Further Pure 1 Complex numbers Maths

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Further Pure 1

Complex Numbers

Page 1

Further Pure 1 Complex Numbers

The EDEXCEL syllabus says that candidates should:

a) understand the idea of a complex number, recall the meaning of the terms real part, imaginary part, modulus, argument, conjugate, and use the fact that two complex numbers are equal if and only if both real and imaginary parts are equal;

b) be able to carry out operations of addition, subtraction, multiplication and division of two complex numbers;

c) be able to use the result that, for a polynomial equation with real coefficients, any non-real roots occur in conjugate pairs;

n d) be able to represent complex numbers geometrically by means of an Argand diagram, and understand the io geometrical effects of conjugating a complex number and of adding and subtracting two complex numbers; Kumar's Maths Revis e) find the two square roots of a complex number;

Further Pure 1

Complex Numbers

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Section 1: Introduction to complex numbers

Suppose we wished to solve the quadratic equation x2 + 4x + 5 =0 . Using the quadratic formula, the solutions would be:

=x -= 4 ? 42 - 4 ?1? 5 -4 ? -4 .

2

2

We notice a problem however since -4 is not a real number. So the equation x2 + 4x + 5 =0 does not

have any real roots.

However, suppose we introduced the symbol i to represent -1 . We could then find expressions for the

solutions of the quadratic:

n =x -= 4 ? -4 -4 ?= 4? -1 -4 ? 2i .

2

2

2

io So the equation has two solutions: x = -2 + i or x = -2 ? i.

is These two solutions are called complex numbers.

v 1.1 Some definitions

e Suppose that z is a complex number. Let z = a + ib. R The real part of z, written Re(z), is a.

The imaginary part of z, written Im(z), is b.

s The complex conjugate of z, written z* or z , is z* = a ? ib.

th Example 1: Let z = 5 ? 2i, w = -2 + i and u = 7i.

Then:

a Re(z) = 5

Im(z) = -2 z* = 5 + 2i

Re(w) = -2 Im(w) = 1 w* = -2 ? i.

M Re(u) = 0

Im(u) = 7

u* = -7i.

r's 1.2 Solving quadratic equations

a Example 2: Solve the quadratic equation 2x2 + 4x + 5 =0 , giving your answers as complex

numbers in surd form.

um Solution: Using the quadratic formula: K =x -= 4 ? 16 - 4? 2? 5 -4 ? -24

4

4

= -= 4 ? i 24 -4 ? 2 6 i

4

4

So the solutions are x =-1+

6 2

i

or

x =-1-

6 2

i

.

Notice that the two solutions are complex conjugates of each other. The solutions form a conjugate pair. This leads to this very important result:

Consider the equation ax2 + bx + c =0 where a, b and c are real numbers. If the equation has complex roots, then the two roots are always conjugates of each other.

Further Pure 1

Complex Numbers

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Note: If a quadratic equation has any complex coefficients then this result doesn't apply.

Example: If z = 1 - i is one solution of the quadratic equation z2 - 2z + 2 =0 , then the second solution must be the complex conjugate (as the quadratic has real coefficients). So the second solution is z = 1 + i.

Section 2: Calculating with complex numbers

2.1 Adding and subtracting

Two complex numbers are added or subtracted by collecting together their real and imaginary parts.

So (x + iy) + (u + iv) = (x + u) + i( y + v)

n and io (x + iy) - (u + iv) = (x - u) + i(y - v)

is We can also easily multiply a complex number by a real number: v k(x + iy) =kx + iky

e Example: If z = 4 + 2i and w = 3 ? i, then

R z + w = (4 + 2i) + (3 ? i) = 7 + i

ths z ? w = (4 + 2i) ? (3 ? i) = 1 + 3i (being careful with the negative signs!)

a 3z + 2w = 3(4 + 2i) + 2(3 ? i) = (12 + 6i) + (6 ? 2i) = 18 + 4i

M 2w ? z* = 2(3 ? i) ? (4 ? 2i) = (6 ? 2i) ? (4 ? 2i) = 2

's 2.2 Multiplying r Complex numbers can be multiplied using the general method for expanding brackets.

a Examples: m (2 + 5i)(4 ? 3i) = 8 ? 6i + 20i ? 15i2

= 8 ? 6i + 20i ? 15(-1)

Ku= 23 + 14i

Remember: i2 = -1

(3 + 2i)2 = (3 + 2i)(3 + 2i) = 9 + 6i + 6i + 4i2

= 9 + 12i + 4(-1)

= 5 + 12i

Further Pure 1

Complex Numbers

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2.3 Dividing

To divide complex conjugates, you multiply through by the complex conjugate of the denominator:

Example: If z = 3 ? i and w = 1 ? 2i, then

z ? w =3 - i 1- 2i

= 3 - i ? 1+ 2i 1 - 2i 1 + 2i

(as the complex conjugate of w is w* = 1 + 2i)

=

3 + 6i - i - 2i2 1 + 2i - 2i - 4i2

=

3 + 5i - 2(-1) 1 - 4(-1)

(as i2 = -1)

= 5 + 5i 5

n Therefore: io z ? w =1+ i

is Note: When a complex number is multiplied by its complex conjugate the answer is always purely

a real number.

v To show this, suppose z = x + iy. e Then zz* = (x + iy)(x ? iy) = x2 ? 2ixy + 2ixy ?i2y2

