Sample PROGRESSION. Chapter 1 Complex numbers. Edexcel AS and A level Further Mathematics NEW FOR. Core Pure Mathematics.

Transcription

1 11-19 ROGRSSION Sample Chapter 1 NW FOR 017 dexcel AS and A level Further Mathematics Core ure Mathematics Book 1/AS

2 dexcel AS and A level Further Mathematics 1 Sample material Objectives Understand and use the definitions of imaginary and complex numbers page Add and subtract complex numbers pages Multiply complex numbers pages 5 6 Understand the definition of a complex conjugate pages 6 8 Divide complex numbers pages 7 8 Solve quadratic equations that have complex roots pages 8 10 Solve cubic or quartic equations that have complex roots pages contain a real and an imaginary part. ngineers and physicists often describe quantities with two components using a single complex number. This allows them to model complicated situations such as air flow over a cyclist. knowledge check 1 Simplify each of the following: a 50 b _ 108 c _ 180 ure Year 1, Chapter 1 RArior In each case, determine the number of distinct real roots of the equation f(x) = 0. a f(x) = x + 8x + 10 b f(x) = x 9x + 7 c f(x) = x + 1x + 9 ure Year 1, Chapter DRA FT Find the solutions of x 8x + 6 = 0, giving your answers in the form a ± b where a and b are integers. ure Year 1, Chapter 7 Write _ in the form p + q where p and q are rational numbers. ure Year 1, Chapter 1 1

3 1.1 Imaginary and complex numbers The quadratic equation a x + bx + c = 0 has Links For the equation a x solutions given by: + bx + c = 0, the x = b ± _ discriminant is b ac. b ac If b ac > 0, there are two distinct real roots. a If b ac = 0, there are two equal real roots. If the expression under the square root is negative, If b ac < 0, there are no real roots. ure Year 1, Section.5 there are no real solutions. You can find solutions to the equation in all cases by extending the number system to include 1. Since there is no real number that squares to produce 1, the number 1 is called an imaginary number, and is represented using the letter i. Sums of real and imaginary numbers, for example + i, are known as complex numbers. i = 1 Notation The set of all complex numbers is An imaginary number is a number of the written as C. form bi, where b R. For the complex number z = a + bi: Re(z) = a is the real part A complex number is written in the Im(z) = b is the imaginary part form a + bi, where a, b R. xample 1 Write each of the following in terms of i: a _ 6 b _ You can use the rules of surds to manipulate 8 imaginary numbers. _ a 6 = _ 6 ( 1) = = 6i Watch out b 8 = _ 8 ( 1) = An alternative way of writing ( 7 ) i 8 = ( is i 7. Avoid writing 7 i as this can easily be )i confused with 7i. In a complex number, the real part and the imaginary part cannot be combined to form a single term. can ord DR( DR 1 1 = 7 be added or subtracted by adding or subtracting their real parts and adding or subtracting their imaginary parts. Dir DR You can multiply a real number by a complex number by multiplying out the brackets in the usual way. xample Simplify each of the following, giving your answers in the form a + bi, where a, b R. _ i a ( + 5i) + (7 + i) b ( 5i) (5 11i) c (5 8i) d a ( + 5i) + (7 + i) = ( + 7) + (5 + )i = 9 + 8i b ( 5i) (5 11i) = ( 5) + ( 5 ( 11))i = + 6i T z T= a DRiRAF DR( DR DR7 DR) ) i Add the real parts and add the imaginary parts. Subtract the real parts and subtract the imaginary parts. c (5 8i) = ( 5) ( 8)i = 10 16i d i _ xercise = i = 5 + i 1A Do not use your calculator in this exercise. 1 Write each of the following in the form bi where b is a real number. a 9 b 9 c 11 d e 5 f 5 g 1 h 5 i 00 j 17 Simplify, giving your answers in the form a + bi, where bft a R and b R. a (5 + i) + (8 + 9i) b ( + 10i) + (1 8i) c (7 + 6i) + ( 5i) d ( 1_ + 1_ i) + ( _ 5 + 5_ i) e (0 + 1i) (11 + i) f ( i) ( 5 + i) g ( 6i) ( 8 8i) h ( + i) ( i) i ( 7i) + (1 + i) ( 1 + i) j (18 + 5i) (15 i) ( + 7i) Simplify, giving your answers in the form a + bi i, where a R and b R. a (7 + i) b (8 i) c ( + i) + ( + i) d 5( + i) ( 1 + i) e 6 i i f 5 _ g i +D+D 8 + i 7_ i h Write in the form a + bi, where a and b simplified surds. a i 6i b Given that z = 7 6i and w =D= 7D7 + 6i, find, Notation are often in the form a + bi, where a, b DRare R represented by the letter z or the letter w. a z w b w + z 6 Given that z 1 = a + 9i, z = + bi and z z 1 = 7 + i, find a and b where a, b R. ( marks) 7 Given that z 1 = + i and z = 7 i, find, in the form a + bi, where a, b R a z 1 z b z c z z 8 Given that z = a + bi and w = a bi, show that: a z + w is always real b z w is always imaginary (5 8i) can also be written as (5 8i) + (5 8i). First separate into a real part and an imaginary part. a AFT RA You can use complex numbers to find solutions to any quadratic equation with real coefficients. If b ac < 0 then the quadratic equation ax + bx + c = 0 has two distinct complex roots.

