem to show that one of the square roots of 1 + i is
2^{1/4}[cos(π/8) + i_{ }sin(π/8)]
[Hint: First write 1 + i in polar form.]
Figure 4 See Answer T2.
Answer T2
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) + i_{ }sin(π /4)].
Setting n = 1/2 in Demoivre’s theorem we have
(cos_{ }θ + i_{ }sin_{ }θ)^{1/2} = cos(θ/2) + i_{ }sin(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 (cos_{ }θ + i_{ }sin_{ }θ)^{ n} by writing out the n + 1 individual terms in its binomial expansion, we know that the answer must be cos(nθ) + i_{ }sin(nθ). To emphasize the advantage of Demoivre’s theorem, consider the evaluation of z^{8} where
z = 2_{ }[cos(π/8) + i_{ }sin(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
z^{8} = (x + iy)^{8} = x^{8} + 8ix^{7}y − 28x^{6}y^{2} − 56ix^{5}y^{3} + 70x^{4}y^{4} + 56ix^{3}y^{5} − 28x^{2}y^{6} − 8ixy^{7} + y^{8} i
and, after a considerable amount of arithmetic, we would obtain the approximate answer
z^{8} ≈ −256.0 − 1.53 × 10^{−4} i
Compare this brute force approach with the elegance of Demoivre’s theorem which gives the exact answer
z^{8} = {2_{ }[cos(π/8) + i_{ }sin(π/8)]}^{8} = 2^{8}[cos(π) + i_{ }sin(π)] = −2^{8} = −256
Demoivre’s theorem can be used to obtain a variety of useful identities involving cos^{n}1θ, sin^{n}1θ, cos(nθ) and sin(nθ). The trick is to let z = e^{iθ}, from which we obtain z^{−1} = e^{−iθ} 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 z^{ n} = e^{inθ} and z^{−n} = e^{−inθ} 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
cos_{ }^{n}_{ }θ or sin^{n}_{ }θ in terms of cos(mθ) or sin(mθ) for various values of m
or
cos(nθ) or sin(nθ) in terms of cos^{m}_{ }θ or sin^{m}_{ }θ 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.
Express cos^{2}_{ }θ in terms of the cosines of multiples of θ.
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 cos^{2}_{ }θ = 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^{iθ} and then expand (z + 1/z)^{3}.] i
Answer T3
If we let z = e^{iθ}, 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) $\cos^3\theta = \dfrac{\cos(3\theta)+3\cos\theta}{4}$
Similar techniques can be used to express an odd power of sin_{ }θ in terms of the sines of multiples of θ.
Express sin^{7}_{ }θ in terms of the sines of multiples of θ.
Starting from the identity for sin_{ }θ 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
$\sin^7\theta = \dfrac{35\sin\theta21\sin(3\theta)+7\sin(5\theta)\sin(7\theta)}{2^6}$(12)
✦ Express sin^{5}_{ }θ in terms of the sines of multiples of θ. i
✧ Let z = e^{iθ} 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)
Answer T4
If we let z = e^{iθ} we can write
$(2i)^3\sin^3\theta = \left(z\dfrac1z\right)^3 = \left(z\dfrac1z\right)\left(z^22+\dfrac{1}{z^2}\right)$
$\phantom{(2i)^3\sin^3\theta }= \left(z^32z+\dfrac1z\right) + \left(z+\dfrac2z\dfrac{1}{z^3}\right)$
$\phantom{(2i)^3\sin^3\theta }= \left(z\dfrac1z\right)  3\left(z\dfrac1z\right)$
$\phantom{(2i)^3\sin^3\theta }= 2i \sin(3\theta)  3 \times 2i\sin\theta$
which can be rearranged to give the required result $\sin^3\theta = \dfrac{3\sin\theta\sin(3\theta)}{4}$
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 cos_{ }θ and sin_{ }θ. (In fact, we can derive two identities at the same time.) We start by using Demoivre’s theorem with n = 2
cos(2θ) + i_{ }sin(2θ) = (cos_{ }θ + i_{ }sin_{ }θ)^{2}
and then expand the right–hand side to give
cos(2θ) + i_{ }sin(2θ) = cos^{2}_{ }θ + 2_{ }i_{ }sin_{ }θ_{ }cos_{ }θ − sin^{2}_{ }θ
We can now equate the real and imaginary parts to obtain two identities
cos(2θ) = cos^{2}_{ }θ − sin^{2}_{ }θ(14)
sin(2θ) = 2_{ }sin_{ }θ_{ }cos_{ }θ(15)
which are the required results.
✦ Use Demoivre’s theorem to obtain identities for cos(5θ) and sin(5θ) in terms of cos_{ }θ and sin_{ }θ.
