rther complex numbe Introduction This chapter extends on

rther complex numbe

Introduction • This chapter extends on what you have learnt in FP 1 • You will learn how to find the complex roots of numbers • You will learn how to use De Moivre’s theorem in solving equations • You will see how to plot the loci of points following a rule on an Argand diagram • You will see how to solve problems involving transforming a set of values in one plane into another plane

achings for exercise 3

Further complex numbers y You can express a complex number in the form z = r(cosθ + isinθ) z (x, y) r You should hopefully remember the modulus-argument form of a complex number z = x + iy from FP 1 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… y θ x r is the modulus of z, its absolute value x By GCSE trigonometry, length x = rcosθ and length y = rsinθ This can be calculated using Pythagoras’ Theorem Replace x and y using the values above Factorise by taking out r 3 A

Further complex numbers y You can express a complex number in the form z = r(cosθ + isinθ) Express the following complex number in the modulus-argument form: Pay attention to the directions The ‘x’ part is negative so will go in the negative direction 1 horizontally This is the argument r θ x √ 3 The ‘y’ part is positive so will go upwards 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 Replace r and θ Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry Calculate Remember that the argument is measured from the positive x-axis! Inverse Tan Subtract from π 3 A

Further complex numbers You can express a complex number in the form z = r(cosθ + isinθ) Express the following complex number in the modulus-argument form: Remember that the argument is not unique z = -√ 3 + i We could add 2π to them and the result would be the same, because 2π radians is a complete turn 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 A

Further complex numbers y You can express a complex number in the form z = r(cosθ + isinθ) Express the following complex number in the modulus-argument form: 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 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 Replace r and θ 1 θ r x 1 Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry Calculate Inverse Tan Negative as below the x -axis 3 A

Further complex numbers You can express a complex number in the form z = reiθ In chapter 6 you will meet series expansions of cosθ 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 θ is the argument This form is known as the ‘exponential form’ 3 A

Further complex numbers y You can express a complex number in the form z = reiθ Express the following complex number in the form reiθ, where -π < θ ≤ π z = 2 – 3 i As with the modulus-argument form, you should start by sketching an Argand diagram and use it to find r and θ 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 r x 3 Once sketched you can then find the modulus and argument using GCSE Pythagoras and Trigonometry Calculate Replace r and θ 2 θ Inverse Tan Negative as below the xaxis 3 A

Further complex numbers y You can express a complex number in the form z = reiθ In Core 2, you will have seen the following: cos(-θ) = cosθ y = cosθ 1 -360º -270º -θ -180º 0 -90º -θ θ θ 90º 180º 270º θ -1 You can see that cos(-θ) = cosθ anywhere on the graph y sin(-θ) = -sinθ y = sinθ 1 -θ -360º -270º -180º -90º 0 θ 90º 180º 270º θ -1 You can see that sin(-θ) = -sinθ anywhere on the graph 3 A

Further complex numbers You can express a complex number in the form z = reiθ Express the following in the form z = reiθ where –π < θ ≤ π You can see from the form that r = √ 2 You can see from the form that θ = π/10 Replace r and θ 3 A

Further complex numbers You can express a complex number in the form z = reiθ Express the following in the form z = reiθ where –π < θ ≤ π 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… 3 A

Further complex numbers You can express a complex number in the form z = reiθ Apply cosθ = cos(-θ) Apply sin(-θ) = -sin(θ) Express the following in the form z = reiθ where –π < θ ≤ π You can see from the form that θ = -π/8 that r = 5 Replace r and θ 3 A

Further complex numbers • You can see from the form that r = √ 2 This means that x and y have to be real numbers (ie not complex) You can see from the form that θ = 3π/4 Replace r and θ You can calculate all of this! Leave the second part in terms of i 3 A

Further complex numbers You can express a complex number in the form z = reiθ Express the following in the form r(cosθ + isinθ), where –π < θ ≤ π You can see from the form that r = 2 You can see from the form that θ = 23π/5 The value of θ is not in the range we want. We can keep subtracting 2π until it is! Subtract 2π Replace r and θ 3 A