= x2 + y2

R Worked examination question: s a) Express in the form a + ib,

th (i) (3 + i)2

(ii) (2 + 4i)(3 + i).

a b) The quadratic equation z2 - (2 + 4i)z + 8i - 6 =0

M has roots z1 and z2 . 's i) Verify that z1 = 3 + i is a root of the equation.

ii) By considering the coefficients of the quadratic, write down the sum of the roots;

r iii) Explain why z1*, the complex conjugate of z1, is not a root of the quadratic; a iv) Find the other root, z2, in the form a + ib.

m Solution: Ku a) (i)

(3 + i)2 = (3 + i)(3 + i) = 9 + 3i + 3i + i2 = 9 + 6i ? 1 = 8 + 6i

Remember: i2 = -1

(ii) (2 + 4i)(3 + i) = 6 + 2i + 12i + 4i2

= 6 + 14i ? 4 = 2 + 14i

b) (i) We substitute z = 3 + i into the expression z2 - (2 + 4i)z + 8i - 6 to check that it gives

an answer of 0.

(3 + i)2 ? (2 + 4i)(3 + i) + 8i - 6

= 8 + 6i ? (2 + 14i) + 8i ? 6

(using the results from (a))

= 8 + 6i ? 2 ? 14i + 8i ? 6

= 0 + 0i = 0 (as required)

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Complex Numbers

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(ii) In a quadratic equation, the sum of roots is given by the expression ?b/a.

So here the sum of the roots is (2 + 4i)

(since b = -(2 + 4i) and a = 1)

(iii) In a quadratic equation with real coefficients, any complex roots form a conjugate pair. However, this quadratic does not have real coefficients so the roots are not complex conjugates of each other.

(iv) From (ii), z1 + z2 = 2 + 4i So 3 + i + z2 = 2 + 4i i.e. z2 = 2 + 4i ? (3 + i) = -1 + 3i.

Examination Question:

Given that z = 1+ i , find z in the form a + ib. 1- 2i

ths Revision Examination Question 2:

z1 =-3 + 4i

z2 = 1 + 2i

Kumar's Ma Express

z1z2

and

z1 z 2

each in the form a + ib, where

a,b .

Further Pure 1

Complex Numbers

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Examination Question 3:

The complex numbers z and w are such that

z =-2 + 5i

zw= 14 + 23i

Find w in the form p + qi, where p and q are real.

ion Examination Question 4:

Given that z =-2 + 2 3i , show that z2 + 4z is real.

aths Revis Examination Question 5 M (a) Show that (3 ? i)2 = 8 ? 6i.

r's (b) The quadratic equation

az2 + bz + 10i = 0,

where a and b are real, has a root 3 ? i.

a (i) Show that a = 3 and find the value of b.

(ii) Determine the other root of the equation, giving your answer in the form p + iq.

Kum (Hint for (ii): Use the fact that the sum of the roots in any quadratic is ?b/a).

Further Pure 1

Complex Numbers

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2.4 Equivalence of two complex numbers

Two complex numbers are equal to each other if and only if their real parts are equal and their imaginary parts are equal, i.e

If z = x + iy and w = u + iv, then z = w if and only if x = u and y = v.

2.5 Solving linear equations with complex coefficients

Linear equations can be solved by substituting z = x + iy.

Example 1: Solve 4z ? 2 + 5i = 6 ? 7i

Solution: Let z = x + iy.

Then:

4(x + iy) ? 2 + 5i = 6 ? 7i

n 4x + 4iy ? 2 + 5i = 6 ? 7i

(4x ? 2) + (4y + 5)i = 6 ? 7i

io Therefore, comparing real and imaginary parts:

4x ? 2 = 6 i.e. x = 2

is 4y + 5 = -7 i.e. y = -3

(expanding bracket)

v So the solution to the original equation must be z = 2 ? 3i.

e Example 2: Find z when 2z - 5z* = 9 + 14i.

R Solution: Let z = x + iy. Then z* = x ? iy. s So:

th 2(x + iy) ? 5(x ? iy) = 9 + 14i

2x + 2iy ? 5x + 5iy = 9 + 14i

a i.e. -3x + 7iy = 9 + 14i.

Comparing real and imaginary parts, we see that x = -3 and y = 2.

M So the solution is z = -3 + 2i.

r's Example 3: Solve (4 + 2i)z + (3 ? 2i) = 9 ? 4i

a Solution: Write z = x + iy.

m Then

(4 + 2i)(x + iy) + (3 - 2i) =9 - 4i

u So

(4x + 4iy + 2ix + 2i2 y) + (3 - 2i) =9 - 4i

K i.e.

(4x - 2 y + 4iy + 2ix) + (3 - 2i) =9 - 4i

(using i2 = -1)

Collecting the real and imaginary terms together on the left hand side:

(4x - 2 y + 3) + i(4 y + 2x - 2) =9 - 4i

Comparing real and imaginary parts on both sides, we get the equations:

4x ? 2y + 3 = 9

i.e. 4x ? 2y = 6 or 2x ? y = 3 (1)

and

4y + 2x ? 2 = -4

i.e. 2x + 4y = -2 or x + 2y = -1 (2)

Equations (1) and (2) can be solved simultaneously:

5x = 5

(2?(1) + (2))

i.e. x = 1

From equation (1), y = 2x ? 3 = 2 ? 3 = -1.

Further Pure 1

Complex Numbers

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