Further complex numbers - PowerPoint Presentation

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Further complex numbers - PPT Presentation

Introduction This chapter extends on what you have learnt in FP1 You will learn how to find the complex roots of numbers You will learn how to use De Moivres theorem in solving equations You will see how to plot the loci of points following a rule on an ID: 579207

numbers complex locus argand complex numbers argand locus plane points diagram form point argument modulus theorem moivre

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Slide1

Further complex numbersSlide2

IntroductionThis chapter extends on what you have learnt in FP1You will learn how to find the complex roots of numbersYou will learn how to use De Moivre’s theorem in solving equationsYou will see how to plot the loci of points following a rule on an

Argand diagramYou will see how to solve problems involving transforming a set of values in one plane into another planeSlide3

Teachings for exercise 3ASlide4

Further complex numbersYou can express a complex number in the form z = r(cosθ + isinθ)

You should hopefully remember the modulus-argument form of a complex number z = x + iy from FP1

The value r is the modulus of the complex number, its distance from the origin (0,0)

The argument is the angle the complex number makes with the positive x-axis, where:-π

< θ ≤ π

To show this visually…

z (

x,y

)

r is the

modulus

of z, its absolute value

 This can be calculated using Pythagoras’ Theorem

By GCSE trigonometry, length x =

rcos

and length y =

rsin

Replace x and y using the values above

Factorise

by taking out r

Slide5

Further complex numbersYou can express a complex number in the form z = r(cosθ + isinθ

)Express the following complex number in the modulus-argument form:

z = -√3 + i

To do this you need to find both the argument and the modulus of the complex number

 Start by sketching it on an Argand diagram

√3

Pay attention to the directions

The ‘x’ part is negative so will go in the negative direction horizontally

The ‘y’ part is positive so will go upwards

 Once sketched you can then find the modulus and argument using GCSE

ythagoras and Trigonometry

Calculate

Inverse Tan

Subtract from

Remember that the argument is measured from the

positive

x-axis!

This is the argument

Replace r and

θSlide6

Further complex numbersYou can express a complex number in the form z = r(cosθ + isinθ)

Express the following complex number in the modulus-argument form:z = -√3 + i

To do this you need to find both the argument and the modulus of the complex number

 Start by sketching it on an Argand diagram

Remember that the argument is

not

unique

We could add 2

to them and the result would be the same

, because 2

radians is a complete turnSlide7

Further complex numbersYou can express a complex number in the form z = r(cosθ + isinθ)

Express the following complex number in the modulus-argument form:z = 1 - i

To do this you need to find both the argument and the modulus of the complex number

 Start by sketching it on an Argand diagram

Pay attention to the directions

The ‘x’ part is positive so will go in the positive direction horizontally

The ‘y’ part is negative so will go downwards

 Once sketched you can then find the modulus and argument using GCSE

ythagoras and Trigonometry

Calculate

Inverse Tan

Negative

as below the x-axis

Replace r and

θSlide8

Further complex numbersYou can express a complex number in the form z = reiθIn chapter 6 you will meet series expansions of cos

θ and sinθ

This can be used to prove the following result (which we will do when we come to chapter 6)If z = x +

iy then the complex number can also be written in this way:z =

reiθ

As before, r is the modulus of the complex number and

is the argument

 This form is known as the ‘exponential form’

Slide9

Further complex numbersYou can express a complex number in the form z = reiθExpress the following complex number in the form re

iθ, where -

π < θ ≤ π

z = 2 – 3i

As with the modulus-argument form, you should start by sketching an Argand diagram and use it to find r and

Pay attention to the directions

The ‘x’ part is positive so will go in the positive direction horizontally

The ‘y’ part is negative so will go downwards

 Once sketched you can then find the modulus and argument using GCSE

ythagoras and Trigonometry

Calculate

Inverse Tan

Negative as below the x-axis

Replace r and

θSlide10

Further complex numbersYou can express a complex number in the form z = reiθIn Core 2, you will have seen the following:

cos(-θ) =

cosθ

sin(-θ

) = -sinθ

-1

90º

180º

270º

90º

180º

270º

-90º

-180º

-360º

-1

-270º

y = sin

y = cos

You can see that

cos

) =

cos

anywhere on the graph

You can see that sin(-

) = -sin

anywhere on the graph

-90º

-270º

-360º

-180ºSlide11

Further complex numbersYou can express a complex number in the form z = reiθExpress the following in the form z =

reiθ

where –π < θ ≤

You can see from the form that r = √2

You can see from the form that

Replace r and

Slide12

Further complex numbersYou can express a complex number in the form z = reiθExpress the following in the form z =

reiθ

where –π < θ ≤

We need to adjust this first

The sign in the

centre

is negative, we need it to be positive for the ‘rules’ to work

We also need both angles to be identical. In this case we can apply the rules we saw a moment ago…Slide13

Further complex numbers

You can express a complex number in the form z = reiθ

Express the following in the form z =

reiθ where –

≤

Apply

cos

)

Apply sin(-

) = -sin(

)

You can see from the form that r =

You can see from the form that

= -

Replace r and

Slide14

Further complex numbersYou can express a complex number in the form z = reiθExpress the following in the form

z = x + iy where

and

You can see from the form that r = √2

You can see from the form that

Replace r and

You can calculate all of this! Leave the second part in terms of

This means that x and y have to be

real

numbers (

not complex)Slide15

Further complex numbersYou can express a complex number in the form z = reiθExpress the following in the form r(cos

θ + isinθ), where –

π < θ ≤

You can see from the form that r = 2

You can see from the form that

The value of

is not in the range we want. We can keep subtracting 2

until it is!

