FP1: Chapter 1�Complex Numbers

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4:: Modulus and argument

|z₁z₂| and arg (z₁z₂)

6:: Find complex solutions to cubic and quartic equations.

3:: Argand Diagram

X axis is real axis, Y axis is imaginary axis

5:: Find complex solutions to quadratic equations.

1:: Imaginary and complex Numbers

Here’s something that someone has almost certainly spoiled for you already…

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Imaginary part

Real part

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We tend to ensure real and imaginary components are grouped together.

(a, b and c are real constants)

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a + 3i – 4 + bi = a – 4 + (3 + b)i

2a – 3bi + 3 – 6ci = 2a+3 – 3(b + 2c)i

Just like we’d write 6π instead of π6, the i appears after any real constants, so we might write 5ki or πi.

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An exception is when we involve a function.

e.g. i sin θ and i√3

Why? This avoids ambiguity over whether the function is being applied to the i.

Convention 1

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Convention 2

In the same way that we often initially use x and y as real-valued variables, we often use z first to represent complex values, then w.

Complex numbers were originally introduced by the Italian mathematician Cardano in the 1500s to allow him to represent the roots of polynomials which weren’t ‘real’. They can also be used to represent outputs of functions for inputs not in the usual valid domain, e.g. Logs of negative numbers, or even the factorial of negative numbers!

Some other major applications of Complex Numbers:

Fractals

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A Mandelbrot Set is the most popular ‘fractal’. For each possible complex number c, we see if z_n+1 = z_n² + c is not divergent (using z₀ = 0), leading to the diagram on the right. Coloured diagrams can be obtained by seeing how quickly divergence occurs for each complex c (if divergent).

This is called an Argand Diagram. We’ll use it later.

Analytic Number Theory

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Number Theory is the study of integers. Analytic Number Theory treats integers as reals/complex numbers to use other (‘analytic’) methods to study them. For example, the Riemann Zeta Function allows complex numbers as inputs, and is closely related to the distribution of prime numbers.

Physics and Engineering

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Used in Signal Analysis, Quantum Mechanics, Fluid Dynamics, Relativity, Control Theory...

Simplify (8 + i)(3 – 2i)

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= 24 – 16i + 3i – 2i²

= 24 – 16i + 3i + 2

= 26 – 13i

Here’s a flavour of some of things you’ll be initially expected to do...

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Example: Let z₁ = 1 + 3i and z₂ = 5 – 2i, find z₁z₂.

(1 + 3i)(5 – 2i) = 5 – 2i + 5i – 6i²

= 5 – 2i + 5i + 6

= 11 + 3i

Calculate z² for the following z.

The quick way to think about this is that i² reverses the sign.

1. z = 1 + i z² = 2i
2. z = 1 – i z² = -2i
3. z = 3 + 2i z² = 5 + 12i
4. z = 7 – 4i z² = 33 – 56i
5. z = -3 + 3i z² = -18i
6. z = a + bi z² = a2 – b² + 2abi

Calculate z₁z₂.

1. z₁ = 1 + i z₂ = 1 – i z₁z₂ = 2
2. z₁ = 2 + i z₂ = 2 – 2i z₁z₂ = 6 – 2i
3. z₁ = 3 + 2i z₂ = 4 – 3i z₁z₂ = 18 – i
4. z₁ = -3 + i z₂ = 5 + 7i z₁z₂ = -22 – 16i
5. z₁ = 7 – 3i z₂ = 3 – 7i z₁z₂ = -58i
6. z₁ = -1 – i z₂ = 9 + 8i z₁z₂ = -1 – 17i

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By using a Binomial Expansion, determine (1 + i)⁵.

= 1 + 5i + 10i² +10i³ + 5i⁴ + i⁵

= 1 + 5i – 10 – 10i + 5+ i

= -4 – 4i

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What is the value of:

1. i¹⁰⁰ = (i⁴)²⁵ = 1²⁵ = 1
2. i²⁰⁰³ = i²⁰⁰⁰ x i³ = 1 x i³ = -i

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The same trick works with complex numbers.

🖉 If z = x + yi, then we define z* = x – yi.

z* is known as the complex conjugate of z.

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z z* = x² + y²

z + z* = 2x

which are both real.

The fact the first is real will help us with dividing complex numbers.

We’ll see the significance of the second later.

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Given z, determine z z*.

z = 3 + 2i z z* = 13

z = 3 – 2i z z* = 13

z = 5 + 4i z z* = 41

z = 1 + i z z* = 2

z = 4 – 2i z z* = 20

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__26__

2 + 3i

(2 – 3i)

( )(2 – 3i)

=

52 – 78i

13

=

4 – 6i

Click to Brosolve

1 - i

1 + i

Put all the following in the form a + bi.

_10_

3 + i

_1_

1 + i

8 – 4i

1 – 3i

= 3 - i

= -i

Note that mark schemes permit the single fraction.