4 xample Solve the equation z + 9 = 0. z = 9 _ z = ± 9 = ± 9 1 = ± z = +i, z = i xample 9 Solve the equation z + 6z + 5 = 0. 1 = ±i Method 1 (Completing the square) z + 6z = (z + ) 9 z + 6z + 5 = (z + ) = (z + ) + 16 (z + ) + 16 = 0 (z + ) = 16 _ z + = ± 16 = ±i z = ± i z = + i, z = i Method (Quadratic formula) _ z = 6 ± = 6 ± 6 6 0= ±8i 6 ± 8i z = = ± i z = + i, z = i xercise 1B Do not use your calculator in this exercise. Note that just as z = 9 has two roots + and, z = 9 also has two roots +i and i. Because (z + ) = (z z + )(z + ) = z + 6z + 9 _ 16 ( 1) 6 = 16 1 = i equation quickly calculator. FT1) = FT _ FT 16 FT16 FT T _ 1 T 1 Solve this using your = b ± b ac Using zaz a Online AFTz FT= _ 6 = 6 ( 1) = 6 1 = 8i DRAF RAFT T 1 Solve each of the following equations. Write your answers in the form ±bi. a z + 11 = 0 b z + 0 = 0 c z + 10 = 0 d z = 8 z e z + 0 = z 66 f 6z + 1 = z Solve each of the following equations. Write your answers in the form a ± bi. a z + z + 5 = 0 b z z + 10 = 0 c z + z + 9 = 0 d z + 10z + 6 = 0 e z + 5z + 5 = 0 f z + z + 5 = 0 Solve each of the following equations. Write your answers in the form a ± bi. a z + 5z + = 0 b 7z z + = 0 c 5z z + = 0 5 The solutions to the quadratic equation z 8z + 1 = 0 are z 1 and z. Find z 1 and z, giving each answer in the form a ± i b. 6 The equation z + bz + 11 = 0, where b R, has distinct complex roots. Find the range of possible values of b. 1. Multiplying complex numbers xample 5 xpress each of the following in the form a + bi, where AFb a and b are real numbers. a ( + i)( + 5i) b (7 i) a ( + i)( + 5i) = ( + 5i) + i( the two brackets as you would with real = i + 1i numbers. AFare DR + 5i) Multiply + 15i = i + 1i 15 Use the fact that i = 1. = (8 15) + (10i + 1i) = i Add real parts and add imaginary parts. RAnum b (7 i) = (7 i)(7 i) Multiply out the two brackets as you would with = 7(7 i)dr i) i(7 i) real numbers. = 9 8i 8i + = 9 8i 8i D+D8i D16i 16D16 D Use the fact that i = 1. = (9 16) + ( 8i 8i) = 56i Add real parts and add imaginary parts. xample 6 Simplify: a i b i c (i) 5 ( marks) You can multiply complex numbers using the same technique that you use for multiplying brackets in algebra. You can use the fact that i = 1 to si mplify powers of i. i = 1 DRAF RAFT Solve each of the following equations. Write your answers in the form a ± bi. a (z ) 9 = 16 b (z 7) + 0 = 6 c 16 (z + 1) + 11 = Hint The left-hand side of each equation is in completed square form already. Use inverse operations to find the values of z. a i = i i i = i i = i b i = i i i i = i i = ( 1) ( 1) = 1 c (i) 5 = i i i i i = (i i i i i) = (i i i) = ( 1) ( 1) i = i i = 1 (i) 5 = 5 i 5 First work out 5 =. 5

5 xercise Do not use your calculator in this exercise. 1 Simplify each of the following, giving your answers in the form a + bi. a (5 + i)( + i) b (6 + i)(7 + i) c (5 i)(1 + 5i) d (1 i)( 8i) e ( i)( + 7i) f (8 + 5i) g ( 9i) h (1 + i)( + i)( + i) Hint For part h, begin by multiplying i ( i)(5 + i)( i) j ( + i) the first pair of brackets. a Simplify ( + 5i)( 5i), giving your answer in the form a + bi. b Simplify (7 i)(7 + i), giving your answer in the form a + bi. c Comment on your answers to parts a and b. d rove that (a + bi)(a bi) is a real number for any real numbers a and b. Given that (a + i)(1 + bi) = 1 8i, find two possible pairs of values for a and b. Write each of the following in its simplest form. a i 6 b (i) c i 5 + i i 5 xpress (1 + i) 6 in the form a bi, where a and found. Ta FTd d 6 Find the value of the real part of ( i). roblem-solving 7 f(z) = z z + 8 FT(i) FT RA b are integers to be idr 01DR DR You can use the binomial theorem to Find: a f(i) b f( 6i) expand (a + b) n. ure Year 1, Section 8. 