✧ From Demoivre’s theorem, with n = 5, we have
cos(5θ) + i_{ }sin(5θ) = (cos_{ }θ + i_{ }sin_{ }θ)^{5}
cos(5θ) + i_{ }sin(5θ) = cos^{5}_{ }θ + 5_{ }i_{ }cos^{4}_{ }θ_{ }sin_{ }θ − 10_{ }cos^{3}_{ }θ_{ }sin^{2}_{ }θ − 10_{ }i_{ }cos^{2}_{ }θ_{ }sin^{3}_{ }θ + 5_{ }cos_{ }θ _{ }sin^{4}_{ }θ + i_{ }sin^{5}_{ }θ i
and equating real and imaginary parts gives us
cos(5θ) = cos^{5}_{ }θ − 10_{ }cos^{3}_{ }θ_{ }sin^{2}_{ }θ + 5_{ }cosθ_{ }sin^{4}_{ }θ and sin(5θ) = 5_{ }cos^{4}_{ }θ_{ }sin_{ }θ − 10_{ }cos^{2}_{ }θ_{ }sin^{3}_{ }θ + sin^{5}_{ }θ
Question T5
Use Demoivre’s theorem to obtain identities for cos(3θ) and sin(3θ) in terms of cos_{ }θ and sin_{ }θ.
Answer T5
We start by using Demoivre’s theorem with n = 3
cos(3θ) + i_{ }sin(3θ) = (cos_{ }θ + i_{ }sin_{ }θ)^{2}
and then expand the right–hand side to obtain
cos(3θ) + i_{ }sin(3θ) = cos^{2}_{ }θ + 3_{ }cosθ_{ }(i_{ }sin_{ }θ)^{2} + 3_{ }i_{ }cos^{2}_{ }θ_{ }sin_{ }θ + (i_{ }sin_{ }θ)^{2}
cos(3θ) + i_{ }sin(3θ) = cos^{2}_{ }θ − 3_{ }cosθ_{ }sin^{2}_{ }θ + i_{ }(3_{ }cos^{2}_{ }θ_{ }sin_{ }θ − sin^{2}_{ }θ)
We can now equate the real and imaginary parts to obtain two identities
cos(3θ) = cos^{2}_{ }θ − 3_{ }cosθ_{ }sin^{2}_{ }θ
sin(3θ) = 3_{ }sin_{ }θ_{ }cos^{2}_{ }θ − sin^{2}_{ }θ
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 = e^{0 i} or e^{2π i} or e^{4π i} or e^{6π i} and so on
so it follows that
1^{1/2} = e^{i} or e^{π i} or e^{2π i} or e^{3π i} 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 = e^{0 i} = e^{2π i} = e^{4π i} ..., whereas −1 = e^{π i} = e^{3π i} = e^{5π i} ..., 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
z^{ n} − 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 = 1^{1/n}). To find these roots we use the fact that we can write the number 1 as i
1 = e^{2πk i} = cos(2πk) + i_{ }sin(2πk)(17)
where k is any integer (positive, negative or zero).
So z is given by
z = 1^{1/n} = [cos(2πk) + i_{ }sin(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.
Figure 1 An Argand diagram showing the two roots of z^{2} − 1 = 0.
$\cos\left(\dfrac{2\pi k}{n}+2\pi\right) = \cos\left(\dfrac{2\pi k}{n}\right)$(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) + i_{ }sin(πk)
where k = 0, 1 or, more explicitly
z_{0} = cos(0) + i_{ }sin(0) = 1
z_{1} = cos(π) + i_{ }sin(π) = −1
And once again we have the familiar result that 1^{1/2} = ±1 which is plotted on an Argand diagram in Figure 1.
Figure 2 An Argand diagram showing the three roots of z^{3} − 1 = 0.
Find the three cube roots of 1.
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) + i_{ }sin(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 z_{1} and z_{2} are as given above, find: (a) _{ }z_{1}_{ } and _{ }z_{2}_{ }, (b) _{ }z_{1}_{ }^{3} and _{ }z_{2}^{3}_{ }.
Answer T6
(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 _{ }z_{1}_{ } = 1, so that _{ }z_{2}_{ }^{3} = 1. The hard way to do this part (and not recommended) is to calculate z^{3} explicitly, and then find its modulus.
Since
$\left(\dfrac12i\dfrac{\sqrt{3\os}}{2}\right)^2 = \left(\dfrac12i\dfrac{\sqrt{3\os}}{2}\right)\left(\dfrac12i\dfrac{\sqrt{3\os}}{2}\right) = \dfrac12+i\dfrac{\sqrt{3\os}}{2}$
it follows that
$z_2^3 = \left(\dfrac12i\dfrac{\sqrt{3\os}}{2}\right)^3 = \left(\dfrac12i\dfrac{\sqrt{3\os}}{2}\right)\left(\dfrac12+i\dfrac{\sqrt{3\os}}{2}\right) = \dfrac14+\dfrac34 = 1$
and therefore _{ }z_{2}^{3}_{ } = 1.
