9231. FP2. Complex Numbers

From earlier work you should already know the following:

Complex numbers in rectangular form: x+yi;
Complex numbers in polar form (trigonometric & exponential);
Arithmetic with complex numbers (remember that multiplication in polar involves multiplying the moduli and adding the arguments);
Fundamental Theorem of Algebra: A degree n equation has n roots, all either real or in complex conjugate pairs.

Demoivre’s Theorem

If z=cos𝜽+isin𝜽, then zⁿ=cos(n𝜽) + isin(n𝜽)

This can be proved using induction.

We can use DeMoivre’s theorem to calculate cos(n𝜽) or sin(n𝜽) as a sum of powers of cos𝜽 or sin𝜽, by using the Binomial Theorem to expand zⁿ, using Pythagoras’ Theorem to change to just one trigonometric function, and then comparing real and imaginary parts.

This can also be used to solve certain equations in x by replacing x with a trigonometric function.

Worked Example 1 – Using DeMoivre’s Theorem

Find the value of:

(a) $(cos \frac{ \pi}{4} + is in \frac{ \pi}{4} )^8$

(b) $\frac{1}{ (cos( - \frac{ 3 \pi }{4} ) + i sin ( - \frac{ 3 \pi}{4} ))^6}$

(d) $( \sqrt{3} - i)^{18}$

Worked Example 2 – Expressing Trigonometric Functions of Multiple Angles

Express cos5𝜽 in terms of cos𝜽.

Express sin3𝜽 in terms of sin𝜽.

Express tan4𝜽 in terms of tan𝜽.

Worked Example 3 – Solving Polynomial Equations using Substitution

By considering the form of tan3𝜽, solve the cubic equation $3t^3 + 6t^2 -9t -2 = 0$

Exercise 1

1.) Find the values of the following complex numbers: (a) $( - \sqrt{3} - i )^{14}$ , (b) $( -1+i )^9$ .

2.) Write cos4xcosx in terms of powers of cos x.

3.) Find the values of the following expressions, giving your answer as an exact value: (a) $(cos \frac{ \pi }{12} + i sin \frac{ \pi }{12} )^8$ , (b) $( \sqrt{3} + i)^{12}$ , (c) $( \frac{1}{1+i})^6$

4.) Using de Moivre’s theorem, or otherwise, find $cos 7 \theta$ in terms of $cos \theta$

5.) Find $tan 3 \theta$ in terms of $tan \theta$ and hence solve the equation $2x^3 + 9x^2 - 3 = 6x$

6.) Show that $\frac{(cos \theta + i sin \theta)^6}{ (cos \theta + i sin \theta)^8 }$ can be written as $cos \theta (2cos \theta - sec \theta - 2i sin \theta)$ .

7.) Find $\frac{ sin 6 \theta}{ cos \theta}$ in terms of $sin \theta$

8.) Show that $tan 4 \theta = \frac{ 4tan \theta - 4 tan^3 \theta}{1 - 6 tan^2 \theta + tan^4 \theta}$ . Hence determine the solutions to the equation $x^4 - 4x^3 - 6x^2 + 4x = -1$

9.) By considering that $tan 5 \theta = \frac{ 5tan \theta - 10 tan^3 \theta + tan^5 \theta}{ 1 - 10 tan^2 \theta + 5 tan^4 \theta}$ , find all solutions for the polynomial $2x^5 - 5x^4 - 20x^3 +10x^2 +10x = 1$

Answers

1.) (a) $8192 + 8192i \sqrt{3}$ , (b) -16 + 16i

2.) cos(x) – 8cos(x³) + 8 cos (x⁵)

3.) (a) $- \frac{1}{2} + \frac{ \sqrt{3} }{2} i$ , (b) 4096, (c) $\frac{i}{8}$

4.) $64 cos^7 \theta - 112 cos^5 \theta + 56 cos^3 \theta - 7 cos \theta$

5.) $\frac{3 tan \theta - tan^3 \theta}{ 1 - 3 tan^2 \theta}$ , x = -5.04, -0.340, 0.876