Further complex numbers You can express a complex number in the form z = reiθ Let θ = -θ Use the relationships above to rewrite Use: To show that: 1) 2) Add 1 and 2 Divide by 2 3 A

achings for exercise 3

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To 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 3 Multiplying a complex number z 1 by another complex number z 2, both in the modulus-argument form 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 ‘i’ out So when multiplying two complex numbers in the modulusargument form: Multiply the moduli Add the arguments together The form of the answer is the same 3 B

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To 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 3 Multiplying a complex number z 1 by another complex number z 2, 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 3 B

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To 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 3 Dividing a complex number z 1 by another complex number z 2, both in the modulus-argument form Multiply to cancel terms on the denominator Multiply out Remove i 2 Group real and complex 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 Rewrite terms Rewrite (again!) 3 B

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number To 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 3 Dividing a complex number z 1 by another complex number z 2, 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 3 B

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Combine using one of the rules above Multiply the moduli Add the arguments Express the following calculation in the form x + iy: Simplify terms Calculate the cos and sin parts (in terms of i where needed) Multiply out 3 B

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number The cos and sin terms must be added for this to work! Rewrite using the rules you saw in 3 A Combine using a rule from above Express the following calculation in the form x + iy: Simplify cos(-θ) = cosθ Calculate the cos and sin parts sin(-θ) = -sinθ Multiply out 3 B

Further complex numbers You need to know how multiplying and dividing affects both the modulus and argument of the resulting complex number Combine using one of the rules above Divide the moduli Subtract the arguments Express the following calculation in the form x + iy: Simplify You can work out the sin and cos parts Multiply out 3 B

achings for exercise 3

Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n Let: z = r(cosθ + isinθ) Use the modulusargument form Multiply the moduli, add the arguments This is De Moivre’s Theorem You need to be able to prove this De Moivre = ‘De Mwavre’ (pronunciation) 3 C

Further complex numbers You 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 FP 1 Proving that: is true for all positive integers BASIS Show that the statement is true for n = 1 Sub in n=1 Basis – show the statement is true for n=1 Simplify each side 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 ASSUMPTION Assume that the statement is true for n = k Replace n with k 3 C

Further complex numbers You 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 FP 1 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 Proving that: is true for all positive integers INDUCTIVE 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 3 B Multiply the moduli, add the arguments 3 C

Further complex numbers You 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 FP 1 Basis – show the statement is true for n=1 Proving that: is true for all positive integers CONCLUSION Explain why this shows it is true… We showed the statement is true for n = 1 We then assumed the following: Assumption – assume the statement is true for n = k Using the assumption, we showed that: Inductive – show that if true for n = k, then the statement is also true for n = k+1 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… 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 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! 3 C

Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n We have just proved theorem for n = k where k 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 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 You can see that the answer has followed the same pattern as De Moivre’s theorem! Multiply out like quadratics – the bottom is the difference of two squares i 2 = -1 You cancel the denominator as it is equal to 1 Use cos(-θ) = cos(θ) and sin(-θ) = -sinθ 3 C

Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(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 cos 0 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! 3 C

Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(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! 3 C

Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(cos(nθ + isinnθ) for any integer n The denominator has to have the ‘+’ sign in the middle Simplify the following: Apply cos(-θ) = cosθ and sin(-θ) = -sinθ 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 3 B 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 Simplify 3 C

Further complex numbers y You need to be able to prove that [r(cosθ + isinθ)]n = rn(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 Pay attention to the directions The ‘x’ part is positive so will go in the positive direction horizontally r √ 3 θ x 1 The ‘y’ part is positive so will go upwards Calculate Inverse Tan 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 Sub in r and θ 3 C