Subtract 2

Replace r and

θSlide16

Further complex numbersYou can express a complex number in the form z = reiθUse:

To show that:

Let

= -

Use the relationships above to rewrite

Add 1 and 2

Divide by 2Slide17

Teachings for exercise 3BSlide18

Further complex numbersYou need to know how multiplying and dividing affects both the modulus and argument of the resulting complex numberTo be able to do this you need to be able to use the identities for sine and cosine of two angles added or subtracted – you will have seen these in Core 33B

Multiplying a complex number z

by another complex number z

, both in the

modulus-argument

form

So when multiplying two complex numbers in the modulus-argument form:

Multiply the moduli

Add the arguments together

The form of the answer is the same

Rewrite

Now you can expand the double bracket as you would with a quadratic

Group terms using the identities to the left

 You can also

factorise

the ‘

’ out

Slide19

Multiplying a complex number z

by another complex number z

, both in the

exponential

form

Rewrite

 Remember you add the powers in this situation

You can

factorise

the power

You can see that in this form the process is essentially the same as for the modulus-argument form:

Multiply the moduli together

Add the arguments together

The answer is in the same form

Slide20

Dividing a complex number z

by another complex number z

, both in the

modulus-argument

form

Multiply to

cancel terms

the denominator

Multiply out

Remove i

Group real and complex

Rewrite terms

Rewrite (again!)

So when dividing two complex numbers in the modulus-argument form:

Divide the moduli

Subtract the arguments

The form of the answer is the same

Slide21

Dividing a complex number z

by another complex number z

, both in the

exponential

form

Rewrite terms

 The denominator can be written with a negative power

Multiplying so add the powers

Factorise

the power

You can see that in this form the process is essentially the same as for the modulus-argument form:

Divide the moduli

Subtract the arguments

The answer is in the same form

Slide22

Further complex numbersYou need to know how multiplying and dividing affects both the modulus and argument of the resulting complex numberExpress the following calculation in the form x + iy:

Combine using one of the rules above

Multiply the moduli

Add the arguments

Simplify terms

Calculate the

cos

and sin parts (in terms of

where needed)

Multiply outSlide23

Further complex numbersYou need to know how multiplying and dividing affects both the modulus and argument of the resulting complex numberExpress the following calculation in the form x + iy:

cos

(-θ) = cosθ

sin(-θ

) = -sinθ

The

cos

and sin terms must be

added

for this to work!

 Rewrite using the rules you saw in 3A

Combine using a rule from above

Simplify

Calculate the

cos

and sin parts

Multiply outSlide24

Further complex numbersYou need to know how multiplying and dividing affects both the modulus and argument of the resulting complex numberExpress the following calculation in the form x + iy:

Combine using one of the rules above

Divide the moduli

Subtract the arguments

Simplify

You can work out the sin and

cos

parts

Multiply outSlide25

Teachings for exercise 3CSlide26

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

Let:z

= r(cosθ + isin

)

Use the modulus-argument form

Multiply the moduli, add the arguments

Use the modulus-argument form

Multiply the moduli, add the arguments

Use the modulus-argument form

Multiply the moduli, add the arguments

This is De

Moivre’s

Theorem

 You need to be able to

prove

this

Moivre

= ‘De

Mwavre

’ (pronunciation)Slide27

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n =

rn(cos(nθ

+ isinnθ) for any integer n

De Moivre’s theorem can be proved using the method of proof by induction from FP1

Basis – show the statement is true for n = 1

Assumption

– assume the statement is true for n = k

Inductive

– show that if true for n = k, then the statement is also true for n = k + 1

Conclusion

– because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n

BASIS

Proving that:

s true for all positive integers

 Show that the statement is true for n = 1

Sub in n = 1

Simplify each side

ASSUMPTION

 Assume that the statement is true for n = k

Replace n with kSlide28

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n =

rn(cos(nθ

+ isinnθ) for any integer n

De Moivre’s theorem can be proved using the method of proof by induction from FP1

Basis – show the statement is true for n = 1

Assumption

– assume the statement is true for n = k

Inductive

– show that if true for n = k, then the statement is also true for n = k + 1

Conclusion

– because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n

INDUCTIVE

Proving that:

s true for all positive integers

 Show that if true for n = k, the statement is also true for n = k + 1

Sub in n = k + 1

You can write this as two separate parts as the powers are added together

We can rewrite the first part based on the assumption step, and the second based on the basis step

Using the multiplication rules from 3B

 Multiply the moduli, add the arguments

Slide29

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n =

rn(cos(nθ

+ isinnθ) for any integer n

De Moivre’s theorem can be proved using the method of proof by induction from FP1

Basis – show the statement is true for n = 1

Assumption

– assume the statement is true for n = k

Inductive

– show that if true for n = k, then the statement is also true for n = k + 1

Conclusion

– because the statement is true for n = 1 and also true if any value is, then the statement is true for all values of n

CONCLUSION

Proving that:

s true for all positive integers

Explain why this shows it is true…

We showed the statement is true for n = 1

We then assumed the following:

Using the assumption, we showed that:

As all the ‘k’ terms have become ‘k + 1’ terms, if the statement is true for one term, it must be true for the next, and so on…

 The statement was true for 1, so must be true for 2, and therefore 3, and so on…

 We have therefore proven the statement for all positive integers!Slide30

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

We have just proved the theorem for n = k where k is a positive integer

Now we need to show it is also true for any negative integer…If n is a negative integer, it can be written as ‘-m’, where m is a positive integer

Write using a positive power instead

Use De

Moivre’s

theorem for a positive number (which we have proved)

Multiply to change some terms in the fraction

Multiply out like quadratics – the bottom is the difference of two squares

= -1

You can cancel the denominator as it is equal to 1

Use

cos

) =

cos

(

) and sin(-

) = -sin

You can see that the answer has followed the same pattern as De

Moivre’s

theorem!Slide31

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

Having now proved that De Moivre’s theorem works for both positive and negative integers, there is only one left

We need to prove it is true for 0!