8 f(z) = z z + 17 Show that z = 1 i is a solution to f(z) = 0. ( marks) 9 a Given that i 1 = i and i = 1 1, write i and i in their simplest forms. b Write i 5, i 6, i 7 and i 8 in their simplest forms. c Write down the value of: i i 100 ii i 5 iii i C Challenge a xpand (a + bi). b Hence, or otherwise, find 0 i, giving your answer in the form a bi, where a and b are positive integers. 1. Complex conjugation For any complex number z = a + bi, the complex conjugate of the number is defined as z* = a bi. xample 7 FT 01D AF Notation The principal square root of a complex number, z, has a positive real part. Notation Together z and z* are called a complex conjugate pair. Given that z = 7i: a write down z* b find the value of z + z* c find the value of zz* a z* = + 7i b z + z* = ( 7i) + ( + 7i) = ( + ) + ( 7 + 7)i = c zz* = ( 7i)( + 7i) = ( + 7i) 7i( + 7i) = + 1i 1i 9i = + 9 = 5 For any complex number z, the product of z and z* is a real number. You can use this property to divide two complex numbers. To do this, you multiply both the numerator and the denominator by the complex conjugate of the denominator and then simplify the result. xample i Write in the form a + bi. i 5 + i i = 5 + i i + i The complex conjugate of the denominator is + i. Multiply both the numerator and the + i denominator by the complex conjugate. (5 + i) ( + i) = ( i) ( + i) is real, so ( i)( + i) will be a real number. (5 + i)( + i) = 5( + i) + i( + i) = i + 8i + 1i = + i Online Divide complex numbers ( i)( + i) = ( + i) i( + i) quickly using your calculator. = + 6i 6i 9i = 1 RAzz* RAis xercise Do not use your calculator in this exercise. 1 Write down the complex conjugate z p for: _ a z = 8 + i b z = 6 5i c z = 1_ i d z = 5 + i 10 Find z + z p and zz p for: a z = 6 i b z = i c z = _ + 1_ i d z = 5 i 5 Write each of the following in the form a + bi. a 5i _ 1 + i _ 5 + i i 1D D1D idr DR = _ + i = D i Divide each term in the numerator by 1. b DR + 5i _ 6 8i c 8 i 1 i Change the sign of the imaginary part from to +. Note that z + z* is real. Remember i = 1. Note that zz* is real. Links The method used to divide complex numbers is similar to the method used to rationalise a denominator when simplifying surds ure Year 1, Section 1.6 DRAF DR DR +DR RAF RAFT D=DR DR 9iDR DR 6i 9i D DRi) d + i _ 1 + i 7

6 Write ( i) in the form x + iy where x, y R. 1 + i 5 Given that z 1 = 1 + i, z = + i and z = + i, write each of the following in the form a + bi. a _ z 1z _ (z ) z 1 + 5z z b z 1 c _ z _ 6 Given that 5 + z i = i, find z in the form a + bi. ( marks) _ 6 + 8i 7 Simplify 1 + i + _ 6 + 8i, giving your answer in the form a + bi. 1 i 8 w = 8 i xpress w in the form a + bi, where a and b are rational numbers. 9 w = 1 9i 1 xpress w in the form a + bi, where a and b are rational numbers. 10 z = i Use algebra to express z + z in the form p + qi, where p and q are rational numbers. 11 The complex number z satisfies the equation ( + i)(z z i) = 6 i. Find z, giving your answer in the form a + bi where a and b are rational numbers. 1 The complex numbers z 1 and z are given by z 1 = p AF= 7i and z = + 5i where p is an integer. z 1 Find z in the form a + bi where a and b are rational, and are given in terms of p. 1 z = 5 + i z* is the complex conjugate of z.. z Show that z = a + bi * 5, where a and b are rational numbers to be found. p + 5i 1 The complex number z is defined by z = p i, p R, p > 0. 