A slightly easier method would be to use Demoivre’s theorem and write
z_{2}^{3} = [cos(4π/3) + i_{ }sin(4π/3)]^{2} = cos(4π) + i_{ }sin(4π)
so that_{ }z_{2}^{3}_{ } = cos^{2}_{ }(4π) + sin^{2}_{ }(4π) = 1
The easiest way is to notice that _{ }z_{2}^{3}_{ } = _{ }z_{2}_{ }^{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 x^{1/2} 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 3^{1/2}, 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, z^{1/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
z_{0} = 2^{1/6}_{ }e^{iπ/12}_{ }e^{0 i} = 2^{1/6}_{ }e^{iπ/12},
z_{1} = 2^{1/6}_{ }e^{iπ/12}_{ }e^{2iπ/3} = 2^{1/6}_{ }e^{3iπ/4},
and
z_{2} = 2^{1/6}_{ }e^{iπ/12}_{ }e^{4iπ/3} = 2^{1/6}_{ }e^{17iπ/12}. i
Question T7
Find all roots of the equation z^{6} − 1 = 0 and plot your results on an Argand diagram.
Figure 5 See Answer T7.
Answer T7
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
$z_0 = \cos(0) + i\sin(0) = 1$
$z_1 = \cos(\pi/3) + i\sin(\pi/3) = \dfrac12+i\dfrac{\sqrt{3\os}}{2}$
$z_2 = \cos(2\pi/3) + i\sin(2\pi/3) = \dfrac12+i\dfrac{\sqrt{3\os}}{2}$
$z_3 = \cos(\pi) + i\sin(\pi) = 1$
$z_4 = \cos(4\pi/3) + i\sin(4\pi/3) = \dfrac12+i\dfrac{\sqrt{3\os}}{2}$
$z_5 = \cos(5\pi/3) + i\sin(5\pi/3) = \dfrac12i\dfrac{\sqrt{3\os}}{2}$
The values obtained for the six roots of z^{6} − 1 = 0 are plotted on an Argand diagram in Figure 6.
Question T8
Show that all roots of the equation z^{ n} − 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.
Answer T8
The n roots of z^{ n} − 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 _{ }z_{k}_{ } = 1, and the roots therefore lie on a circle of radius 1 centred at the origin.
In this section we consider complex algebra; that is the manipulation of expressions involving complex variables.
Finding the n^{th} 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.
Solve the following equation
−z^{2} + izγ + ω^{2} = 0(20) i
where γ and ω are real constants, which occurs in the theory of damped oscillations.
We can solve for z by using the well–known formula for the roots of the quadratic equation
az^{2} + bz + c = 0
namely
$z = \dfrac{b\pm\sqrt{b^24ac}}{2a}$
In our particular example we have a = −1, b = iγ and c = ω^{2} and therefore
$z = \dfrac{i\gamma\pm\sqrt{\smash[b]{(i\gamma)^24(1)\omega^2}}}{2}$
$z = \dfrac{i\gamma\pm\sqrt{\smash[b]{4\omega^2\gamma^2}\us}}{2} = \dfrac{i\gamma}{2}\pm\sqrt{\smash[b]{\omega^2\gamma^2/4}\us}$
✦ Find the roots of the equation z^{2} + 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,
$z = \dfrac{b\pm\sqrt{b^24ac}}{2a} = \dfrac{1\pm\sqrt{14i\os}}{2}$
However, we can find a square root of 1 − 4_{ }i by writing 1 − 4_{ }i = 4.123_{ }e^{−1.3258 i} so that
$\sqrt{14i\os} \approx (4.1231)^{1/2}{\rm e}^{0.6629} \approx 2.0305[\cos(0.6629)  i\sin(0.6629)] \approx 1.6005  1.2496i$
and therefore the roots of the equation are $\dfrac{1\pm(1.60051.2496i)}{2}$
which gives 0.300 − 0.625_{ }i and −1.300 + 0.625_{ }i 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 2z^{2} − 11z_{ }i − 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.
Answer T9
For the equation az^{2} + bz + c = 0 the solutions are given by
$z = \dfrac{b\pm\sqrt{b^24ac}}{2a}$
In this case we have a = 2, b = −11_{ }i and c = −5 and therefore
$z=\dfrac{11i\pm\sqrt{\smash[b]{(11)^2+4\times 2\times 5}}}{4} = \dfrac{11i\pm9i}{4}$ = 5_{ }i 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
2_{ }iz − w = 5_{ }i − 1(23)
2_{ }iz − 6w = 2_{ }i(24)
Subtracting Equation 24 from Equation 23 we obtain
5w = 3_{ }i − 1 so that $w = \dfrac{3i1}{5}$
and substituting this value for w into Equation 22 we have
$z = 3iw + 1 = 1  3i\left(\dfrac{3i1}{5}\right) = \dfrac{3i+14}{5}$
Suppose we want to solve the pair of equations
I_{1}(Z_{1} + Z_{2}) + I_{2}Z_{2} = ε_{1} i
I_{1}Z_{2} + I_{2}Z_{2} = ε_{2}
for I_{1} and I_{2} where all the variables are complex.