6.) Proof

7.) $32sin^5 \theta - 32sin^3 \theta + 6sin \theta$

8.) Proof, x = -1.50, -0.199, 0.668, 5.03

9.) -2.32, -0.593, 0.0930, 0.879, 4.44

Exercise 1 Worked Solutions

Further Uses of DeMoivre’s Theorem

We can use Demoivre’s theorem to write $z+ \frac{1}{z}$ or $z- \frac{1}{z}$ as purely real or purely imaginary.

We can do the same for zⁿ+z^-n or zⁿ-z^-n.

By comparing this with the binomial expansion of these expressions, we can rewrite trigonometric functions raised to a power as trigonometric functions of multiple angles, allowing us to integrate them.

N.B. The above can also be usefully written in exponential form.

Worked Example

Evaluate $\int sin^6 \theta d \theta$

Express cos⁶𝜽 – sin⁶𝜽 in the form pcosq𝜽 + rcoss𝜽, where p, q, r and s are constants to be determined.

In summary, for your reference, given that $z = cos \theta + i sin \theta$ and $\frac{1}{z} = cos \theta - i sin \theta$ , then:

$z + \frac{1}{z} = 2 cos \theta$

$z - \frac{1}{z} = 2i sin \theta$

$z^n + \frac{1}{z^n} = 2cos n \theta$

$z^n - \frac{1}{z^n} = 2i sin n \theta$

Exercise 2

Answers

1.) $\frac{15}{8} + \frac{3}{2} cos2 \theta + \frac{5}{8} cos 4 \theta$

2.) $\frac{43}{120} \sqrt{2}$

3.) 1

4.) $\frac{1}{4} cos 5 \theta + \frac{5}{4} cos 3 \theta + \frac{5}{2} cos \theta$

5.) $- \frac{1}{32} sin 6 \theta + \frac{3}{32} sin 2 \theta$

6.) $\frac{1}{16} e^{ 5i \theta} + \frac{5}{16} e^ {-3i \theta} + \frac{5}{8} e^{i \theta}$

7.) $\frac{1}{64} cos 8 \theta + \frac{7}{16} cos 4 \theta + \frac{35}{64}$

8.) $\frac{16}{35}$

Exercise 2 Worked Solutions

Roots of Unity

The Fundamental Theorem of Algebra tells us that equation z³=1 has 3 roots.

We can find these roots by rearranging it into the form z³-1 = 0, noticing that z = 1 is one root, then using this fact to factor z³-1

We could also find them using deMoivre’s theorem.

To do this we rewrite 1 in its general polar form, and then can express the solutions generally as:

(cos(0+2kπ) + isin(0+2kπ))^1/3

We apply deMoivre’s theorem to this and then draw the roots on an Argand diagram. Notice that as the angles increase by equal increments, the diagram shows a regular shape.

We can also notice that the roots sum to zero – this can also seen if we represent the three complex numbers as vectors.

This use of deMoivre’s theorem to find roots can be used with any complex number on the right hand side of the equation, not just 1.

Worked Example 1

Find, in exponential form, the 7th roots of unity.

Worked Example 2

Given that z³ – 1 = 0, find the solutions in x + iy form, naming them $\omega_0$ , $\omega_1$ and $\omega_2$ .

Show that the square of one complex root is the other complex root and hence name the three roots $1$ , $\omega$ , $\omega^2$

Hence find the value of $\omega ^6 + \omega ^7 + \omega ^8$ .

Worked Example 3

Solve $z^6 = 4 + 4 \sqrt{3} i$ , giving all answers in the form $z = re^{i \theta}$ . Show all your solutions on an Argand diagram.

Exercise 3

1.) Find the 6th roots of unity, giving your answers in integer or exponential form.