1_ Given that the real part of z is, a find the value of p DRp DR= DR_, whdi, wh b write z in the form a + bi, where a and b are real. 1. Roots of quadratic equations For real numbers a, b and c, if the roots of the quadratic equation az + bz + c = 0 are complex, then they occur as a conjugate pair. Another way of stating this is that for a real-valued quadratic function f(z), if z 1 is a root of f(z) = 0 then z 1 * is also a root. You can use this fact to find one root if you know the other, or to find the original equation. If the roots of a quadratic equation are α and β, then you can write the equation as (z α)(z β) = 0 or z (α + β)z + αβ = 0 AFz AF = isdr,dr DRp + DR+R_ f fdr zdr DR 1_DR 1DR Notation Roots of complex-valued polynomials are often written using Greek letters such as α (alpha), β (beta) and γ (gamma). xample 9 Given that α = 7 + i is one of the roots of a quadratic equation with real coefficients, a state the value of the other root, β b find the quadratic equation c find the values of α + β and αβ and interpret the results. a β = 7 i b (z α)(z β) = 0 (z (7 + i))(z (7 i)) = 0 z z(7 i) z(7 + i) + (7 + i)(7 i) = 0 z 7z + iz 7z iz + 9 1i + 1i i = 0 z 1z = 0 z 1z + 5 = 0 c α + β = (7 + i) + (7 i) = (7 + 7) + ( + ( ))i = 1 The coefficient of z in the above equation is (α + β). αβ = (7 + i)(7 i) = 9 1i + 1i i = 9 + = 5 The constant term in the above equation is αβ.. xercise 1 1 The roots of the quadratic equation z + z z + 6 = 0 are α and β. Find: a α and β b DRβ α + β c αβ The roots of the quadratic equation Rc DR8 z 8z + 5 = 0 are α and β. Find: a α and β b α + β c αβ Given that + i is one of roots of a quadratic equation with real coefficients, a write down the other root of Dthe the equation b find the quadratic equation, giving your answer in the form az + bz + c = 0 where a, b and c are real constants. Given that 5 i is a root of the equation z + pz + q = 0, where p and q are real constants, a write down the other root of the equation b find the value of p and the value of q. 6RA αdr +DR βdr + b α + α and β will always be a complex conjugate pair. roblem-solving 5 Given that z 1 = 5 + i is one of the roots of the quadratic equation z + bz + c = 0, where b and c are real constants, find the values of b and c. The quadratic equation with roots α and β is (z α)(z β) = 0 Collect like terms. Use the fact that i = 1. zz*ft For z = a + bi, you should learn the results: z + z* * = a zz* * = a + b You can use these to find the quadratic equation FTbi, FTi, quickly. +FT zft a + bdr 6 Given that 1 + i is one of the roots of a quadratic equation with real coefficients, find the equation giving your answer in the form z + bz + c = 0 where b and c are integers to be found. ( marks) ( marks) 8 9

7 7 Given that 5i is one of the roots of a quadratic equation with real coefficients, find the equation giving your answer in the form z + bz + c = 0 where b and c are real constants. 8 z = 5 _ i a Find z in the form a + bi, where a and b are real constants. Given that z is a complex root of the quadratic equation x + px + q = 0, where p and q are real integers, b find the value of p and the value of q. 9 Given that z = 5 + qi is a root of the equation z pz + = 0, where p and q are positive real constants, find the value of p and the value of q. 1.5 Solving cubic and quartic equations You can generalise the rule for the roots of quadratic equations to any polynomial with real coefficients. If f(z) is a polynomial with real coefficients, and z 1 is a root of f(z) = 0, Hint Note that if z 1 is real, then z 1 * = z 1. then z 1 * is also a root of f(z) = 0. You can use this property to find roots of cubic and quartic equations with real