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
I_{1}Z_{1} = ε_{1}−ε_{2}
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 I_{2}Z_{1}Z_{2} = ε_{2}Z_{1} − (ε_{1} − ε_{2})Z_{2} = ε_{2}(Z_{1} + Z_{2}) − ε_{1}Z_{2}
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 I_{1} and we are told that
Z_{1} = Z_{2} = R +iX, ε_{1} = ε and ε_{2} = ε_{ }e^{iφ} i
where R, X, ε and φ are all real. I_{1} is given by
$I_1 = \dfrac{\varepsilon_1\varepsilon_2}{R+iX} = \dfrac{ [\overbrace{\vphantom{0} \varepsilon}^{\large\color{purple}{\varepsilon_1}}\overbrace{\varepsilon(\cos\phi+i\sin\phi)}^{\large\color{purple}{\varepsilon_2}}]}{R+iX} \times \dfrac{RiX}{R+iX}$
$\phantom{I_1 }=\dfrac{\varepsilon(1\cos\phii\sin\phi)}{R^2+X^2}(RiX)$
$\phantom{I_1 }=\dfrac{\varepsilon}{R^2+X^2}\left\{[R(1\cos\phi)X\sin\phi]+i[R\sin\phiX(1\cos\phi)]\right\}$
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(I_{2}) and Im(I_{2}) in Example 5, i.e. $I_2 = \dfrac{\varepsilon_2(Z_1+Z_2)\varepsilon_1 Z_2}{Z_1 Z_2}$ and Z_{1} = Z_{2} = R + iX
Answer T10
From the text we have Z_{1} = Z_{2} = R + iX so that
$I_2 = \dfrac{\varepsilon_2(Z_1+Z_2)\varepsilon_1 Z_2}{Z_1 Z_2} = \dfrac{\varepsilon_2(Z_1+Z_1)\varepsilon_1 Z_1}{Z_1^2} = \dfrac{2\varepsilon_2\varepsilon_1}{Z_1} = \dfrac{\varepsilon}{R+iX}(\cos\phi +2i\sin\phi1)$
$\phantom{I_2 }= \dfrac{\varepsilon}{R^2+X^2}(\cos\phi +2i\sin\phi1)(riX)$
$\phantom{I_2 }= \dfrac{\varepsilon}{R^2+X^2}\left\{[R(\cos\phi1) +2X\sin\phi]) + i[2R\sin\phi + X(12\cos\phi)]\right\}$
which gives
${\rm{Re}}(I_2) = \dfrac{\varepsilon}{R^2+X^2}[R(\cos\phi1)+2X\sin\phi]$
and${\rm{Re}}(I_2) = \dfrac{\varepsilon}{R^2+X^2}[R\sin\phi+X(12\cos\phi)]$
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
z^{2} + 1 = (z − i)(z + i)
and soz^{2} + 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
z^{2} + (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 z^{2} + (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
z^{ n} + a_{n−1}z^{ n−1} + a_{n−2}z^{ n−2} + ... + a_{0}
for some (possibly quite large) positive integer, n, then the n roots, z_{1}, z_{2}, ... , z_{n}, of the equation
z^{ n} + a_{n−1}z^{ n−1} + a_{n−2}z^{ n−2} + ... + a_{0} = 0
correspond to the factorization
z^{ n} + a_{n−1}z^{ n−1} + a_{n−2}z^{ n−2} + ... + a_{0} = (z − z_{1})(z − z_{2}) ... (z − z_{n})
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.
Factorize 3z^{2} + 27.
Notice that the expression is zero for z = ±3_{ }i and therefore (z − 3_{ }i) and (z + 3_{ }i) are factors. Since the highest power of z in the polynomial 3z^{2} + 27 is z^{2}, the fundamental theorem of algebra tells us that these are the only roots, and therefore
3z^{2} + 27 = k_{ }(z − 3_{ }i)(z + 3_{ }i)
for some constant k. Comparing the coefficients on each side of this expression for any particular power of z (z^{2} is the most convenient in this case) we obtain k = 3, so that
3z^{2} + 27 = 3_{ }(z − 3_{ }i)(z + 3_{ }i)
Question T11
Factorize the expression 2z^{2} + 32.
Answer T11
First solve the equation 2z^{2} + 32 = 0 as follows: z^{2} = −16 and therefore z = ±4_{ }i. So we have
2z^{2} + 32 = k_{ }(z + 4_{ }i)(z − 4_{ }i)
for some constant k, and comparison of the coefficient of any particular power, say z^{2}, gives the factorization
2z^{2} + 32 = 2_{ }(z + 4_{ }i)(z − 4_{ }i)
Factorize the expression −z^{2} + izγ + w^{2}
where z is regarded as the ‘unknown variable’.
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 z^{2} (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 z^{2} + iz + 2.
Answer T12
Inspection shows that the expression is zero for z = i, −2_{ }i (or, alternatively, use the formula for the roots of a quadratic equation). Therefore we have
z^{2} + iz + 2 = k_{ }(z − i)(z + 2_{ }i)
for some constant k, and, equating coefficients, the final result is
z^{2} + iz + 2 = (z − i)(z + 2_{ }i)
We simplify expressions involving complex numbers in much the same way that we simplify expressions involving real numbers, except that every occurrence of i^{2} 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.