This is straightforward. As it is just a single value, we can substitute it in to see what happens…

Sub in n = 0

Left side = 1 as anything to the power 0 is 1

 You can find cos0 and sin 0 as well

‘Calculate’

So we have shown that De

Moivre’s

Theorem is true for all positive integers, all negative integers and 0’

 It is therefore true for all integers!Slide32

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

It is important to note that De Moivre’s theorem can also be used in exponential form.

Both parts will be raised to the power ‘n’

You can remove the bracket!

This is De

Moivre’s

theorem in exponential form!

Slide33

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

Simplify the following:

The denominator has to have the ‘+’ sign in the middle

 Apply

cos

) =

cos

and sin(-

) = -sin

Apply De

Moivre’s

theorem (there is no modulus value to worry about here!)

 Just multiply the arguments by the power

Apply the rules from 3B for the division of complex numbers

 Divide the moduli and subtract the arguments

Simplify the sin and

cos

terms

Calculate the sin and

cos

terms

SimplifySlide34

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

Express the following in the form x + iy

where x Є R and y Є R

You need to write this in one of the forms above, and you can then use De

Moivre’s

theorem

This is easier than raising a bracket to the power 7!

Start with an

argand

diagram to help find the modulus and argument of the part in the bracket

√3

Pay attention to the directions

The ‘x’ part is positive so will go in the positive direction horizontally

The ‘y’ part is positive so will go upwards

Calculate

Inverse Tan

Sub in r and

Slide35

Further complex numbersYou need to be able to prove that [r(cosθ + isinθ)]n = r

n(cos(nθ

+ isinnθ) for any integer n

Express the following in the form x + iy

where x Є R and y Є R

You need to write this in one of the forms above, and you can then use De

Moivre’s

theorem

This is easier than raising a bracket to the power 7!

Start with an

argand

diagram to help find the modulus and argument of the part in the bracket

Rewrite using the different form we worked out before

Use De

Moivre’s

Theorem

as above

Calculate the

cos

and sin parts

Multiply out and simplifySlide36

Teachings for exercise 3DSlide37

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesThis involves changing expressions involving a function of θ into one without.

For example changing a cos6θ

into powers of cosθ

You will need to use the binomial expansion for C2 in this section

Remember

is a function you can find on your calculator

The first term has the full power of n

 As you move across you slowly swap the powers over to the second term until it has the full power of n

For example:

Follow the pattern above

You can work out the

parts

Slide38

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesExpress cos3θ using powers of

cosθ.

This type of question involves making a comparison between two processesOne which will give you a ‘cos3

θ’ term – you will use De Moivre’s Theorem for this

One which will give you an expression in terms of cos

– you will use the binomial expansion for this

You have to think logically and decide where to start

If we apply De

Moivre’s

theorem to this, we will end up with a ‘cos3

’ term

If we apply the binomial expansion to it, we will end up with some terms with

cos

So this expression is a good starting point!Slide39

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesExpress cos3θ using powers of

cosθ.

This type of question involves making a comparison between two processesOne which will give you a ‘cos3

θ’ term – you will use De Moivre’s Theorem for this

One which will give you an expression in terms of cos

– you will use the binomial expansion for this

You have to think logically and decide where to start

Apply De

Moivre’s

theorem

Follow the rules you know

Apply the Binomial expansion

Write out

‘Tidy up’

Replace i

parts with -1

The two expressions we have made must be equal

Therefore the real parts in each and the imaginary parts in each must be the same

Equate the real parts

Slide40

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesExpress cos3θ using powers of

cosθ.

This type of question involves making a comparison between two processesOne which will give you a ‘cos3

θ’ term – you will use De Moivre’s Theorem for this

One which will give you an expression in terms of cos

– you will use the binomial expansion for this

You have to think logically and decide where to start

Replace sin

with an expression in cos

Expand the bracket

Simplify

We have successfully expressed cos3

as posers of

cos

!Slide41

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesExpress the following as powers of cos

θ:

So we need something that will give us sin6θ using De

Moivre’s theorem

It also needs to give us terms of cosθ from the binomial expansion

If we apply De

Moivre’s

theorem to this, we will end up with a ‘sin6

’ term

If we apply the binomial expansion to it, we will end up with some terms with

cos

So this expression is a good starting point!

(and yes you will have to do some expansions larger than powers of 3 or 4!)Slide42

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesExpress the following as powers of cos

θ:

Apply De

Moivre’s

theorem

Follow the rules you know

Apply the Binomial expansion

Replace terms:

and i

= -1

= 1

So the two expressions created from De

Moivre

and the Binomial Expansion must be equal

The real parts will be the same, as will the imaginary parts

This time we have to equate the

imaginary

parts as this has sin6

Divide all by

iSlide43

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesExpress the following as powers of cos

θ:

Divide all terms by sin

Replace sin

terms with (1 – cos

) terms

Expand the first bracket, square the second

Expand the second bracket

Group together the like terms

So we have changed the expression we were given into powers of

cos

!Slide44

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesYou also need to be able to work this type of question in a different way:

For example, you might have a power or cos or sin and need to express it using several linear terms instead

Eg) Changing sin6θ to

sinaθ + sinbθ

where a and b are integers

 To do this we need to know some other patterns first!

Let:

Write as ‘1 over’



or with a power of -1

Use De

Moivre’s

theorem

Use

cos

) =

cos

and sin(-

) = -sin

We can add our two results together:

SimplifySlide45

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesYou also need to be able to work this type of question in a different way:

For example, you might have a power or cos or sin and need to express it using several linear terms instead

Eg) Changing sin6θ to

sinaθ + sinbθ

where a and b are integers

 To do this we need to know some other patterns first!