coefficients. An equation of the form az + bz + cz + d = 0 is called a cubic equation, and has three roots. For a cubic equation with real coefficients, cients, either Watch out A real-valued cubic all three roots are real, or equation might have two, or three, one root is real and the other two roots form a repeated real roots. complex conjugate pair. xample 10 DfindD Given that 1 is a root of the equation z z + z + k = 0, a find the value of k bdb find the other two roots of the equation. a If 1 is a root, ( 1) ( 1) + ( 1) + k = k = 0 k = 5 b 1 is a root of the equation, so z + 1 is a factor of z z + z + 5. _ z z + 5 z + 1 ) z z + z + 5 z + z z + z z z 5z + 5 5z roblem-solving Use the factor theorem to help: if f(α) = 0, then α is a root of the polynomial and z α is a factor of the polynomial. Use long division (or another method) to find the quadratic factor. z z + z + 5 = (z + 1)(z z + 5) = 0 Solving z z + 5 = 0 z z = (z 1) 1 z z + 5 = (z 1) = (z 1) + (z 1) + = 0 (z 1) = z 1 = ± = ±i z = 1 ± i z = 1 + i, z = 1 i So the other two roots of the equation are 1 + i and 1 i. An equation of the form az + bz + cz + dz + e = 0 is called a quartic equation, and has four roots. For a quartic equation with real coefficients, either FTA Watch out A real-valued quartic all four root are real, or equation might have repeated real two roots are real and the other two roots form a roots or repeated complex roots. complex conjugate pair, or two roots form a complex conjugate pair and the other two roots also form a complex conjugate pair. xample 11 DR Given that + i is a root of the quartic equation z 9z + 10z 50 = 0, solve the equation completely. Complex roots occur in Another root is i. conjugate pairs. So (z ( + i))(z ( of z z 9z RAz A i)) is a factor + 10z z 50 If α and β are roots of f(z) = 0, (z ( + i))(z ( i)) = z z( i) z( ( + i) + ( + i)( i) then (z α)(z β ) is a factor = z 6zDz +D+ 10 of f(z). So z 6z + 10 is a factor of z z 9z + 10z 50. D D6D DDD10 You can work this out quickly by (z 6z + 10)(az + bz + c) = z z 9z + 10z 50 noting that Consider z : (z (a + bi))(z (a bi)) The only z term in the expansion is z az, so a =. = z az + a + b (z 6z + 10)(z + bz + c) = z z 9z + 10z 50 z(dr zdr DDRA i)dr DRAz RA RA ( i) zdz Consider z : The z terms in the expansion are z bz and 6z z, so bz 1z = z b 1 = so b = 9 (z 6z + 10)(z + 9z + c) = z z 9z + 10z 50 The other two roots are found by solving the quadratic equation. Solve by completing the square. Alternatively, you could use the quadratic formula. The quadratic equation has complex roots, which must be a conjugate pair. You could write the equation as (z + 1)(z (1 + i))(z (1 i)) = 0 roblem-solving It is possible to factorise a polynomial without using a formal algebraic method. Here, the polynomial is factorised by inspection. By considering each term of the quartic separately, it is possible to work out the missing coefficients

8 Consider 50: The only constant term in the expansion is 10 c, so c = 5. z z 9z + 10z 50 = (z 6z + 10)(z + 9z 5) Solving z + 9z 5 = 0: (z 1)(z + 5) = 0 z = 1_, z = 5 So the roots of z z 9z + 10z 50 = 0 are 1_, 5, + i and i xample 1 Show that z + is a factor of z z + 1z 8z Hence solve the equation z z + 1z 8z + 68 = 0. You can check this by considering the z and z terms in the expansion. Using long division: Alternatively, the quartic can be _ z z + 17 factorized by inspection: z + ) z z + 1z 8z + 68 z z + 1 z 8z + 68 z + z ( z + ) (a z + bz + c) z + 17z 8z a = 1, as the leading coefficient z 8z is 1. 17z z The only z + 68 term is formed by z bz so b =. 