Further complex numbers You need to be able to prove that [r(cosθ + isinθ)]n = rn(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! 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 simplify Start with an argand diagram to help find the modulus and argument of the part in the bracket 3 C

achings for exercise 3

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities This involves changing expressions involving a function of θ into one without. For example changing a cos 6θ into powers of cosθ You will need to use the binomial expansion for C 2 in this section Remember n. Cr 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 n. Cr parts 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express cos 3θ using powers of cosθ. This type of question involves making a comparison between two processes One which will give you a ‘cos 3θ’ term – you will use De Moivre’s Theorem for this If we apply De Moivre’s theorem to this, we will end up with a ‘cos 3θ’ term If we apply the binomial expansion to it, we will end up with some terms with cosθ in So this expression is a good starting point! One which will give you an expression in terms of cosθ – you will use the binomial expansion for this You have to think logically and decide where to start 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express cos 3θ using powers of cosθ. This type of question involves making a comparison between two processes Apply De Moivre’s theorem Follow the rules you know Apply the Binomial expansion Write out One which will give you a ‘cos 3θ’ term – you will use De Moivre’s Theorem for this ‘Tidy up’ Replace i 2 One which will give you an parts with -1 expression in terms of cosθ – you will use the binomial expansion The two expressions we have made must be equal for this Therefore the real parts in each and the imaginary parts in each must You have to think logically and decide where to start be the same Equate the real parts 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express cos 3θ using powers of cosθ. This type of question involves making a comparison between two processes One which will give you a ‘cos 3θ’ term – you will use De Moivre’s Theorem for this Replace sin 2θ with an expression in cos 2θ Expand the bracket Simplify We have successfully expressed cos 3θ as posers of cosθ! One which will give you an expression in terms of cosθ – you will use the binomial expansion for this You have to think logically and decide where to start 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express the following as powers of cosθ: So we need something that will give us sin 6θ 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 ‘sin 6θ’ term If we apply the binomial expansion to it, we will end up with some terms with cosθ in So this expression is a good starting point! (and yes you will have to do some expansions larger than powers of 3 or 4!) 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express the following as powers of cosθ: Apply De Moivre’s theorem Follow the rules you know Apply the Binomial expansion Replace terms: i 2 and i 6 = -1 i 4 = 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 sin 6θ in Divide all by i 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Express the following as powers of cosθ: 2 So we have changed the expression we were given into powers of cosθ! 4 Divide all terms by sinθ Replace sin 2θ terms with (1 – cos 2θ) terms Expand the first bracket, square the second Expand the second bracket Group together the like terms 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities You 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 sin 6θ 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(-θ) = cosθ and sin(-θ) = -sinθ We can add our two results together: Simplify 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities You 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 sin 6θ 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(-θ) = cosθ and sin(-θ) = -sinθ We could also subtract our two results: Simplify 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities You 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 sin 6θ 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(-θ) = cosθ and sin(-θ) = -sinθ We could add our two results together: Simplify You can also apply the rules we just saw to powers of z 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities You 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 sin 6θ 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(-θ) = cosθ and sin(-θ) = -sinθ We could also subtract our two results: Simplify You can also apply the rules we just saw to powers of z 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Let’s now see how we can use these ‘patterns’ in solving problems: Creating a cos 5θ term Using the Identity above Creating the other cos terms – use the Binomial expansion! Express cos 5θ in the form acos 5θ + bcos 3θ + ccosθ Where a, b and c are constants to be found. When working this way round you need to use the identities above to express both cos 5θ and terms with cos 5θ, cos 3θ and cosθ. You can then set the expressions equal to each other Use the B. E. Cancel some z terms Group up terms with the same power Rewrite using an identity above Simplify 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Let’s now see how we can use these ‘patterns’ in solving problems: Using the two expressions These two expressions must be equal to each other Express cos 5θ in the form acos 5θ + bcos 3θ + ccosθ Where a, b and c are constants to be found. When working this way round you need to use the identities above to express both cos 5θ and terms with cos 5θ, cos 3θ and cosθ. Divide both sides by 32 So we have written cos 5θ using cos 5θ, cos 3θ and cosθ You can then set the expressions equal to each other 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Using the Identity above Creating a sin 3θ term Show that: Creating the other sin terms – use the Binomial expansion! Use the B. E. This is similar to the previous question. You need to use rules above to find a way to create a sin 3θ expression, and an expression containing sin 3θ and sinθ Cancel some z terms Group up terms with the same power Rewrite using an identity above Simplify 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Show that: This is similar to the previous question. You need to use rules above to find a way to create a sin 3θ expression, and an expression containing sin 3θ and sinθ Using the two expressions These two expressions must be equal to each other Divide both sides by i Divide both sides by -8 So we have written sin 3θ using sin 3θ and sinθ! 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities Creating a sin 4θ term Using the Identity above a) Express sin 4θ in the form: Creating the cos terms – use the Binomial expansion! Where d, e and f are constants to be found. Use the B. E. b) Hence, find the exact value of the following integral: Cancel some z terms Group up terms with the same power (use positive values in the brackets so we get cos terms) Start exactly as with the previous questions, by finding an expression with sin 4θ and one with cos 4θ, cos 2θ and a number Replace using an identity above Simplify 3 D