Let:

Write as ‘1 over’



or with a power of -1

Use De

Moivre’s

theorem

Use

cos

) =

cos

and sin(-

) = -sin

We could also subtract our two results:

Simplify

Slide46

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesYou also need to be able to work this type of question in a different way:

For example, you might have a power or cos or sin and need to express it using several linear terms instead

Eg) Changing sin6θ to

sinaθ + sinbθ

where a and b are integers

To do this we need to know some other patterns first!

You can also apply the rules we just saw to powers of z

Let:

Write as ‘1 over’



or with a power of -1

Use De

Moivre’s

theorem

Use

cos

) =

cos

and sin(-

) = -sin

We could add our two results together:

Simplify

Slide47

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesYou also need to be able to work this type of question in a different way:

For example, you might have a power or cos or sin and need to express it using several linear terms instead

Eg) Changing sin6θ to

sinaθ + sinbθ

where a and b are integers

To do this we need to know some other patterns first!

You can also apply the rules we just saw to powers of z

Let:

Write as ‘1 over’



or with a power of -1

Use De

Moivre’s

theorem

Use

cos

) =

cos

and sin(-

) = -sin

We could also subtract our two results:

Simplify

Slide48

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesLet’s now see how we can use these ‘patterns’ in solving problems:

Express cos5θ in the form a

cos5θ + bcos3

θ + ccosθ

Where a

and

are constants to be found.

When working this way round you need to use the identities above to express both cos

θ and terms with cos5

θ, cos3

θ and cosθ

You can then set the expressions equal to each other

Creating a cos

term

Creating the other cos terms – use the Binomial expansion!

Use the B.E.

Cancel some z terms

Using the Identity above

Group up terms with the same power

Rewrite using an identity above

SimplifySlide49

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesLet’s now see how we can use these ‘patterns’ in solving problems:

Express cos5θ in the form a

cos5θ + bcos3

θ + ccosθ

Where a

and

are constants to be found.

When working this way round you need to use the identities above to express both cos

θ and terms with cos5

θ, cos3

θ and cosθ

You can then set the expressions equal to each other

Using the two expressions

These two expressions must be equal to each other

Divide both sides by 32

So we have written cos

using cos5

, cos3

and cos

θSlide50

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesShow that:

 This is similar to the previous question. You need to use rules above to find a way to create a sin3

θ expression, and an expression containing sin3θ

and sinθ

Creating a sin

term

Creating the other sin terms – use the Binomial expansion!

Use the B.E.

Cancel some z terms

Using the Identity above

Group up terms with the same power

Rewrite using an identity above

Simplify

Slide51

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesShow that:

 This is similar to the previous question. You need to use rules above to find a way to create a sin3

θ expression, and an expression containing sin3θ

and sinθ

Using the two expressions

These two expressions must be equal to each other

Divide both sides by

So we have written sin

using sin3

and sin

Divide both sides by -8

Slide52

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesa) Express sin4θ

in the form:Where d, e and f are constants to be found.

b) Hence, find the exact value of the following integral:

 Start exactly as with the previous questions, by finding an expression with sin

4θ and one with cos4θ

, cos2

and a number

Creating a sin

term

Creating the

cos

terms – use the Binomial expansion!

Use the B.E.

Cancel some z terms

Using the Identity above

Group up terms with the same power (use positive values in the brackets so we get

cos

terms)

Replace using an identity above

SimplifySlide53

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesa) Express sin4θ

in the form:Where d, e and f are constants to be found.

b) Hence, find the exact value of the following integral:

 Start exactly as with the previous questions, by finding an expression with sin

4θ and one with cos4θ

, cos2

and a number

Using the two expressions

These two expressions must be equal to each other

Divide both sides by 16

So we have written sin

using cos4

and cos2

Slide54

Further complex numbersYou can apply De Moivre’s theorem to trigonometric identitiesa) Express sin4θ

in the form:Where d, e and f are constants to be found.

b) Hence, find the exact value of the following integral:

 Start exactly as with the previous questions, by finding an expression with sin

4θ and one with cos4θ

, cos2

and a number

Replace with an equivalent expression

Cosine Integrals (in C4)

Integrate each term with respect to

, using knowledge from C4

Sub in limits

Work outSlide55

Teachings for exercise 3ESlide56

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberYou already know how to find real roots of a number, but now we need to find both real roots and imaginary roots!

We need to apply the following results:If:

Then: where k is an integer

This is because we can add multiples of 2π to the argument as it will end up in the same place (2

π = 360º)2) De

Moivre’s

theorem

Slide57

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberSolve the equation z3 = 1 and represent your solutions on an

Argand diagram.First you need to express z in the modulus-argument form. Use an

Argand diagram.

Now we know r and θ we can set z

3 equal to this expression, when written in the modulus-argument form

We can then find an expression for z in terms of k

We can then solve this to find the roots of the equation above

In this case the modulus and argument are simple to find!

Apply the rule above

Cube root (use a relevant power)

Apply De

Moivre’s

theoremSlide58

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberSolve the equation z3 = 1 and represent your solutions on an

Argand diagram.We now just need to choose different values for k until we have found all the roots

The values of k you choose should keep the argument within the range:

-π < θ ≤

π3E

k = 0

k = 1

k = -1

Sub k = 0 in and calculate the cosine and sine parts

Sub k = 1 in and calculate the cosine and sine parts

Sub k = -1 in and calculate the cosine and sine parts



(k = 2 would cause the argument to be outside the range)

So the roots of z

= 1 are:

Slide59

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberSolve the equation z3 = 1 and represent your solutions on an

Argand diagram.We now just need to choose different values for k until we have found all the roots

The values of k you choose should keep the argument within the range:

-π < θ ≤

π3E

So the roots of z

= 1 are:

The solutions will all the same distance from the origin

The angles between them will also be the same

The sum of the roots is always equal to 0Slide60

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberSolve the equation z4

- 2√3i = 2Give your answers in both the modulus-argument and exponential forms.