2.) The roots of $z^3-1=0$ are $1$ , $\omega$ , $\omega^2$ . Determine the value of $\omega ^3 + \omega ^4 + \omega ^5$ .

3.) Find the roots of the equation $z^4 = 8 + 8 \sqrt{3} i$ , giving your answers in exponential form.

4.) Find the 4th roots of -16, giving each root in an exponential form.

5.) A polynomial equation is given as $z^5 -1 = 0$ . (a) Show that 1 is one root, then factorise your polynomial into the form (z-1)f(z) = 0, where f(z) is to be stated. (b) Using the four complex solutions, show that $cos( \frac{2 \pi}{5}) + cos( \frac{4 \pi}{5}) = - \frac{1}{2}$

6.) Solve the equation $x^8 = -8 \sqrt{3} + 8i$ . Show all of these solutions on an Argand diagram.

7.) Find the roots of z¹² = -1. Give your answers in exponential form.

8.) Show that $(z - e^{ i \theta} )( z - e^{-i \theta} ) = z^2 - 2z cos \theta + 1$

9.) (a) Write down the sixth roots of unity in exponential form, (b) State the two real roots, (c) The polynomial $z^6 - 1 = 0$ can be written as the product of two linear factors and two quadratic factors. Show that one of the quadratic factors is z²-z+1 and find the other one.

Answers

Exercise 3 Worked Solutions

More advanced problems, including series

Our formulae for z+z^-1 and z-z^-1can often help us when tackling problems, by making it easier to write an expression with its real and imaginary parts separated.

Please make careful note that we are now solving purely trigonometric problems, by working with the complex number and then considering real and imaginary parts. This is often the easiest method.

Complex summations

Let’s try and find the value of $\sum_{n=0}^{N-1} cos n \theta$ .

We could start listing out the terms, i.e. $\sum_{n=0}^{N-1} cos n \theta = 1 + cos \theta + cos 2 \theta + ...$ , but this doesn’t seem to help.

As $cos n \theta$ is the real part of zⁿ, we can instead consider $\sum_{n=0}^{N-1} z^n$

$\sum_{n=0}^{N-1} z^n = 1 + z + z^2 + ...$ , which is a geometric sum equal to $\frac{z^N - 1}{z-1}$ .

Replacing z with $e^{ i \theta }$ , we have $\sum_{n=0}^{N-1} z^n = \frac{ e^{Ni \theta} - 1}{e^{i \theta } -1 }$

We can multiply through by $e^{- \frac{1}{2} i \theta}$ to give $\sum_{n=0}^{N-1} z^n = \frac{ e^{(N- \frac{1}{2}) i \theta} - e^{- \frac{1}{2} i \theta}}{e^{ \frac{1}{2}i \theta} - e^{ - \frac{1}{2} i \theta }}$ .

The denominator can now be written as the single term $2i sin \frac{1}{2} \theta$ , which will allow us to split the fraction into real and imaginary parts. Writing it out in full gives:

$\sum_{n=0}^{N-1} z^n = \frac{cos(N- \frac{1}{2}) \theta + isin (N - \frac{1}{2}) \theta - cos \frac{1}{2} \theta + i sin \frac{1}{2} \theta }{ 2i sin \frac{1}{2} \theta }$

We only need the real part of this fraction, so we have: $\sum_{n=0}^{N-1} cos n \theta = \frac { sin(N- \frac{1}{2}) \theta + sin \frac{1}{2} \theta }{ 2 sin \frac{1}{2} \theta }$ = $\frac{1}{2} [sin( N - \frac{1}{2}) \theta cosec \frac{1}{2} \theta + 1]$

Worked Example 1

In the following example the denominator is not symmetric, meaning the technique used above to make the denominator whole real or wholly imaginary will not work. In cases where the denominator is of the form $az-1$ , we can multiply thorough by $az^{-1} -1$ , utilising the formulae learned above.