Further Pure — Complex numbers: A-level Further Maths OCR revision notes

4.02

Complex numbers and their arithmetic

Complex numbers and their arithmetic 4.02

Definitions

Imaginary unit: $i$ , defined by $i^{2} = - 1$ . Powers of $i$ cycle with period four: $i, - 1, - i, 1, i, \dots$
Complex number: $z = x + i y$ with $x, y$ real; $x = Re (z)$ is the real part and $y = Im (z)$ the imaginary part (a real number, without the $i$ ).
Complex conjugate: $z^{*} = x - i y$ (also written $\overset{z}{ˉ}$ ).

Key results

Equality: $a + ib = c + i d$ if and only if $a = c$ and $b = d$ . One complex equation is two real equations.
$(a + ib) (c + i d) = (a c - b d) + i (a d + b c)$ .
$z z^{*} = x^{2} + y^{2}$ , which is real and non-negative; $z + z^{*} = 2 Re (z)$ and $z - z^{*} = 2 i Im (z)$ .
$(z_{1} z_{2})^{*} = z_{1}^{*} z_{2}^{*}$ and $(z_{1} + z_{2})^{*} = z_{1}^{*} + z_{2}^{*}$ .

Method

Add and subtract by collecting real and imaginary parts separately.
Multiply by expanding the brackets as usual, then replace $i^{2}$ by $- 1$ .
To divide, multiply numerator and denominator by the conjugate of the denominator, which makes the denominator real: $\frac{3 + i}{1 - 2 i} = \frac{( 3 + i ) ( 1 + 2 i )}{( 1 - 2 i ) ( 1 + 2 i )} = \frac{1 + 7 i}{5}$ .

Notes

Equating real and imaginary parts is the most useful idea in the topic: it turns one equation in $z$ into two real simultaneous equations.
$Im (3 - 4 i) = - 4$ , not $- 4 i$ . Writing the $i$ into the imaginary part is a common slip.

Quadratics, square roots and solving for z 4.02

Key results

A real quadratic $a z^{2} + b z + c = 0$ with $b^{2} - 4 a c < 0$ has two non-real roots $z = \frac{- b \pm i 4 a c - b ^{2}}{2 a}$ , which are conjugates of each other.
Every non-zero complex number has exactly two square roots, each the negative of the other.

Method

Square root of $w = a + ib$ : let $(x + i y)^{2} = a + ib$ . Equate parts: $x^{2} - y^{2} = a$ and $2 x y = b$ . Substitute $y = \frac{b}{2 x}$ into the first equation, solve the resulting quadratic in $x^{2}$ , and keep only the positive value of $x^{2}$ (since $x$ is real).
Equations such as $2 z + i z^{*} = 6 + 3 i$ : write $z = x + i y$ , so $z^{*} = x - i y$ , expand, equate real and imaginary parts and solve the two real equations.

Notes

Example: $(x + i y)^{2} = 5 + 12 i$ gives $x^{2} - y^{2} = 5$ and $x y = 6$ , so $x^{4} - 5 x^{2} - 36 = 0$ , $(x^{2} - 9) (x^{2} + 4) = 0$ , $x^{2} = 9$ , and the square roots are $\pm (3 + 2 i)$ .
The rejected root $x^{2} = - 4$ is not an error to apologise for: it appears because $x$ was assumed real, and is discarded for exactly that reason.

Worked example

Find the complex number $z$ satisfying:

2 z + i z^{*} = 6 + 3 i

Show worked solution

Let $z = x + i y$ , so $z^{*} = x - i y$ and $i z^{*} = y + i x$ .

Then:

2 x + 2 i y + y + i x = 6 + 3 i

Real parts: $2 x + y = 6$ .

Imaginary parts: $x + 2 y = 3$ .

Solving, $x = 3$ and $y = 0$ , so $z = 3$ .

Check:

2 (3) + i (3) = 6 + 3 i

4.02

The Argand diagram and loci

The Argand diagram, modulus and argument 4.02

Definitions

Argand diagram: the plane in which $z = x + i y$ is the point $(x, y)$ ; the horizontal axis is the real axis and the vertical axis the imaginary axis.
Modulus: $∣ z ∣ = r = x^{2} + y^{2}$ , the distance from the origin to $z$ .
Argument: $ar g z = θ$ , the angle from the positive real axis to the line from O to $z$ , measured anticlockwise. The principal argument satisfies $- π < θ \leq π$ .

Key results

Modulus–argument form: $z = r (cos θ + i sin θ)$ , with $x = r cos θ$ , $y = r sin θ$ .
Exponential form: $z = r e^{i θ}$ , using Euler's relation $e^{i θ} = cos θ + i sin θ$ .
$∣ z^{*} ∣ = ∣ z ∣$ and $ar g z^{*} = - ar g z$ : the conjugate is the reflection in the real axis.
$∣ z_{1} - z_{2} ∣$ is the distance between the points $z_{1}$ and $z_{2}$ .