By rearranging…z4

= 2 + 2√3iAs before, use an

argand diagram to express the equation in the modulus-argument form

Then choose values of k until you have found all the solutions

Find the modulus and argument

Apply the rule above

2√3

Take the 4

root of each side

Moivre’s

Theorem

Work out the power at the frontSlide61

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberSolve the equation z4

- 2√3i = 2Give your answers in both the modulus-argument and exponential forms.

By rearranging…z4

= 2 + 2√3iAs before, use an

argand diagram to express the equation in the modulus-argument form

Then choose values of k until you have found all the solutions

k = 0

Sub k = 0 in and simplify (you can leave in this form)

k = 1

k = -1

k = -2

Choose values of k that keep the argument between –

and

πSlide62

Further complex numbersYou can use De Moivre’s theorem to find the nth roots of a complex numberSolve the equation z4

- 2√3i = 2Give your answers in both the modulus-argument and exponential forms.

By rearranging…z4

= 2 + 2√3iAs before, use an

argand diagram to express the equation in the modulus-argument form

Then choose values of k until you have found all the solutions

Solutions in the modulus-argument form

Solutions in the exponential form

Slide63

Teachings for exercise 3FSlide64

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramA locus a set of points which obey a rule

 You will need to be able to understand Loci based on Argand diagrams

The locus of points a given distance from a point O is a circle

The locus of points equidistant from two fixed points A and B is the perpendicular bisector of line AB

BSlide65

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramz = x + iy

represents a variable point P(x,y) on an Argand diagram

z1 = x

1 + iy1 represents a fixed point A(x

1,y1) on an

Argand

diagram

What is represented by:

 It represents the distance between the fixed point A(x

) and the variable point P(

x,y)

x,y

)

A(x

)

z - z

If we want to get from the fixed point A to the variable point P, we need to travel back along z1

and then out along z(-z1 + z)

This can be written as a vector, z – z1

So |z – z1| represents the distance between the fixed point and the variable point!

Is the distance between the variable point z and the fixed point z

when they are represented on a Argand diagramSlide66

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y) which is represented by z on an Argand

diagram3F

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Leave z as it is – this is the variable point

Put this part in a bracket

- This is the fixed point

So we want the locus where the distance between the variable point z and the fixed point (5,3) is equal to 3

A(5,3)

x,y

)

This will be a circle of radius 3 units, centre (5,3)Slide67

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Use an algebraic method to find a Cartesian equation of the locus of z

So you have to do this without using the graph you drew

We will quickly remind ourselves of something that will be useful for this!

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

x,y

)

If:

Then:

(By Pythagoras’ Theorem)

|z|

Slide68

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Use an algebraic method to find a Cartesian equation of the locus of z

Now we can find the equation of the locus algebraically…

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Replace z with ‘x +

’

Group the real and imaginary terms

Use the rule above to remove the modulus

Square both sides

You (hopefully) recognise that this is the equation of a circle, radius 3 and with centre (5,3)!Slide69

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagram

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

As a general rule, the locus of:

a circle of radius r and

centre

) where z

= x1

+ iy1Slide70

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramGive a geometrical interpretation of each of the following loci of z:

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

 Circle radius r and centre (x

)

 Circle, centre (0,3) radius 4

 Circle, centre (2,3) radius 5

 Circle, centre (-3,5) radius 2

Put the ‘fixed’ part in a bracket

‘Factorise’ the part inside the modulus

You can write this as 2 moduli multiplied

|-1| = 1, put the ‘fixed’ part in a bracket

 Circle, centre (2,-5) radius 3

Effectively for d), you just swap the signs of everything in the modulus, its value will not change



|10 - 8| = |-10 + 8|Slide71

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramSketch the locus of P(x,y) which is represented by z on an

Argand diagram, if:

We therefore need the set of points that are the same distance from (0,0) and (0,6)

This will be the bisector of the line joining the two co-ordinatesYou can see that it is the line with equation y = 3

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

This is the distance of P(

x,y

) from the origin (0,0)

This is the distance of P(

x,y

) from (0,6)

(0,0)

(0,6)

y = 3Slide72

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramSketch the locus of P(x,y) which is represented by z on an

Argand diagram, if:

Show that the locus is y = 3 using an algebraic method

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Replace z with x +

Factorise the ‘

’ terms on the right side

Use the rule above to remove the moduli

Square both sides

Expand the bracket

Cancel terms on each side

Add 12y

Divide by 12Slide73

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramUse an algebraic method to find the Cartesian equation of the locus of z if:

Represent the locus of z on an

Argand diagram

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Replace z with x +

Group real and imaginary parts

Use the rule above to remove the moduli

Square both sides

Expand brackets

Cancel terms

Subtract 1

Divide by 2Slide74

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramUse an algebraic method to find the Cartesian equation of the locus of z if:

Represent the locus of z on an

Argand diagram

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

(0,-1)

(3,0)

y = -3x + 4

Distance from (3,0)

Distance from (0,-1)

(0,4)Slide75

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Use algebra to show that the locus of z is a circle, stating its centre and radius

b) Sketch the locus of z on an Argand diagram

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Replace z with ‘x +

’

Group real and imaginary parts

Replace the moduli using the rule above

Square both sides (remember the ‘2’)

Expand some brackets

Expand another bracket

Group all terms on one side

Divide by 3

Completing the square

Simplify

Add 100

Circle, centre (-10,12) and radius 10Slide76

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Use algebra to show that the locus of z is a circle, stating its centre and radius

b) Sketch the locus of z on an Argand diagram

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Circle, centre (-10,12) and radius 10

(-10,12)

The distance from (6,0)

