Figure 2 An Argand diagram showing the three roots of z³ − 1 = 0.

Answer F1

We can write the number, 1, as

1 = e^2πk = 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.

Answer F2

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

Answer F3

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 + i4cos² θsinθ − 6cos²θsin²θ − i4cosθsin²θ + 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

Answer F4

Z can be written as

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

and therefore

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

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

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

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

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(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^iθ, where r and θ are again real. Using Euler’s formula

e^iθ = cosθ + isinθ

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 = re^iθ then this corresponds to z* = re^−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 = re^iθ, the inverse is given by z⁻¹ = r⁻¹e^−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.

Answer R2

Figure 4 See Answer R2.

(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 = rρ[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.

Figure 5 See Answer R3.

Answer R3

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.

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

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

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

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

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

Consult complex roots in the Glossary for further information.

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.

Answer R6

(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.

Answer T1

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.

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) + 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

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

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

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

Answer T4

If we let z = e^iθ we can write

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

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

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

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

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

Answer T5

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

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 |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).

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

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

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

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

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

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

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

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

Answer T8

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

Answer T9

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

Answer T10

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

Answer T11

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)

Answer T12

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(z − i)(z + 2i)

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

z² + iz + 2 = (z − i)(z + 2i)

Answer T13

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.

Answer T14

(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

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

Answer T15

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

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

Answer T16

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

Answer T17

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

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

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

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

Answer E1

Demoivre’s theorem states that

(cosθ + isinθ)ⁿ = cos(nθ) + isin(nθ)

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

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

(since cos(π) = −1 and sin(π) = 0).

(Reread Subsection 2.1 if you had difficulty with this question.)

Answer E2

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

(Reread Subsection 2.3 if you had difficulty with this question.)

Answer E3

(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 − 36cos⁷ θsin²θ + 126cos⁵θsin⁴θ − 84cos³ θsin⁶θ + 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.

Answer E4

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

(Reread Section 3 if you had difficulty with this question.)

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 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 z − xi/2 and z − 5xi are the only factors and therefore

2z² − 11izx − 5x² = k(z − xi/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 = (2z − xi)(z − 5xi)

(Reread Subsection 3.2 if you had difficulty with this question.)

Answer E6

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

The roots of the equation 1 + z + z² + ... + z⁸ = 0 correspond to zeros of the numerator 1 − z⁹, 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 + z² + ... + z⁸ = 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 + z² + ... + z⁸ = 0.

(Reread Subsection 3.2 if you had difficulty with this question.)