Further complex numbers You can apply De Moivre’s theorem to trigonometric identities a) Express sin 4θ in the form: Using the two expressions These two expressions must be equal to each other 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 cos 4θ, cos 2θ and a number Divide both sides by 16 So we have written sin 4θ using cos 4θ and cos 2θ! 3 D

Further complex numbers Cosine Integrals (in C 4) You can apply De Moivre’s theorem to trigonometric identities a) Express sin 4θ in the form: Where d, e and f are constants to be found. b) Hence, find the exact value of the following integral: Replace with an equivalent expression Integrate each term with respect to θ, using knowledge from C 4 Sub in limits Start exactly as with the previous questions, by finding an expression with sin 4θ and one with cos 4θ, cos 2θ and a number Work out 3 D

achings for exercise 3

Further complex numbers You can use De Moivre’s theorem to find the nth roots of a complex number You 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: 1) 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 (2π = 360º) 2) De Moivre’s theorem 3 E

Further complex numbers You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z 3 = 1 and represent your solutions on an Argand diagram. In this case the modulus and argument are simple to find! y 1 x 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 Apply the rule above Cube root (use a relevant power) Apply De Moivre’s theorem 3 E

Further complex numbers You can use De Moivre’s theorem to find the nth roots of a complex number k=0 Solve the equation z 3 = 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 Sub k = 0 in and calculate the cosine and sine parts k=1 Sub k = 1 in and calculate the cosine and sine parts The values of k you choose should keep the argument within the range: -π < θ ≤ π So the roots of z 3 = 1 are: k = -1 Sub k = -1 in and calculate the cosine and sine parts (k = 2 would cause the argument to be outside the range) 3 E

Further complex numbers y You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z 3 = 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 x The values of k you choose should keep the argument within the range: -π < θ ≤ π So the roots of z 3 = 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 0 3 E

Further complex numbers You can use De Moivre’s theorem to find the nth roots of a complex number Solve the equation z 4 - 2√ 3 i = 2 Give your answers in both the modulusargument and exponential forms. By rearranging… z 4 = 2 + 2√ 3 i As before, use an argand diagram to express the equation in the modulusargument form Then choose values of k until you have found all the solutions Find the modulus and argument y r 2√ 3 θ 2 x Apply the rule above Take the 4 th root of each side De Moivre’s Theorem Work out the power at the front 3 E

Further complex numbers You can use De Moivre’s theorem to find the nth roots of a complex number k=0 Solve the equation z 4 - 2√ 3 i = 2 Give your answers in both the modulusargument and exponential forms. k=1 By rearranging… z 4 = 2 + 2√ 3 i As before, use an argand diagram to express the equation in the modulusargument form Then choose values of k until you have found all the solutions Sub k = 0 in and simplify (you can leave in this form) k = -1 Choose values of k that keep the argument between –π and π k = -2 3 E

Further complex numbers You can use De Moivre’s theorem to find the nth roots of a complex number Solutions in the modulus-argument form Solve the equation z 4 - 2√ 3 i = 2 Give your answers in both the modulusargument and exponential forms. Solutions in the exponential form By rearranging… z 4 = 2 + 2√ 3 i As before, use an argand diagram to express the equation in the modulusargument form Then choose values of k until you have found all the solutions 3 E