Method

Plot the point first and decide its quadrant.
Find the acute angle $α = arctan \frac{y}{x}$ .
Adjust: first quadrant $θ = α$ ; second $θ = π - α$ ; third $θ = - (π - α)$ ; fourth $θ = - α$ .

Notes

A calculator's $tan^{- 1}$ only returns angles between $- \frac{π}{2}$ and $\frac{π}{2}$ , so it gives the wrong argument for any point with a negative real part. The sketch is what catches this.
Addition in the Argand diagram is vector addition: $z_{1} + z_{2}$ is the fourth vertex of the parallelogram on O, $z_{1}$ and $z_{2}$ .

Multiplying and dividing in modulus–argument form 4.02

Key results

$∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ and $ar g (z_{1} z_{2}) = ar g z_{1} + ar g z_{2}$ (adjusted by $\pm 2 π$ into the principal range if needed).
$\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$ and $ar g \frac{z _{1}}{z _{2}} = ar g z_{1} - ar g z_{2}$ .
In exponential form these are the laws of indices: $r_{1} e^{i θ_{1}} \cdot r_{2} e^{i θ_{2}} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ .

Notes

Geometrically, multiplying by $w$ enlarges by scale factor $∣ w ∣$ and rotates anticlockwise about O through $ar g w$ . Multiplying by $i$ is a quarter turn.
The proof uses the compound-angle formulae: $(cos α + i sin α) (cos β + i sin β) = cos (α + β) + i sin (α + β)$ .
Shown on the sheet: $z_{1} = 1 + i$ and $z_{2} = 1 + 3 i$ multiply to a number of modulus $22$ and argument $\frac{π}{4} + \frac{π}{3} = \frac{7 π}{12}$ ; expanding directly gives $(1 - 3) + (1 + 3) i$ , which agrees.

Loci in the Argand diagram 4.02

Key results

$∣ z - a ∣ = r$ : the circle with centre $a$ and radius $r$ . $∣ z - a ∣ < r$ is the inside of the circle.
$∣ z - a ∣ = ∣ z - b ∣$ : the perpendicular bisector of the line segment joining $a$ and $b$ .
$ar g (z - a) = α$ : the half-line from $a$ (not including $a$ ) making angle $α$ with the positive real direction.

Method

Rewrite each condition as a distance or an angle measured from a fixed point: $∣ z + 2 - i ∣$ is $∣ z - (- 2 + i) ∣$ , the distance from $- 2 + i$ .
For a region, sketch each boundary, decide whether it is included (solid line for $\leq$ , dashed for $<$ ), and shade the points satisfying every condition.

Notes

To find a locus algebraically, substitute $z = x + i y$ : $∣ z - (2 + i) ∣ = 1.5$ becomes $(x - 2)^{2} + (y - 1)^{2} = 2.25$ .
Questions often ask for the greatest or least value of $∣ z ∣$ or $ar g z$ on a circular locus: draw the line from O through the centre (for the modulus) or the tangents from O (for the argument).

Worked examples

Worked example

Express $z = - 1 + 3 i$ in the form $r e^{i θ}$ with $- π < θ \leq π$ , and hence find $z^{6}$ .

Show worked solution

$r = 1 + 3 = 2$ .

The point is in the second quadrant; the acute angle is:

arctan 3 = \frac{π}{3}

so:

θ = π - \frac{π}{3} = \frac{2 π}{3}

and $z = 2 e^{2 πi /3}$ .

Then:

z^{6} = 2^{6} e^{4 πi} = 64

Worked example

The locus of $z$ is:

∣ z - 3 - 4 i ∣ = 2

Find the greatest and least values of $∣ z ∣$ .

Show worked solution

This is the circle with centre $3 + 4 i$ and radius 2.

The centre is $∣3 + 4 i ∣ = 5$ from O.

Along the line through O and the centre, the nearest point is $5 - 2 = 3$ from O and the furthest $5 + 2 = 7$ .

So $3 \leq ∣ z ∣ \leq 7$ .

4.02

De Moivre's theorem and exponential form

De Moivre's theorem 4.02

Key results

For every integer $n$ : $(cos θ + i sin θ)^{n} = cos n θ + i sin n θ$ .
Hence $[r (cos θ + i sin θ)]^{n} = r^{n} (cos n θ + i sin n θ)$ : raise the modulus to the power $n$ and multiply the argument by $n$ .
If $z = cos θ + i sin θ$ , then $z^{n} + \frac{1}{z ^{n}} = 2 cos n θ$ and $z^{n} - \frac{1}{z ^{n}} = 2 i sin n θ$ .