Simplify the following expression
Z = (3R + 2iX) − (R + iX) + (7R + 3iX)
To simplify such an expression we treat the real and imaginary parts separately to obtain i
Z = 9R + 4iX
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.
To find the real and imaginary parts of Z we rationalize the complex quotients
$Z = R_1 + \dfrac{R_2+i\omega L}{1+i\omega C(R_2+i\omega L)}$
$\phantom{Z }= R_1 + \dfrac{R_2+i\omega L}{1+i\omega C(R_2+i\omega L)} \times \dfrac{1i\omega C(R_2+i\omega L)}{1i\omega C(R_2+i\omega L)}$ 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 $\dfrac{3+2i}{1+i}\dfrac{1}{1+2i}$
✧ $\dfrac{3+2i}{1+i}\dfrac{1}{1+2i} = \dfrac{3+2i}{1+i}\times \dfrac{1i}{1i}  \dfrac{1}{1+2i}\times \dfrac{12i}{12i} = \dfrac{5i}{2} \dfrac{12i}{5} = \dfrac{23i}{10}$
Question T13
Simplify the expression $\dfrac{1}{1i}+\dfrac{1+i}{2}$
Answer T13
One way to simplify $\dfrac{1}{1i} + \dfrac{1+i}{2}$ is to put everything over a common denominator and then to rationalize the resulting expression as in
$\dfrac{1}{1i} + \dfrac{1+i}{2} = \dfrac{1}{2(1i)}[2+(1+i)(1i)]$
$\phantom{\dfrac{1}{1i} + \dfrac{1+i}{2} }= \dfrac{1}{2(1i)}[2+1+1] = \dfrac{2(1+i)}{(1i)(1+i)} = 1+i$
Alternatively we can rationalize the first term and then simplify it, as in the following
$\dfrac{1}{1i} + \dfrac{1+i}{2} = \dfrac{1+i}{(1i)(1+i)} + \dfrac{1+i}{2}$
$\phantom{\dfrac{1}{1i} + \dfrac{1+i}{2} }= \dfrac{1+i}{2} + \dfrac{1+i}{2} = 1+i$
which is probably slightly easier.
You probably recall that the binomial coefficient, ${}^nC_r$, is defined by
${}^nC_r = \dfrac{n!}{r!(nr)!}$(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} = ^{n}C_{n}a^{ n} + ^{n}C_{n−1}a^{ n−1}b + ... + ^{n}C_{n−r}a^{ n−r}b^{ r} + ... + ^{n}C_{1}ab^{ n−1} + ^{n}C_{0}b^{ n} = $\displaystyle \sum_{r=0}^n{}^nC_{nr}a^{nr}b^r$(28)
Question T14
(a) Evaluate ^{3}C_{r} for r = 0, 1, 2, 3. (b) Prove the general result ^{n}C_{r} = ^{n}C_{n−r}.
Answer T14
(a) Using ${}^nC_r = \dfrac{n!}{(nr)!r!}$ 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
${}^nC_{nr} = \dfrac{n!}{[n(nr)]!(nr)!} = \dfrac{n!}{r!(nr)!} = \dfrac{n!}{(nr)!r!} = {}^nC_r$
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 $\left(z  \dfrac1z\right)^3$
✧ $\displaystyle \left(z\dfrac1z\right)^3 = \sum_{r=0}^3 \left[{}^3C_{3r}z^{3r}\times\left(\dfrac1z\right)^r\right]$
$\phantom{\left(z\dfrac1z\right)^3 }={}^3C_3z^3  {}^3C_2\dfrac{z^2}{2}+{}^3C_1\dfrac{z}{z^2}{}^3C_0\dfrac{1}{z^3} = z^3  3z + \dfrac3z\dfrac{1}{z^3}$
$\phantom{\left(z\dfrac1z\right)^3 }=\left(z^3\dfrac{1}{z^3}\right)  3\left(z\dfrac1z\right)$
✦ Expand and simplify $\left(z  \dfrac1z\right)^7$
✧ $\displaystyle \left(z\dfrac1z\right)^7 = \sum_{r=0}^7 \left[{}^7C_{7r}z^{7r}\times\left(\dfrac1z\right)^r\right]$
$\phantom{\left(z\dfrac1z\right)^7 }= z^7  7z^5 + 21z^3  35z + \dfrac{35}{z}  \dfrac{21}{z^3} + \dfrac{7}{z^5}  \dfrac{1}{z^7}$
$\phantom{\left(z\dfrac1z\right)^7 }=\left(z^7\dfrac{1}{z^7}\right) 7 \left(z^5\dfrac{1}{z^5}\right) + 21\left(z^3\dfrac{1}{z^3}\right)$
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 ^{n}C_{r} = ^{n}C_{n−r}?]