achings for exercise 3

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram A locus a set of points which obey a rule x The locus of points a given distance from a point O is a circle x O x You will need to be able to understand Loci based on Argand diagrams A The locus of points equidistant from two fixed points A and B is the perpendicular bisector of line AB B 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram z = x + iy represents a variable point P(x, y) on an Argand diagram P(x, y) z z 1 = x 1 + iy 1 represents a fixed point A(x 1, y 1) on an Argand diagram What is represented by: z - z 1 A(x 1, y 1) z 1 x It represents the distance between If we want to get from the fixed point A to the variable point the fixed point A(x 1, y 1) and the P, we need to travel back along z 1 and then out along z variable point P(x, y) (-z 1 + z) This can be written as a vector, z – z 1 So |z – z 1| represents the distance between the fixed point and the variable point! 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram P(x, y) If: A(5, 3) x Sketch the locus of P(x, y) which is represented by z on an 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 This will be a circle of radius 3 units, centre (5, 3) 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram y P(x, y) If: |z| Use an algebraic method to find a Cartesian equation of the locus of z x y x 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! If: (By Pythagoras’ Theorem) Then: 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram If: Replace z with ‘x + iy’ Group the real and imaginary terms Use an algebraic method to find a Cartesian equation of the locus of z Use the rule above to remove the modulus Now we can find the equation of the locus algebraically… Square both sides You (hopefully) recognise that this is the equation of a circle, radius 3 and with centre (5, 3)! 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram As a general rule, the locus of: is a circle of radius r and centre (x 1, y 1) where z 1 = x 1 + iy 1 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers Circle radius r and centre (x 1, y 1) You can use complex numbers to represent a locus of points on an Argand diagram Give a geometrical interpretation of each of the following loci of z: a) d) Circle, centre (0, 3) radius 4 |-1| = 1, put the ‘fixed’ part in a bracket b) Circle, centre (2, 3) radius 5 c) ‘Factorise’ the part inside the modulus You can write this as 2 moduli multiplied Put the ‘fixed’ part in a bracket Circle, centre (-3, 5) radius 2 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| 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x, y) which is represented by z on an Argand diagram, if: This is the distance of P(x, y) from the origin (0, 0) This is the distance of P(x, y) from (0, 6) y (0, 6) 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-ordinates y=3 (0, 0) x You can see that it is the line with equation y = 3 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Replace z with x + iy Sketch the locus of P(x, y) which is represented by z on an Argand diagram, if: Factorise the ‘i’ terms on the right side Show that the locus is y = 3 using an algebraic method Square both sides Use the rule above to remove the moduli Expand the bracket Cancel terms on each side Add 12 y Divide by 12 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Replace z with x + iy a) Use an algebraic method to find the Cartesian equation of the locus of z if: Group real and imaginary parts Use the rule above to remove the moduli Square both sides b) Represent the locus of z on an Argand diagram Expand brackets Cancel terms Subtract 1 Divide by 2 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram y (0, 4) a) Use an algebraic method to find the Cartesian equation of the locus of z if: x (3, 0) b) Represent the locus of z on an Argand diagram (0, -1) y = -3 x + 4 Distance from (3, 0) Distance from (0, -1) 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Replace z with ‘x + iy’ If: Group real and imaginary parts Replace the moduli using the rule above a) Use algebra to show that the locus of z is a circle, stating its centre and radius Square both sides (remember the ‘ 2’) Expand some brackets Expand another bracket Group all terms on one side Circle, centre (-10, 12) and radius 10 b) Sketch the locus of z on an Argand diagram Divide by 3 Completing the square Simplify Add 100 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram (-10, 12) If: The circle shows the set of points that are twice as far from (6, 0) as they are from (-6, 9)! (-6, 9) a) Use algebra to show that the locus of z is a circle, stating its centre and radius P(x, y) (6, 0) x Circle, centre (-10, 12) and radius 10 b) Sketch the locus of z on an Argand diagram The distance from (6, 0) Is equal Twice the distance to from (-6, 9) 3 F

Is the distance between the variable point z and the fixed point z 1 when they are represented on a Argand diagram Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram If: a) Use algebra to show that the locus of z is a circle, stating its centre and radius 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!) Circle, centre (-10, 12) and radius 10 b) Sketch the locus of z on an Argand diagram 3 F