Method

Proof by induction for positive integers $n$ . Base case $n = 1$ is immediate. Assume $(cos θ + i sin θ)^{k} = cos k θ + i sin k θ$ . Then $(cos θ + i sin θ)^{k + 1} = (cos k θ + i sin k θ) (cos θ + i sin θ)$ $= (cos k θ cos θ - sin k θ sin θ) + i (sin k θ cos θ + cos k θ sin θ) = cos (k + 1) θ + i sin (k + 1) θ$ .
Negative integers: for $n = - m$ , $(cos θ + i sin θ)^{- m} = \frac{1}{cos m θ + i sin m θ} = cos m θ - i sin m θ = cos (- m θ) + i sin (- m θ)$ .

Notes

Large powers become easy: $(1 + i)^{10} = (2 e^{iπ /4})^{10} = 32 e^{i 5 π /2} = 32 i$ .
The induction proof is a standard request. State the hypothesis for $n = k$ , show the step to $n = k + 1$ , and finish with the concluding sentence.

Trigonometric identities from De Moivre's theorem 4.02

Key results

$cos 3 θ = 4 cos^{3} θ - 3 cos θ$ and $sin 3 θ = 3 sin θ - 4 sin^{3} θ$ .
$cos^{4} θ = \frac{1}{8} (cos 4 θ + 4 cos 2 θ + 3)$ .

Method

Multiple angles in terms of powers ( $cos 3 θ$ in terms of $cos θ$ ): expand $(cos θ + i sin θ)^{3}$ by the binomial theorem, equate real parts with $cos 3 θ$ (imaginary parts with $sin 3 θ$ ), then use $sin^{2} θ = 1 - cos^{2} θ$ .
Powers in terms of multiple angles ( $cos^{4} θ$ in terms of $cos 4 θ$ , $cos 2 θ$ ): write $(2 cos θ)^{4} = (z + \frac{1}{z})^{4}$ , expand, and pair the terms as $z^{n} + z^{- n} = 2 cos n θ$ .

Notes

The second type of identity is the one that matters for integration: $\int cos^{4} θ d θ$ cannot be done directly, but $\frac{1}{8} \int (cos 4 θ + 4 cos 2 θ + 3) d θ$ can.
Check an identity by substituting a value: at $θ = 0$ , $cos^{4} θ = 1$ and $\frac{1}{8} (1 + 4 + 3) = 1$ .

Exponential form and sums of series 4.02

Definitions

Exponential form: $z = r e^{i θ}$ , using Euler's relation $e^{i θ} = cos θ + i sin θ$ .

Key results

$r_{1} e^{i θ_{1}} \times r_{2} e^{i θ_{2}} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ , and De Moivre's theorem reads $(e^{i θ})^{n} = e^{in θ}$ .
For $z = e^{i θ}$ : $z^{n} + z^{- n} = 2 cos n θ$ and $z^{n} - z^{- n} = 2 i sin n θ$ .
With $C = \sum cos k θ$ and $S = \sum sin k θ$ over the same range, $C + i S = \sum e^{ik θ}$ , a geometric series with common ratio $e^{i θ}$ .

Method

Powers in terms of multiple angles: write $(2 cos θ)^{n} = (z + z^{- 1})^{n}$ , expand by the binomial theorem, and pair $z^{k}$ with $z^{- k}$ .
Summing $C$ or $S$ : form $C + i S$ , sum the geometric series, then separate real and imaginary parts, often by multiplying top and bottom by $e^{- i θ /2}$ to make the denominator real or purely imaginary.

Notes

The exponential form makes multiplication, powers and roots quick; Cartesian form makes addition quick. Choose the form that suits the operation.
Example: $(z + z^{- 1})^{3} = z^{3} + 3 z + 3 z^{- 1} + z^{- 3}$ gives $8 cos^{3} θ = 2 cos 3 θ + 6 cos θ$ , so $cos^{3} θ = \frac{1}{4} (cos 3 θ + 3 cos θ)$ .

Worked example

Show that:

cos^{4} θ = \frac{1}{8} (cos 4 θ + 4 cos 2 θ + 3)

and hence find:

\int_{0}^{π /2} cos^{4} θ d θ

Show worked solution

With $z = e^{i θ}$ ,

(2 cos θ)^{4} = (z + z^{- 1})^{4} = z^{4} + 4 z^{2} + 6 + 4 z^{- 2} + z^{- 4} = (z^{4} + z^{- 4}) + 4 (z^{2} + z^{- 2}) + 6 = 2 cos 4 θ + 8 cos 2 θ + 6

Dividing by 16 gives the identity.