Answer T15
$\displaystyle \left(z+\dfrac1z\right)^4 = \sum_{r=0}^4\left[{}^4C_{4r}z^{4r}\times\dfrac{1}{z^r}\right] = {}^4C_4z^4 + {}^4C_3z^2 + {}^4C_2 + {}^4C_1\dfrac{1}{z^2} + {}^4C_0\dfrac{1}{z^4}$
$\phantom{\displaystyle \left(z+\dfrac1z\right)^4 }=z^4+4z^2+6+4\dfrac{1}{z^2}+\dfrac{1}{z^4} = \left(z^4+\dfrac{1}{z^4}\right)+4\left(z^2+\dfrac{1}{z^2}\right)+6$
In Subsection 2.2 we showed how to write powers of trigonometric functions in terms of functions of multiple angles by setting z = e^{iθ} 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 cos^{5}_{ }θ in terms of cosines of
multiples of θ.
Answer T16
(a) Using the binomial theorem we have
$\displaystyle \left(z+\dfrac1z\right)^5 = \sum_{r=0}^5\left[{}^5C_{5r}z^{5r}\times\dfrac{1}{z^r}\right] = {}^5C_5z^5 + {}^5C_4z^3 + {}^5C_3z + {}^5C_2\dfrac1z + {}^5C_1\dfrac{1}{z^3}+ {}^5C_0\dfrac{1}{z^5}$
$\phantom{\displaystyle \left(z+\dfrac1z\right)^5 }= {}^5C_5\left(z^5+\dfrac{1}{z^5}\right) + {}^5C_4\left(z^3 + \dfrac{1}{z^3}\right) + {}^5C_3\left(z +\dfrac1z\right)$
$\phantom{\displaystyle \left(z+\dfrac1z\right)^4 }= \left(z^5+\dfrac{1}{z^5}\right) + 5\left(z^3 + \dfrac{1}{z^3}\right)+ 10\left(z +\dfrac1z\right)$
(b) Defining z = e^{iθ}, 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 $\cos^5\theta = \dfrac{1}{16}[\cos(5\theta) + 5\cos(3\theta) + 10\cos\theta]$
For real numbers a and x, the series a + ax + ax^{2} + ... + ax^{ n} is known as a geometricseries, and its sum is
$a\dfrac{1x^{n+1}}{1x}$
provided that x ≠ 1.
The geometric series for complex numbers a and z is identical in form, and i
$a+az+az^2+\dots+az^n = a\dfrac{1z^{n+1}}{1z}$ if z ≠ 1(29)
✦ Write down the sum of the series 1 + e^{iθ} + e^{2iθ} + e^{3iθ}.
✧ $1+{\rm e}^{i\theta} + {\rm e}^{2i\theta} + {\rm e}^{3i\theta} = \dfrac{1{\rm e}^{4i\theta} }{1{\rm e}^{i\theta} }$
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^{iθ} then
sin(θ) + sin(2θ) + sin(3θ) + ... + sin(9θ) = Im(z + z^{2} + ... + z^{9})
$={\rm {Im}}\left(\dfrac{zz^{10}}{1z}\right) = {\rm {Im}}\left(\dfrac{{\rm e}^{i\theta}{\rm e}^{10i\theta}}{1  {\rm e}^{i\theta}}\right) = {\rm {Im}}\left[\dfrac{({\rm e}^{i\theta}{\rm e}^{10i\theta})(1{\rm e}^{i\theta})}{(1{\rm e}^{i\theta})(1{\rm e}^{i\theta})}\right]$ 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).
Answer T17
cos_{ }θ + cos(2θ) + cos(3θ) + ... + cos(9θ)
$={\rm{Re}}\left(\dfrac{zz^{10}}{1z}\right) = {\rm{Re}}\left(\dfrac{{\rm e}^{i\theta}{\rm e}^{10i\theta}}{1{\rm e}^{i\theta}}\right) = {\rm{Re}}\left[\dfrac{({\rm e}^{i\theta}{\rm e}^{10i\theta})(1{\rm e}^{i\theta})}{(1{\rm e}^{i\theta})(1{\rm e}^{i\theta})}\right]$
$=\dfrac{1}{2(1\cos\theta)}{\rm{Re}}({\rm e}^{i\theta}1{\rm e}^{10i\theta}+{\rm e}^{9i\theta})$
$=\dfrac{\cos\theta1\cos(10\theta)+\cos(9\theta)}{2(1\cos\theta)}$
The identity
(cos_{ }θ + i_{ }sin_{ }θ)^{ n} = cos(nθ) + i_{ }sin(nθ)(Eqn 2)
is known as Section 2Demoivre’s theorem. It is valid for any real value of n.
Defining z = e^{iθ} 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) cos^{n}_{ }θ or sin^{n}_{ }θ in terms of cos(mθ) or sin(mθ);
(b) cos(nθ) or sin(nθ) in terms of cos^{m}_{ }θ and sin^{m}_{ }θ.
The n solutions of the equation z^{ n} − 1 = 0 are known as the n^{ th} 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
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.