Further complex numbers y You 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. Then find the Cartesian equation of this locus algebraically. The line is not extended back downwards It is known as a ‘halfline’ x The locus will be the set of points which start at (0, 0) and make an argument of π/4 with the positive xaxis 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram y If: 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 xaxis x x Replace z with ‘x + iy’ The value of the argument is tan-1(opposite/adjacent) ‘Normal tan’ Calculate the tan part Multiply by x (x > 0) 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: (2, 0) x 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)! 3 F

Further complex numbers y You 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. 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)! y (2, 0) x-2 x Replace z with ‘x + iy’ The value of the argument is tan-1(opposite/adjacent) ‘Normal tan’ Calculate the tan part Multiply by (x – 2) (x > 2) 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram x If: (-3, -2) 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 2 i to z, the argument from (0, 0) and the positive x-axis will be 3π/4 So the line will have to start at -3, -2) ( 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram y+2 x If: x + 3 (-3, -2) Sketch the locus of z on an Argand diagram and use an algebraic method to find the equation of the line. Replace z with ‘x + iy’ The value of the argument is tan-1(opposite/adjacent) When we add 3 and 2 i to z, the argument from (0, 0) and the positive x-axis will be 3π/4 So the line will have to start at -3, -2) ‘Normal tan’ ( Calculate the tan part Multiply by (x + 3) Subtract 2 (x < -3) 3 F

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram So therefore: Is represented by a half line starting at z 1 and making an angle of θ with a line parallel to the x-axis 3 F

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram 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” For the next set of Loci, you need to remember some rules relating to circles Major arc – θ is acute Minor arc – θ is obtuse θ θ A θ Semi-circle – θ is 90° θ B θ x θ B A 2 x A A B If they are joined to a point on the major arc If they are joined to a point on the minor arc If the chord is the diameter of the circle The angle will be acute The angle will be obtuse The angle will be 90° B The angle at the centre is twice the angle at the circumference 3 F

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram If: a) 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 π/4 3 F

Further complex numbers y arg(z – 6) = θ 1 You can use complex numbers to represent a locus of points on an Argand diagram arg(z – 2) = θ 2 If: θ 2 This angle must therefore be θ 1 – θ 2, the difference between the arguments! θ 1 a) Sketch the locus of P(x, y), which is represented by z on an Argand diagram (2, 0) So what we are doing is drawing the locus of points where the difference between these arguments is π/4 However, there are more points that satisfy this rule! θ 1 θ 2 (6, 0) x Imagine drawing both arguments – we will use θ 1 and θ 2 to represent their values Using alternate angles, we can show the angle between the arguments is their difference We want this difference to be π/4 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram If: π/ π/ a) Sketch the locus of P(x, y), which is represented by z on an Argand diagram 4 θ 2 (2, 0) So what we are doing is drawing the locus of points where the difference between these arguments is π/4 Geogebra Example 4 θ 1 θ 2 θ 1 (6, 0) x If we move the point where the lines cross along the major arc of a circle, then the value of π/4 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 3 F

Further complex numbers y “The angle at the centre is twice the angle at the circumference” You can use complex numbers to represent a locus of points on an Argand diagram π/ 4 If: π/ a) Sketch the locus of P(x, y), which is represented by z on an Argand diagram b) Find 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: (2, 0) 2 (6, 0) x We can use this isosceles triangle to find the information we need… Centre Radius 3 F

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Centre Radius If: (4, 2) π/ 4 a) Sketch the locus of P(x, y), which is represented by z on an Argand diagram b) Find the Cartesian equation of this locus We need the centre of the ‘circle’ and its radius Centre (4, 2) Radius 2√ 2 2 π/ 4 (2, 0) 2 (4, 0) (6, 0) Split the triangle in the middle, the smaller angles will both be π/ (45ᵒ) (because the top angle was π/ ) 4 2 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) 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram π/ 4 If: a) Sketch the locus of P(x, y), which is represented by z on an Argand diagram b) Find the Cartesian equation of this locus We need the centre of the ‘circle’ and its radius Centre (4, 2) (2, 0) (6, 0) x The locus is therefore the arc of a circle with the following equation: Radius 2√ 2 3 F

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 semicircle, 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 θ is positive Clockwise if θ is negative If the value we want is positive, then θ 1 > θ 2 If the value we want is negative, then θ 2 > θ 1 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! 3 F