0 AFTAlternativel = So z z + 1z 8z + 68 = (z z + )(z z z + 17) = 0 The constant term is formed by ither z + = 0 or z c, so c = 68, and c = 17. z z + 17 = 0 Solving z + = 0: z = x = ±i Solving z z + 17 = 0: (z 1) + 16 = 0 (z 1) = 16 Solve by completing the square. Alternatively, you could use the quadratic formula. z 1 = ±i z = 1 ± i Watch out You could also use So the roots of z z + 1z 8z + 68 = 0 are i, i, 1 + i and 1 i your calculator to solve z z + 17 = 0. You should still write down the equation you are solving, and both roots. z +RA zra zft xercise 1F 1 f(z) = z 6 z + 1z 6 a Show that f() = 0. b Hence solve f(z) = 0 completely. f(x) = z + 5 z + 9z 6 a Show that f ( 1_ ) = 0. b Hence write f(z) in the form (z 1)(z + bz + c), where b and c are real constants to be found. c Use algebra to solve f(z) = 0 completely. g(x) = z z 5z Given that z = is a root of the equation g(z) = 0, solve g(z) = 0 completely. p(z) = z + z 15z 68 Given that z = + i is a solution to the equation p(z) = 0, a show that z + 8z + 17 is a factor of p(z) b hence solve p(z) = 0 completely. 5 f(z) = z + 9 z + z + 5 Given that f(z) = (z + 1)(z + az + b), where a and b are real constants, a find the value of a and the value of b b find the three roots of f(z) = 0 c find the sum of the three e roots of f(z) = 0. 6 g(z) = z 1 z + cz + d = 0, where c, d R Given that 6 and equation g(z) = 0, a write down the of the equation b find the value of c and of d. 7 h(z) = z + z + z + 1 DRc, DRR + i are roots of the other complex root the value Dnd Given that z + 1 is a factor of h(z), find the three roots of h(z) = 0. 8 f(z) = z 6 z + 8z + k Given that f() = 0, a find the value of k b find the other two roots of the equation. 0.RA DR, DRd DR d R ( marks) ( marks) ( marks) ( marks) ( marks) ( marks) 9 Find the four roots of the equation z 16 = f(z) = z 1 z + 1 z + 108z 60 a Write f(z) in the form (z 9)(z + bz + c), where b and c are real constants to be found. b Hence find all the solutions to f(z) = 0. ( marks) ( marks) 1 1

9 11 g(z) = z + z z + 8z + 10 Given that g( + i) = 0, find all the roots of g(z) = 0. 1 f(z) = z 10 z + 71 z + Qz +, where Q is a real constant. Given that z = i is a root of the equation f(z) = 0, a show that z 6z + is a factor of f(z) b find the value of Q c solve completely the equation f(z) = 0. Challenge Three of the roots of the equation az 5 + bz + cz + dz + ez + f = 0 are, i and 1 + i. Find the values of a, b, c, d, e and f. Mixed exercise 1 1 Given that z 1 = 8 i and z = + i, find, in the form a + bi, where a, b R, a z 1 + z b z c 6z 1 z FTa, ara The equation z + bz + 1 = 0, where b R has no real roots. Find the range of possible values of b.. The solutions to the quadratic equation z 6z z + 1 = 0 are z 1 and z. Find z 1 and z, giving each answer in the form a ± i b. ( marks) ( marks) By using the binomial ial expansion, or otherwise, show that (1 + i) 5 = 1 8i. ( marks) 5 f(z) = z 6z + 10 Show that z = + i is a solution to f(z) = 0. 6 z 1 = + i, z = + i Dolution xpress, in the form a + bi, where a, b R, z 1 z a z * 1 b z 1 z c 7 Write (7 i) 1 + i in the form x + iy where x, y R. ( marks) 7i 8 Given that z = + i, find z in the form a + bi, where a, b R. ( marks) _ 1 9 z = + i xpress in the form a + bi, where a, b R a z b z 1 z FT, FTb FT b R 6zRA 10 Given that z = a + bi, show that z z * = ( a b a + b ) + ( ab a + b ) i 11 The complex number z is defined by z = 1 Given that the real part of z is 1, a find the possible values of q + qi q 5i, where q R. b write the possible values of z in the form a + bi, where a and b are real constants. 