Is equal to

Twice the distance from (-6,9)

(6,0)

(-6,9)

x,y

)

The circle shows the set of points that are twice as far from (6,0) as they are from (-6,9)! Slide77

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Use algebra to show that the locus of z is a circle, stating its centre and radius

b) Sketch the locus of z on an Argand diagram

Is the distance between the variable point z and the fixed point z

when they are represented on a

Argand

diagram

Circle, centre (-10,12) and radius 10

Therefore:

When



An Algebraic method will most likely be the best way to find the equation of the locus of z



You will probably have to use completing the square (sometimes with fractions as well!)Slide78

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y) which is represented by z on an Argand

diagram. Then find the Cartesian equation of this locus algebraically. The locus will be the set of points which start at (0,0) and make an argument of

π/4

with the positive x-axis

The line is

not

extended back downwards

 It is known as a ‘half-line’Slide79

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y) which is represented by z on an Argand

diagram. Then find the Cartesian equation of this locus algebraically. The locus will be the set of points which start at (0,0) and make an argument of

π/4

with the positive x-axis

Replace z with ‘x +

’

The value of the argument is tan

-1

(

opposite

adjacent

)

‘Normal tan’

Calculate the tan part

Multiply by x

(x > 0)Slide80

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y) which is represented by z on an Argand

diagram. Then find the Cartesian equation of this locus algebraically.The locus will be the set of values that,

when we subtract 2 from them, make an angle of π/

3 with the origin

The locus must therefore start at (2,0) rather than (0,0)!

(2,0)Slide81

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y) which is represented by z on an Argand

diagram. Then find the Cartesian equation of this locus algebraically.The locus will be the set of values that,

when we subtract 2 from them, make an angle of π/

3 with the origin

The locus must therefore start at (2,0) rather than (0,0)!

(2,0)

x - 2

Replace z with ‘x +

’

The value of the argument is tan

-1

(

opposite

adjacent

)

‘Normal tan’

Calculate the tan part

Multiply by (x – 2)

(x > 2)Slide82

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line.

When we add 3 and 2i to z, the argument from (0,0) and the positive x-axis will be 3

π/4

So the line will have to start at (-3,-2)

(-3,-2)Slide83

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line.

When we add 3 and 2i to z, the argument from (0,0) and the positive x-axis will be 3

π/4

So the line will have to start at (-3,-2)

(-3,-2)

Replace z with ‘x +

’

The value of the argument is tan

-1

(

opposite

adjacent

)

‘Normal tan’

Calculate the tan part

Multiply by (x +

)

(x < -3)

x + 3

y + 2

Subtract 2Slide84

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramSo therefore:

Is represented by a half line starting at z1 and making an angle of

θ with a line parallel to the x-axis

Slide85

Further complex numbers

You can use complex numbers to represent a locus of points on an

Argand diagram

For the next set of Loci, you need to remember some rules relating to circles

Major arc –

is acute

Minor arc –

is obtuse

Semi-circle –

is 90°

Joining the ends of a chord to different points on the circumference will always create the same angle, if the points are in the same sector

“Angles in the same sector are equal”

If they are joined to a point on the

major

arc

 The angle will be

acute

If they are joined to a point on the

minor

arc

 The angle will be

obtuse

If the chord is the

diameter

of the circle

 The angle will be

90°

 The angle at the centre is twice the angle at the circumference

θSlide86

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y), which is represented by z on an

Argand diagram The argument above can be rewritten using this rule:

So what we are doing is drawing the locus of points where the

difference

between these arguments is

Slide87

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y), which is represented by z on an

Argand diagram

So what we are doing is drawing the locus of points where the

difference

between these arguments is

(2,0)

arg

(z – 6) =

arg

(z – 2) =

Imagine drawing both arguments – we will use

and

to represent their values

Using alternate angles, we can show the angle between the arguments is their differenceWe want this difference to be

π/4

θ2θ

This angle must therefore be θ

1 – θ2

, the difference between the arguments!

However, there are more points that satisfy this rule!

(6,0)

1Slide88

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagram

If:

Sketch the locus of P(x,y), which is represented by z on an Argand

diagram

So what we are doing is drawing the locus of points where the

difference

between these arguments is

(6,0)

(2,0)

If we move the point where the lines cross along the major arc of a circle, then the value of

will remain the same

The arguments

will

change but this

doesn’t matter

, it is the difference that matters!

So the locus of a difference between arguments is always given by an arc of a circle

Geogebra

ExampleSlide89

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y), which is represented by z on an

Argand diagramFind the Cartesian equation of this locus

We need the centre of the ‘circle’ and its radius

 We need to use another of the rules we saw:

(6,0)

(2,0)

“The angle at the

centre

is twice the angle at the circumference”

We can use this isosceles triangle to find the information we need…

Centre

Radius

RadiusSlide90

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y), which is represented by z on an

Argand diagramFind the Cartesian equation of this locus

We need the centre of the ‘circle’ and its radius

Centre

Radius

(6,0)

(2,0)

(4,0)

(4,2)

Split the triangle in the middle, the smaller angles will both be

(45

ᵒ

) (because the top angle was

)

The middle of the base will be (4,0), and you can work out the side lengths from this

The top will therefore be at (4,2)

Use Pythagoras’ Theorem to find the diagonal (the radius)

Centre (4,2)

Radius 2√2Slide91

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramIf:

Sketch the locus of P(x,y), which is represented by z on an

Argand diagramFind the Cartesian equation of this locus

We need the centre of the ‘circle’ and its radius

Centre (4,2)

Radius 2√2

(6,0)

(2,0)

The locus is therefore the arc of a circle with the following equation:

Slide92

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramGenerally, for this type of question, you need to follow 3 steps:

Step 1: Mark on the Argand diagram the two points where the arguments start

Step 2: Decide whether the arc is going to be major, minor, or a semi-circle, by considering the angle

Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator pointAnti-clockwise if

θ is positiveClockwise if

negative

If the value we want is positive, then

If the value we want is negative, then

Drawing in the direction indicated in step 3 means you will ensure the arguments are correct to give a positive or negative answer

 As we do some examples we will refer to this!