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x, y) on an Argand diagram if: y 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 semicircle, 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 θ is positive Clockwise if θ is negative (0, 4) (0, 0) and (0, 4) The angle to make is π/2 A semi-circle (0, 0) x Θ is positive, so draw anti -clockwise from (0, 0) (numerator point) to (0, 4) (denominator point) 3 F

Further complex numbers You can use complex numbers to In step 3 we had to choose whether to draw the diagram clockwise or anti -clockwise from the numerator point to the denominator point represent a locus of points on an Argand diagram Lets show why this is correct! y y (0, 4) θ 2 θ 2 θ 1 θ 1 x (0, 0) We drew the angle anticlockwise from (0, 0) to (0, 4) However, as θ 2 is actually negative, the sum is really θ 1 + (-θ 2) Using the alternate angles, the angle between the arguments is θ 1 + θ 2 = θ 1 – θ 2 This angle is therefore what we were wanting! Basically, always use the rule in step 3! x (0, 0) If we drew the arc the other way clockwise from (0. 0) to (0, 4) Using the alternate angles, the on the outside is θ 1 + θ 2 However, as θ 2 is actually negative, the sum is really θ 1 + (-θ 2) = θ 1 – θ 2 But of course it is on the wrong side of the arc so we do not want this part of the circle! 3 F

Further complex numbers You can use complex numbers to represent a locus of points on an Argand diagram Sketch the locus of P(x, y) on an Argand diagram if: y 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 semicircle, 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 θ is positive Clockwise if θ is negative (0, -3) and (0, 2) The angle to make is π/3 A major arc Θ is positive, so draw anticlockwise from (0, -3) (numerator point) to (0, 2) (denominator point) x (0, -3) 3 F

Further complex numbers y You can use complex numbers to represent a locus of points on an Argand diagram 3 Given that the complex number z = x + iy satisfies the equation: Find the minimum and maximum values of |z| 3 (12, 5) 13 x Start by drawing this on an Argand The smallest and largest values for |z| will be on the same straight line diagram through the circle’s centre It is a circle, centre (12, 5) radius 3 units 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 3 F

achings for exercise 3

Further complex numbers y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci (4, 2) The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required x 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 3 G

Further complex numbers y 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 (4, 0) (6, 0) x 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’ The distance to |z – 4| must be less than the distance to |z – 6| Shade the region closest to (4, 0) 3 G

Further complex numbers y You can use complex numbers to represent regions on a Argand diagram This is very similar to what you have been doing with loci (2, 2) The only extra part is that once you have drawn the locus representing the point, you need to indicate the area required x Shade on an Argand diagram the region indicated by: The argument must be between these two values Start by drawing the limits of the argument from the point (2, 2) Shade the region between the two arguments 3 G

Further complex numbers y 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 x Shade on an Argand diagram the region indicated by: Imagine all the regions were on the same diagram and The region we want will have to satisfy all of these at the same time! 3 G

achings for exercise 3

Further complex numbers y You 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 + iv The z-plane (uses x and y) x Effectively 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! Transformation from one plane to the next! v The w-plane (uses u and v) u 3 H

Further complex numbers y You 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 + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T 1 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 T 1 is given by: We will work out the new set of points algebraically… v x The z-plane u The w-plane Circle centre (0, 0), radius 2 Circle centre (-2, 4), radius 2 To start with, make z the subject Add 2, subtract 4 i The modulus of each side must be the same We know |z| from the question Circle, centre (-2, 4), radius 2 3 H

Further complex numbers You 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 + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T 2 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 T 2 is given by: We will work out the new set of points algebraically… The z-plane y The w-plane v x Circle centre (0, 0), radius 2 u Circle centre (0, 0), radius 6 To start with, make z the subject Divide by 3 Modulus of both sides |z|= 2 Split the modulus up |3|=3 so multiply by 3 3 H

Further complex numbers You 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 + iv The point P represents the complex number z on an Argand diagram where |z| = 2. T 2 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 T 3 is given by: The z-plane y The w-plane v x Circle centre (0, 0), radius 2 u Circle centre (0, 1), radius 1 To start with, make z the subject Subtract i Leaving z like this can make the problem easier! (rather than rearranging completely) Modulus of both sides You can split the modulus on the right We will work out the new set of points algebraically… |z| = 2 Simplify the right side 3 H