If we determine the roots, z^{1}, z^{2}, ..., z^{ n}, of the equation
z^{ n} + a_{n−1}z^{ n−1} + a_{n−2}z^{ n−2} + ... + a_{0} = 0
then we have the following Subsection 3.2factorization
z^{ n} + a_{n−1}z^{ n−1} + a_{n−2}z^{ n−2} + ... + a_{0} = (z − z_{1})(z − z_{2}) ... (z − z_{n})
The complex binomial expansion is (a + b)^{ n} = ^{n}C_{n}a^{ n} + ^{n}C_{n−1}a^{ n−1}b + ... + ^{n}C_{n−r}a^{ n−r}b^{ r} + ... + ^{n}C_{1}ab^{ n−1} + ^{n}C_{0}b^{ n}
(a + b)^{ n} = ^{n}C_{n}a^{ n} + ^{n}C_{n−1}a^{ n−1}b + ... + ^{n}C_{n−r}a^{ n−r}b^{ r} + ... + ^{n}C_{1}ab^{ n−1} + ^{n}C_{0}b^{ n} = $\\ \displaystyle \sum_{r=0}^n{}^nC_{nr}a^{nr}b^r$(Eqn 28)
where
${}^nC_r = \dfrac{n!}{r!(nr)!}$(Eqn 26)
and n! = n_{ }(n − 1)(n − 2)(n − 3) ... 2 × 1, for n ≥ 1, with 0! = 1
The complex form of the Subsection 3.5geometric series is
$a+az+az^2+\dots+az^n = a\dfrac{1z^{n+1}}{1z}$ 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.
Having completed this module, you should be able to:
A1 Define the terms that are emboldened and flagged in the margins of the module.
State and derive Demoivre’s theorem.
Use Demoivre’s theorem to derive trigonometric identities.
Explain what is meant by the n^{th} roots of unity and find all such roots.
Describe the relevance of the fundamental theorem of algebra to the solution of equations and the factorization of polynomials.
Solve quadratic equations and linear equations involving complex variables.
Factorize simple polynomials.
Simplify complex expressions.
State and apply the complex binomial expansion.
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.
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 z^{ n} where
z = 3_{ }[cos(π/n) + i_{ }sin(π/n)]
for any integer, n.
Answer E1
Demoivre’s theorem states that
(cos_{ }θ + i_{ }sin_{ }θ)^{ n} = cos(nθ) + i_{ }sin(nθ)
for any real number, n. Using this theorem we have
z^{ n} = 3^{n}[cos(π/n) + i_{ }sin(π/n)]^{ n} = 3^{n}[cos(nπ/n) + i_{ }sin(nπ/n)] = −3^{n}
(since cos(π) = −1 and sin(π) = 0).
(Reread Subsection 2.1 if you had difficulty with this question.)
Question E2 (A2 and A4)
Use Demoivre’s theorem to find an expression for the roots of the equation z^{ n} − 1 = 0. Hence find the simple form of the solutions of this equation when n = 4.
Answer E2
The roots of z^{ n} − 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
z_{0} = cos(0) + i_{ }sin(0) = 1
z_{1} = cos(2π/4) + i_{ }sin(2π/4) = cos(π/2) + i_{ }sin(π/2) = i
z_{2} = cos(4π/4) + i_{ }sin(4π/4) = cos(π) + i_{ }sin(π) = −1
z_{3} = cos(6π/4) + i_{ }sin(6π/4) = cos(3π/2) + i_{ }sin(3π/2) = −i
(Reread Subsection 2.3 if you had difficulty with this question.)
Question E3 (A2, A3 and A9)
Expand the expression (cos_{ }θ + i_{ }sin_{ }θ)^{2} by removing the brackets, and then rewrite it using Demoivre’s theorem. Hence show that cos(2θ) = cos^{2}_{ }θ − sin^{2}_{ }θ and sin(2θ) = 2_{ }sin_{ }θ_{ }cos_{ }θ. 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 cos_{ }θ and sin_{ }θ.