3FSlide93

Further complex numbers

You can use complex numbers to represent a locus of points on an

Argand

diagram

Generally, for this type of question, you need to follow 3 steps:

Step 1: Mark on the

Argand

diagram the two points where the arguments start

Step 2: Decide whether the arc is going to be major, minor, or a semi-circle, by considering the angle

Step 3: Draw the arc between the points. You always draw from the numerator point to the denominator point

Anti-clockwise if

positive

Clockwise if θ

is negative

Sketch the locus of P(

x,y

) on an

Argand diagram if:

(0,0)

(0,4)

(0,0) and (0,4)

The angle to make is

 A semi-circle

is positive, so draw anti-clockwise

from (0,0) (numerator point) to (0,4) (denominator point)

3FSlide94

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagram

step 3 we had to choose whether to draw the diagram clockwise or anti-clockwise from the numerator point to the denominator point

Lets show why this is correct!

(0,0)

(0,4)

(0,0)

(0,4)

 We drew the angle anti-clockwise from (0,0) to (0,4)

 Using the alternate angles, the angle

between

the arguments is

 However, as

is actually negative, the sum is really

+ (-

)

–

 This angle

therefore what we were wanting!

If we drew the arc the other way - clockwise from (0.0) to (0,4)

Using the alternate angles, the on the

outside

 However, as

is actually negative, the sum is really

+ (-

)

–

But of course it is on the wrong side of the arc so we do not want this part of the circle!

Basically, always use the rule in step 3!

3FSlide95

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramGenerally, for this type of question, you need to follow 3 steps:

Step 1: Mark on the Argand diagram the two points where the arguments start

Step 2: Decide whether the arc is going to be major, minor, or a semi-circle, by considering the angleStep 3: Draw the arc between the points. You always draw from the numerator point to the denominator point

Anti-clockwise if θ is positive

Clockwise if θ is negative

Sketch the locus of P(

x,y

) on an

Argand

diagram if:

(0,-3) and (0,2)

The angle to make is

 A

major

arc

positive

, so draw

anti-clockwise

from (0,-3) (numerator point) to (0,2) (denominator point)

(0,-3)

(0,2)Slide96

Further complex numbersYou can use complex numbers to represent a locus of points on an Argand diagramGiven that the complex number z = x + iy satisfies the equation:

Find the minimum and maximum values of |z|

Start by drawing this on an Argand

diagramIt is a circle,

centre (12,5) radius 3 units

(12,5)

The smallest and largest values for |z| will be on the same straight line through the circle’s centre

You can mark the size of the radius on the diagram

Find the distance from (0,0) to (12,5), then add/subtract 3 to find the largest and smallest values

So the largest value of |z| will be 16 and the smallest will be 10

13Slide97

Teachings for exercise 3GSlide98

Further complex numbersYou can use complex numbers to represent regions on a Argand diagram

This is very similar to what you have been doing with lociThe only extra part is that once you have drawn the locus representing the point, you need to indicate the area required

Shade on an Argand

diagram the region indicated by: Start with a circle, centre (4,2) and radius 2 units (as 2 is the ‘limit’)

The region we want is where the absolute value of z is

less

than 2

 This will be the region

inside

the circle

(4,2)Slide99

Further complex numbersYou can use complex numbers to represent regions on a Argand diagram

This is very similar to what you have been doing with lociThe only extra part is that once you have drawn the locus representing the point, you need to indicate the area required

Shade on an Argand diagram the region indicated by:

 Start with the perpendicular bisector between (4,0) and (6,0) as this is the ‘limit’

(4,0)

(6,0)

The distance to |z – 4| must be

less

than the distance to |z – 6|

 Shade the region

closest

to (4,0)

ySlide100

Further complex numbers

You can use complex numbers to represent regions on a

Argand

diagram

This is very similar to what you have been doing with loci

The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required

Shade on an

Argand

diagram the region indicated by:

 Start by drawing the limits of the argument from the point (2,2)

(2,2)

The argument must be between these two values

 Shade the region

between

the two arguments

Slide101

Further complex numbers

You can use complex numbers to represent regions on a

Argand

diagram

This is very similar to what you have been doing with loci

The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required

Shade on an

Argand

diagram the region indicated by:

and

Imagine all the regions were on the same diagram

 The region we want will have to satisfy all of these at the same time!

Slide102

Teachings for exercise 3HSlide103

Further complex numbersYou can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + ivEffectively we take a set of points in the complex plane, transform them all and map them on a new complex plane

 You will need to use Algebraic methods a lot for this as visualising the transformations can be very difficult!

The z-plane

The w-plane

(uses x and y)

(uses u and v)

Transformation from one plane to the next!Slide104

Further complex numbersYou can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + ivThe point P represents the complex number z on an Argand diagram where |z| = 2.