Further complex numbers You 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 + iv For the transformation w = z 2, where z = x + iy and w = u + iv, find the locus of w when z lies on a circle with equation x 2 + y 2 = 16 It is very important for this topic that you draw information on z or |z| from the question The equation x 2 + y 2 = 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 y The w-plane v x Circle centre (0, 0), radius 4 u 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 16 3 H

Further complex numbers You 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 + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Multiply by (z + 1) Expand the bracket Subtract 5 iz and subtract w Factorise the left side Divide by (w – 5 i) 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… Modulus of both sides |z| = 1 Multiply by |w – 5 i| |i - w| = |w – i| 3 H

Further complex numbers Transformation T You 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 + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Circle centre (0, 0), radius 1 The z-plane Perpendicular bisector between (0, 1) and (0, 5) The line v = 3 y The w-plane v x 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. u So a circle can be transformed into a straight line! 3 H

Further complex numbers You 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 + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Multiply by (z + 1) Expand the bracket Subtract wz and add 2 Factorise the right side Divide by (3 – w) Show that the image, under T, of the circle with equation x 2 + y 2 = 4 in the z -plane, is a different circle C in the wplane. Modulus of each side Split up the modulus |z| = 2 Multiply by |3 - w| State the centre and radius of C. |3 - w| = |w - 3| Remember that x 2 + y 2 = 4 is the same as |z| = 2 We now need to find what the equation of this will be! 3 H

We will find the equation as we did in the early part of section 3 F! Further complex numbers You 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 + iv Replace w with ‘u + iv’ Group real/imaginary terms Remove the modulus The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Expand brackets Expand more brackets! Move all to one side Show that the image, under T, of the circle with equation x 2 + y 2 = 4 in the z -plane, is a different circle C in the wplane. Divide by 3 Use completing the square Move the number terms across State the centre and radius of C. Remember that x 2 + y 2 = 4 is the same as |z| = 2 Circle, centre (14/3, 0), radius 10/ 3 3 H

Further complex numbers Transformation T You 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 + iv The transformation T from the z-plane, where z = x + iy, to the w-plane, where w = u + iv, is given by: Circle centre (0, 0), radius 2 The z-plane Show that the image, under T, of the circle with equation x 2 + y 2 = 4 in the z -plane, is a different circle C in the wplane. Circle, centre (14/3, 0), radius 10/3 y The w-plane x v u State the centre and radius of C. 3 H

Further complex numbers You 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 + iv A transformation T of the z-plane to the w-plane is given by: Multiply by (1 – z) Expand the bracket Add 2, Add wz Factorise the right side Divide by (i + w) 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 Argand diagram. Start by rearranging to make z the subject ‘as usual’ Write the other way round (if you feel it is easier!) i+w=w+i 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 + iy’ 3 H

Further complex numbers You 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 + iv Replace w with u + iv Group real and imaginary terms A 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 Argand 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! Multiply by the denominator but with the opposite sign (this will cancel ‘i’ terms on the bottom Simplify i 2 = -1 Separate real and ‘i’ terms As z lies on the x-axis, we know y = 0 Therefore, the imaginary part on the right side must also equal 0 3 H

Further complex numbers You 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 + iv A transformation T of the z-plane to the w-plane is given by: Set the imaginary part equal to 0 Multiply by (u 2 + (v + 1)2) (you will be left with the numerator) Multiply out the double bracket 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 Argand diagram. You can now find an equation for the line in the w-plane Subtract all these terms The ‘uv’ terms cancel out Make v the subject Divide by 2 So the transformation has created this line in the w-plane (remember v is essentially ‘y’ and u is essentially ‘x’) 3 H

Further complex numbers You 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 + iv Transformation T (z lies on the real axis) Straight line, gradient is and y-intercept at (0, -1) 1/ 2 A transformation T of the z-plane to the w-plane is given by: The z-plane 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 Argand diagram. y v The w-plane x -2 u -1 3 H

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