Answer E3
(cos_{ }θ + i_{ }sin_{ }θ) = cos^{2}_{ }θ + 2_{ }i_{ }sin_{ }θ_{ }cosθ − sin^{2}_{ }θ and, from Demoivre’s theorem:
(cos_{ }θ + i_{ }sin_{ }θ)^{2} = cos(2θ) + i_{ }sin(2θ)
Equating real and imaginary parts gives
cos(2θ) = cos^{2}_{ }θ − sin^{2}_{ }θ
andsin(2θ) = 2_{ }sin_{ }θ_{ }cosθ
From Demoivre’s theorem we have
cos(9θ) + i_{ }sin(9θ) = (cos_{ }θ + i_{ }sin_{ }θ)^{9} and from the binomial theorem
(a + b)^{9} = $\displaystyle \sum_{r=0}^9{}^9C_{9r}a^{9r}b^r$
(a + b)^{9} = ^{9}C_{9}a^{9} + ^{9}C_{8}a^{8}b + ^{9}C_{7}a^{7}b^{2} + ... + ^{9}C_{1}ab_{8} + ^{9}C_{0}b^{9}
(a + b)^{9} = a^{9} + 9a^{8}b + 36a^{7}b^{2} + 84a^{6}b^{3} +126a^{5}b^{4} +126a^{4}b^{5} + 84a^{3}b^{6} + 36a^{2}b^{7} + 9ab^{8} + b^{9}
So, putting together the results obtained from Demoivre’s theorem and the binomial theorem (with a = cos_{ }θ and b = i_{ }sin_{ }θ) we find:
cos(9θ) + i_{ }sin(9θ) = cos^{9}_{ }θ + 9_{ }i_{ }cos^{8}_{ }θ_{ }sin_{ }θ − 36_{ }cos^{7}_{ }θ_{ }sin^{2}_{ }θ − 84_{ }i_{ }cos^{6}_{ }θ_{ }sin^{3}_{ }θ + 126_{ }cos^{5}_{ }θ_{ }sin^{4}_{ }θ
cos(9θ) + i_{ }sin(9θ) = + 126_{ }i_{ }cos^{4}_{ }θ_{ }sin^{5}_{ }θ − 84_{ }cos^{3}_{ }θ_{ }sin^{6}_{ }θ −36_{ }i_{ }cos^{2}_{ }θsin^{7}_{ }θ + 9_{ }cosθ_{ }sin^{8}_{ }θ + i_{ }sin^{9}_{ }θ
Comparing the coefficients of the real and imaginary parts on each side, we find
cos(9θ) = cos^{9}_{ }θ − 36_{ }cos^{7} θsin_{ }^{2}_{ }θ + 126_{ }cos^{5}_{ }θ_{ }sin_{ }^{4}_{ }θ − 84_{ }cos^{3} θsin_{ }^{6}_{ }θ + 9_{ }cosθ_{ }sin_{ }^{8}_{ }θ
andsin(9θ) = 9_{ }cos^{8}_{ }θ_{ }sin_{ }θ − 84_{ }cos^{6}_{ }θ_{ }sin_{ }^{3}_{ }θ + 126_{ }cos^{4}_{ }θ_{ }sin_{ }^{5}_{ }θ − 36_{ }cos^{2}_{ }θ_{ }sin_{ }^{7}_{ }θ + sin^{9}_{ }θ
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:
(−ω^{2} − iωβ_{0} + ω_{0}) z_{0}_{ }e^{iωt} = α_{0}_{ }e^{iωt}
where t, α_{0}, ω_{0}, β_{0}, and ω are real and positive, and z_{0} is complex. Find expressions for Re(z_{0}), Im(z_{0}) and _{ }z_{0}_{ }.
Answer E4
The factor, e^{iωt}, is non–zero and may be cancelled to give
$z_0 = \dfrac{\alpha_0}{\omega_0^2\omega^2i\omega\beta_0} = \dfrac{\alpha_0(\omega_0^2\omega^2+i\omega\beta_0)}{(\omega_0^2\omega^2)^2 +\omega^2\beta_0^2}$
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 _{ }z_{0}_{ } 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$\lvert\,z_0\,\rvert^2 = \sqrt{z_0(z_0\cc)} = \dfrac{\alpha_0}{\sqrt{(\omega_0^2\omega^2)^2 +\omega^2\beta_0^2}}$
(Reread Section 3 if you had difficulty with this question.)
Question E5 (A6 and A7)
Factorize the following expression:
2z^{2} − 11_{ }izx − 5x^{2}
Answer E5
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 2z^{2} − 11_{ }izx − 5x^{2} is zero; these are z = x_{ }i/2 and for z = 5x_{ }i. (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 z − x_{ }i/2 and z − 5x_{ }i are the only factors and therefore
2z^{2} − 11_{ }izx − 5x^{2} = k_{ }(z − x_{ }i/2)(z − 5x_{ }i) for some constant k
Comparing the coefficients of z^{2}, we see that the coefficient of proportionality is 2 and therefore
20z^{2} − 11_{ }izx − 5x^{2} = (2z − x_{ }i)(z − 5x_{ }i)
(Reread Subsection 3.2 if you had difficulty with this question.)
Question E6 (A4 and A10)
Write down the sum of the following geometric series
1 + z + z^{2} + ... + z^{8}
and hence show that the roots of the equation 1 + z + z^{2} + ... + z^{8} = 0 lie on a circle with its centre at the origin.
Answer E6
$1+z+z^2+\dots+z^8 = \dfrac{1z^9}{1z}$ if z ≠ 1
The roots of the equation 1 + z + z^{2} + ... + z^{8} = 0 correspond to zeros of the numerator 1 − z^{9}, in other words, the 9th roots of unity, with the exception of z = 1. The value z = 1 is certainly not a root of the equation 1 + z + z^{2} + ... + z^{8} = 0, since for this value the left–hand side is equal to 9. The n^{th} roots of unity lie on a circle centred at the origin with radius 1, so the same must be true of the roots of the equation 1 + z + z^{2} + ... + z^{8} = 0.
(Reread Subsection 3.2 if you had difficulty with this question.)
Study comment This is the final Exit test question. When you have completed the Exit test go back and try the Subsection 1.2Fast track questions if you have not already done so.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.
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.