T1 represents a transformation from the z plane, where z = x +

iy, to the w-plane where w = u + iv.Describe the locus of P under the transformation T

1, when T1 is given by:

 We will work out the new set of points algebraically…

The z-plane

Circle

centre

(0,0), radius 2

 To start with, make z the subject

Add 2, subtract 4i

The modulus of each side must be the same

We know |

z| from

the question

The w-plane

Circle

centre

(-2,4), radius 2

Circle,

centre

(-2,4), radius 2Slide105

T2 represents a transformation from the z plane, where z = x +

iy, to the w-plane where w = u + iv.Describe the locus of P under the transformation T

2, when T2 is given by:

 We will work out the new set of points algebraically…

The z-plane

Circle

centre

(0,0), radius 2

The w-plane

Circle

centre

(0,0), radius 6

Divide by 3

 To start with, make z the subject

Modulus of both sides

|z|= 2

Split the modulus up

|3|=3 so multiply by 3Slide106

T2 represents a transformation from the z plane, where z = x +

iy, to the w-plane where w = u + iv.Describe the locus of P under the transformation T

3, when T3 is given by:

 We will work out the new set of points algebraically…

The z-plane

Circle

centre

(0,0), radius 2

The w-plane

Circle

centre

(0,1), radius 1

 To start with, make z the subject

Subtract

Modulus of both sides

You can split the modulus on the right

|z| = 2

Simplify the right side

Leaving z like this can make the problem easier! (rather than rearranging completely)Slide107

Further complex numbersYou can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + ivFor the transformation w = z2, where z = x +

iy and w = u + iv, find the locus of w when z lies on a circle with equation x2 + y

2 = 16It is very important for this topic that you draw information on z or |z| from the question

The equation x

2 + y2

= 16 is a circle,

centre

(0,0) and radius 4

Therefore |z| = 4

We now proceed as before, by writing the equation linking w and z in such a way that |z| can be replaced

The z-plane

Circle

centre

(0,0), radius 4

The w-plane

Circle

centre

(0,0), radius 16

Modulus of both sides

Split the modulus up

Replace |z| with 4

Calculate

Circle,

centre

(0,0) and radius 16Slide108

Further complex numbersYou can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + ivThe transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by:

Show that the image, under T, of the circle |z| = 1 in the z-plane, is a line l in the w-plane. Sketch l on an

Argand diagram.

Make z the subject!Now eliminate z using what we know…

Multiply by (z + 1)

Expand the bracket

Subtract 5iz and subtract w

Factorise

the left side

Divide by (w – 5i)

Modulus of both sides

|z| = 1

Multiply by |w – 5i|

- w| = |w –

|Slide109

Show that the image, under T, of the circle |z| = 1 in the z-plane, is a line l in the w-plane. Sketch l on an

Argand diagram.

The z-plane

Circle

centre

(0,0), radius 1

The w-plane

Transformation T

Perpendicular bisector between (0,1) and (0,5)



The line v = 3

So a circle can be transformed into a straight line!Slide110

Show that the image, under T, of the circle with equation x

2 + y2 = 4 in the z-plane, is a different circle C in the w-plane.

State the centre and radius of C.

 Remember that x2

+ y

= 4 is the same as |z| = 2

Multiply by (z + 1)

Expand the bracket

Subtract

and add 2

Factorise

the right side

Divide by (3 – w)

Modulus of each side

Split up the modulus

|z| = 2

Multiply by |3 - w|

|3 - w| = |w - 3|

We now need to find what the equation of this will be!Slide111

Show that the image, under T, of the circle with equation x

2 + y2 = 4 in the z-plane, is a different circle C in the w-plane.

State the centre and radius of C.

 Remember that x2

+ y

= 4 is the same as |z| = 2

Replace w with ‘u + iv’

Circle,

centre

(

,0), radius

Group real/imaginary terms

Remove the modulus

We will find the equation as we did in the early part of section 3F!

Expand brackets

Expand more brackets!

Move all to one side

Divide by 3

Use completing the square

Move the number terms acrossSlide112

Show that the image, under T, of the circle with equation x

2 + y2 = 4 in the z-plane, is a different circle C in the w-plane.

State the centre and radius of C.

The z-plane

Circle

centre

(0,0), radius 2

Transformation T

Circle,

centre

(

,0), radius

The w-planeSlide113

Further complex numbersYou can apply transformations that map points on the z-plane to points on the w-plane by applying a formula relating to z = x + iy to w = u + ivA transformation T of the z-plane to the w-plane is given by:

Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an A

rgand diagram. Start by rearranging to make z the subject ‘as usual’

Multiply by (1 – z)

Expand the bracket

Add 2, Add

Factorise

the right side

Divide by (

+ w)

Write the other way round (if you feel it is easier!)

+ w = w +

At this point we have a problem, as we do not know anything about |z|

However, as z lies on the ‘real’ axis, we know that y = 0

Replace z with ‘x +

’

Slide114

Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an A

rgand diagram.Now, you need to rewrite the right side so you can separate all the real and imaginary terms

You must be

extremely careful with positives and negatives here!

Replace w with u + iv

Group real and imaginary terms

Multiply by the denominator but with the opposite sign (this will cancel ‘

’ terms on the bottom

Simplify

= -1

Separate real and ‘

’ terms

As z lies on the x-axis, we know y = 0

 Therefore, the imaginary part on the right side must

also

equal 0Slide115

Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an A

rgand diagram.You can now find an equatio

n for the line in the w-plane

 Set the imaginary part equal to 0

So the transformation has created this line in the w-plane

(remember v is essentially ‘y’ and u is essentially ‘x’)

Multiply by (u

+ (v + 1)

)

(you will be left with the numerator)

Multiply out the double bracket

Subtract all these terms

The ‘

’ terms cancel out

Make v the subject

Divide by 2Slide116

Show that as z lies on the real axis in the z-plane, then w lies on a line l in the w-plane. Find the equation of l and sketch it on an A

rgand diagram.

The z-plane

(z lies on the real axis)

Transformation T

Straight line, gradient is

and y-intercept at (0,-1)

The w-plane

-1

-2

Slide117

SummaryYou have learnt a lot in this chapter!!You have seen proofs of and uses of De Moivre’s theoremYou have found real and complex roots of powersYou have see how to plot Loci and perform transformations of complex functions

Further complex numbers - PowerPoint Presentation

Further complex numbers - PPT Presentation

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