Then:

\int_{0}^{π /2} cos^{4} θ d θ = \frac{1}{8} [\frac{sin 4 θ}{4} + 2 sin 2 θ + 3 θ]_{0}^{π /2} = \frac{1}{8} \cdot \frac{3 π}{2} = \frac{3 π}{16}

4.02

Roots of complex numbers

Roots of unity 4.02

Definitions

$n$ th roots of unity: the $n$ solutions of $z^{n} = 1$ , namely $z = e^{2 kπi / n}$ for $k = 0, 1, \dots, n - 1$ .

Key results

Writing $ω = e^{2 πi / n}$ , the roots are $1, ω, ω^{2}, \dots, ω^{n - 1}$ ; they lie on the unit circle at the vertices of a regular $n$ -gon with one vertex at 1.
For $n \geq 2$ : $1 + ω + ω^{2} + \dots + ω^{n - 1} = 0$ .
The non-real roots occur in conjugate pairs, $ω^{n - k} = (ω^{k})^{*}$ .

Notes

The sum is zero because it is a geometric series: $\frac{ω ^{n} - 1}{ω - 1} = 0$ since $ω^{n} = 1$ and $ω \neq = 1$ . Equally, the roots of $z^{n} - 1$ sum to minus the coefficient of $z^{n - 1}$ , which is zero.
Cube roots of unity: $1$ , $ω = - \frac{1}{2} + \frac{3}{2} i$ , $ω^{2} = - \frac{1}{2} - \frac{3}{2} i$ , with $1 + ω + ω^{2} = 0$ . This identity simplifies many expressions.

Solving zⁿ = w and real polynomials 4.02

Key results

If $w = R e^{i α}$ , the $n$ roots of $z^{n} = w$ are $z = R^{1/ n} e^{i (α + 2 kπ) / n}$ for $k = 0, 1, \dots, n - 1$ : a regular $n$ -gon centred at O with radius $R^{1/ n}$ .
Conjugate root theorem: if a polynomial has real coefficients and $z$ is a root, then $z^{*}$ is a root too.
So a real polynomial of odd degree has at least one real root, and $(z - z_{0}) (z - z_{0}^{*}) = z^{2} - 2 Re (z_{0}) z + ∣ z_{0} ∣^{2}$ is a real quadratic factor.

Method

Given one non-real root $z_{0}$ of a real cubic or quartic: write down $z_{0}^{*}$ , form the real quadratic factor $z^{2} - 2 Re (z_{0}) z + ∣ z_{0} ∣^{2}$ , then find the remaining factor by comparing coefficients.

Notes

Example: $z = 2 + i$ is a root of $z^{3} - 7 z^{2} + 17 z - 15 = 0$ . Then $2 - i$ is also a root, giving the factor $z^{2} - 4 z + 5$ ; comparing coefficients, $z^{3} - 7 z^{2} + 17 z - 15 = (z^{2} - 4 z + 5) (z - 3)$ , so the third root is $3$ .
Finish an $n$ th-roots question by checking that the arguments are in the required range and that there are exactly $n$ of them.

Worked examples

Worked example

Solve:

z^{3} = 42 (1 + i)

giving your answers in the form $r e^{i θ}$ with $- π < θ \leq π$ .

Show worked solution

$42 (1 + i)$ has modulus $42 \cdot 2 = 8$ and argument $\frac{π}{4}$ .

So:

z = 8^{1/3} e^{i (π /4 + 2 kπ) /3} = 2 e^{i (π + 8 kπ) /12}

For $k = 0, 1, - 1$ the arguments are $\frac{π}{12}$ ,

\frac{9 π}{12} = \frac{3 π}{4}

and $- \frac{7 π}{12}$ .

The roots are $2 e^{iπ /12}$ , $2 e^{3 πi /4}$ , $2 e^{- 7 πi /12}$ .

Worked example

Given that $1 - 2 i$ is a root of:

z^{4} - 2 z^{3} + 6 z^{2} - 2 z + 5 = 0

find the other roots.

Show worked solution

The coefficients are real, so $1 + 2 i$ is also a root, and:

(z - 1 + 2 i) (z - 1 - 2 i) = z^{2} - 2 z + 5

is a factor.

Let the quartic be:

(z^{2} - 2 z + 5) (z^{2} + a z + b)

The $z^{3}$ coefficient gives $a - 2 = - 2$ , so $a = 0$ ; the constant gives $5 b = 5$ , so $b = 1$ .

Check the $z^{2}$ coefficient: $b - 2 a + 5 = 6$ , and the $z$ coefficient: $5 a - 2 b = - 2$ .

So the other factor is $z^{2} + 1$ , and the roots are $1 \pm 2 i$ and $\pm i$ .