1 Given that z = x + iy, find the value of x and the value of y such that z + i z * = + 18i where z* is the complex conjugate of z. 1 z = 9 + 6i, w = i z xpress w in the form a + bi, where a and b are real constants. q + i 1 The complex number z is given by z = where q is an integer. + qi xpress z in the form a + bi where a and b are rational and are given in terms of q. 15 Given that 6 i is one of the roots of a quadratic equation with real coefficients, a write down the other root of the equation b find the quadratic equation, giving your answer in the form az + bz + c = 0 where a, b and c are real constants. 16 Given that z = ki is a root of the equation z mz + 5 = 0, where k and m are positive real constants, find the value of k and the value of m. 17 h(z) = z 11z + 0 Given that + i is a root of the equation h(z) = 0, solve h(z) = 0 completely. 18 f(z) = z + 6z + 0 Given that f(1 + i) = 0, solve f(z) = 0 completely. 19 f(z) = z + z + kz + 8 Given that f(i) = 0, a find the value of k b find the other two roots of the equation. 0 f(z) = z z 16 z 7z 60 a Write f(z) in the form (z 5z 6)(z + bz + c), where b and c are real constants to be found. b Hence find all the solutions to f(z) = 0. 1 g(z) = z 6 z + 19 z 6z + 78 Given that g( i) = 0, find all the roots of g(z) = 0. f(z) = z z z + pz + 16 Given that f() = 0, a find the value of p b solve completely the equation f(z) = 0. +AF FT az mz z)dr z)ra ) =DR (5 marks) ( marks) ( marks) ( marks) ( marks) ( marks) (5 marks) 1 15

10 Challenge a xplain why a cubic equation with real coefficients cannot have a repeated complex root. b By means of an example, show that a quartic equation with real coefficients can have a repeated complex root. Summary of key points 1 i = 1 and i = 1 An imaginary number is a number of the form bi, where b R. A complex number is written in the form a + bi, where a, b R. can be added or subtracted by adding or subtracting their real parts and adding or subtracting their imaginary parts. You can multiply a real number by a complex number by multiplying out the brackets in the usual way. βra If b ac < 0 then the quadratic equation ax + bx + c = 0 has two distinct complex roots. For any complex number z = a + bi, the complex conjugate of the number is defined as z* = a bi. 5 For real numbers a, b and c, if the roots of the quadratic equation az + bz + c = 0 are complex, then they occur as a conjugate pair. 6 If the roots of a quadratic equation are α and β, then you can write the equation as (z α)(z β) = 0 or z (α + β)z + = 0 7 If f(z) is a polynomial coefficients, and z 1 is a root of f(z) = 0, then z 1 * is also a root of f(z) = 0. 8 An equation of the form + cz + d = 0 is called a cubic equation, and has three roots. For a cubic equation with real DRβ) DR)z DR+ αβrαβ with real Daz D +D+ bzdbz D coefficients, either all three roots are real, or ald one root is real and the other two roots form a complex conjugate pair. 9 An equation of the form az + bz + cz + dz + e = 0 is called a quartic equation, and has four roots. For a quartic equation with real coefficients, either all four root are real, or DRα DR+ DRz + β + two roots are real and the other two roots form a complex conjugate pair, or two roots form a complex conjugate pair and the other two roots also form a complex conjugate pair. 16

11 Order now! Visit our webpage to see all products in this series, and order now: Speak to us For help and advice you can request a call back or appointment with one of our maths experts: UK W876d earson is committed to reducing its impact on the environment by using responsibly sourced and recycled paper.