1 - i

2

=

10 + 3i

1 + 2i

16 – 17i

5

=

_i_

1 – i

-1 + i

2

=

14 – 5i

3 – 2i

= 4 + i

= 2 + 2i

a + i

a – i

a² – 1 + 2ai

a² + 1

=

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Edexcel June 2013 (R)

Given that (2 + i)(z + 3i) = 10 – 5i, find z, giving your answer in the form a + bi.

z = 3 – 7i (either by dividing 10 – 5i by 2 + i and subtracting 3i, or replacing z with a + bi before expanding and comparing real and imaginary parts).

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Argand Diagrams are a way of geometrically representing complex numbers.

z = x + yi

The x-axis is the real component.

The y-axis is the imaginary component.

Im[z]

Re[z]

-3 -2 -1 1 2 3

3

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2

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1

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-1

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-3

Click to move.

In FP2, you’ll encounter something called ‘polar coordinates’. This is an alternative way of representing coordinate, which instead of using the (x,y) position (known as a Cartesian coordinate), uses the distance from the origin and the angle.

Im[z]

Re[z]

-3 -2 -1 1 2 3

3

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-1

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-2

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-3

z = 2 + 3i

Distance from origin:

(Don’t write anything down yet!)

√(2² + 3²) = √13

This is known as the modulus of z, and we write

|2 + 3i| = √13

Angle: (anticlockwise from the real axis)

Using trigonometry, we have (in radians):

tan^-1(3/2) = 0.983

This is known as the argument of z, and we write:

arg(2 + 3i) = 0.983

θ

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Im[z]

Re[z]

arg z

|z|

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Give exact answers where possible, otherwise to 3dp.

Given that arg(3 + a + 4i) = π/3 and that a is real, determine a.

4/(3 + a) = √3. Thus a = (4/√3) – 3

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Give that arg(5 + i + ai) = π/4 and that a is real, determine a.

(a + 1)/5 = 1. So a = 4

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Given that arg(santa + 3 + 2i) = π/2 and that santa is real, determine santa.

santa = -3

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Im[z]

Re[z]

arg z

|z|

Suppose that r = |z| and θ = arg(z).

How could we express z in terms of r and o?

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z = x + yi

= r cos θ + r i sin θ

= r(cos θ + i sin θ)

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This is known as the modulus-argument form (or the polar form) of z.

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Question 1 from earlier...

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Any cubic (with coefficient of x³ of 1) can be expressed as:

y = (x – α)(x – β)(x – γ)

where α, β and γ are the roots.

However, these 3 roots may not necessarily all be real...

α

β

γ

α

All 3 roots are real.

1 real root, 2 complex roots.

Are there any other possibilities?

No: since cubics have a range of –∞ to +∞, it must cross the x axis. And it can’t cross an even number of times, otherwise the cubic would start and end in the same vertical direction.

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α

β=γ

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3 real roots (one repeated)

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Classic Question:

x = 2 is one of the roots of the polynomial x³ – x – 6 Find the other two roots.

Use polynomial division (as per C2) to divide x³ – x – 6 by (x – 2).

This gives x² + 2x + 3.

Using the quadratic formula, we obtain the roots:

x = -1 ± i√2

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In summary, complex roots of polynomials always come in complex conjugate pairs.

Another classic type of exam question:

-1 + 2i is one of the roots of the cubic x³ – x² – x – 15.

Find the other two roots.

Other complex root is the complex conjugate: -1 – 2i.

Now expand (x – (-1 + 2i))(x – (-1 – 2i))

= x² – (-1 – 2i)x – (-1 + 2i)x + (-1 + 2i)(-1 – 2i)

= x² + 2x + 5

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We can now either use polynomial division or “Factor Theorem with trial and error” to establish the real root as 3.

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Your turn:

2 – i is one of the roots of the cubic x³ – 11x + 20. Find the other two roots.

2 + i is other complex root.

(x – (2 + i))(x – (2 – i)) = x² – 4x + 5.

Dividing we get (x + 4), so real root is -4.

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Any quartic (with coefficient of x⁴ of 1) can be expressed as:

y = (x – α)(x – β)(x – γ)(x – δ)

where α, β, γ and δ are the roots.

α

β

δ

γ

α

β

Possibility 1:

4 real roots (some potentially repeated)

Possibility 2:

2 real roots, pair of complex conjugate roots.

Possibility 3:

No real roots. Two pairs of complex conjugate roots.

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Any other possibilities?

No. On y-axis, both ends of line must either be both positive infinity or both negative. x-axis must therefore be crossed an even number of times, so an even number of real roots.

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Edexcel Jan 2013 (R)

Im[z]

Re[z]

b)

Since real roots appear on the x-axis and complex roots come in conjugate pairs, we know we’ll have symmetry in the line y = 0

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