Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
10 Complex Numbers Assessment statements ___ 1.5 Complex numbers: the number i 5 √ 21 ; the term’s real part, imaginary part, conjugate, modulus and argument. Cartesian form z 5 a 1 ib. Sums, products and quotients of complex numbers. 1.6 Modulus–argument (polar) form z 5 r (cosu 1 i sinu) = rcis(u) = reiu. The complex plane. 1.7 De Moivre’s theorem. Powers and roots of a complex number. 1.8 Conjugate roots of polynomial equations with real coefficients. Introduction You have already met complex numbers in Chapters 1 and 3. This chapter will broaden your understanding to include trigonometric representation of complex numbers and some applications. Fractals can be generated using complex numbers. Solving a linear equation of the form ax 1 b 5 0, with a 0 is a straightforward procedure if we are using the set of real numbers. The situation, as you already know, is different with quadratic equations. For example, as you have seen in Chapter 3, solving the quadratic equation 428 x 2 1 1 5 0 over the set of real numbers is not possible. The square of any real number has to be non-negative, i.e. (x 2 > 0 ⇔ x2 1 1 > 1) ⇒ x 2 1 1 > 0 for any choice of a real number x. This means that x 2 1 1 5 0 is impossible for every real number x. This forces us to introduce a new set where such a solution is possible. The situation with finding a solution to x2 1 1 5 0 is analogous to the following scenario: For a child in the first or second grade, a question such as 5 1 ? 5 9 is manageable. However, a question such as 5 1 ? 5 2 is impossible because the student’s knowledge is restricted to the set of positive integers. However, at a later stage when the same student is faced with the same question, he/she can solve it because their scope has been extended to include negative numbers too. Also, at early stages an equation such as x2 5 5 cannot be solved till the student’s knowledge of sets is extended__to include irrational numbers where he/she can recognize numbers such as x 5 6 √5 . The situation is much the same for x2 1 1 5 0. We extend our number system to ___ include numbers such as √ 21 ; i.e. a number whose square is 21. 10.1 ___ Numbers such as √ 21 are not intuitive and many mathematicians in the past resisted their introduction, so they are called imaginary numbers. Thanks to Euler’s (1707–1783) seminal work on imaginary numbers, they now feature prominently in the number system. Euler skilfully employed them to obtain many interesting results. Later, Gauss (1777–1855) represented them as points in the plane and renamed them as complex numbers, using them to obtain various significant results in number theory. Complex numbers, sums, products and quotients Electronic components like capacitors are used in AC circuits. Their effects are represented using complex numbers. As you have seen in the introduction, the development of complex numbers had its origin in the search for methods of solving polynomial equations. The quadratic formula ________ √ b 2 2 4ac x 5 ___ 2b 6 _________ 2a 2a had been used earlier than the 16th century to solve quadratic equations – in more primitive notations, of course. However, mathematicians stopped short of using it for cases where b 2 2 4ac was negative. The use of the formula in cases where b 2 2 4ac is negative depends on two principles (in 429 10 Complex Numbers addition to the other principles inherent in the set of real numbers, such as associativity and commutativity of multiplication). ___ ___ ___ __ 1. √21 √21 5 21 ___ 2. √2k 5 √ k √21 for any real number k . 0 Example 1 ____ ____ Multiply √ 236 √249 . Solution First we simplify each square root using rule 2. ____ ___ ___ ___ ____ ___ ___ ___ √ 236 5 √ 36 √21 5 6 √21 √ 249 5 √ 49 √21 5 7 √21 And hence using rule 1 with the other obvious rules: ____ ____ ___ ___ ___ ___ √ 236 √ 249 5 6 √21 7 √21 5 42 √21 √21 5 242 To deal with the quadratic formula expressions that consist of combinations of real numbers and square roots of negative numbers, we can apply the rules of binomials to numbers of the form ___ a 1 b √ 21 ___ ___ where a and b are real numbers. For example, to add 5 1 7 √21 to 2 2 3 √21 we combine ‘like’ terms as we do in polynomials: ___ ___ ___ ___ (5 1 7 √21 ) 1 (2 2 3 √21 )5 5 1 2 1 7 √21 2 3 √21 ___ ___ 5 (5 1 2) 1 (7 2 3) √21 5 7 1 4 √ 21 Similarly, to multiply these numbers we use the binomial multiplication procedures: ___ ___ ___ ___ ___ (5 1 7 √21 ) (2 2 3 √21 ) 5 5 2 1___ (7 √21 ) (23 √21 ) 1 5 (23 √21 ) 1 (7 √21 ) 2 ___ ___ ___ 5 10 2 21 ( √ 21 )2 2 15 √ 21 1 14 √21 ___ 5 10 2 21 (21) 1 (215 1 14) √21 ___ 5 31 2 √ 21 ___ Euler introduced the symbol i for √ 21 . A pure imaginary number is a number of the form ki, where k is a real number and i, the imaginary unit, is defined by i 2 5 21. Note: In some cases, especially in engineering sciences, the number i is sometimes denoted as j. Note: With this definition of i, a few interesting results are immediately apparent. For example, i 3 5 i 2 i 5 21 i 5 2i, and i 4 5 i 2 i 2 5 (21) (21) 5 1, and so i 5 5 i 4 i 5 1 i 5 i, and also i 6 5 i 4 i 2 5 i 2 5 21; i 7 5 2i, and finally i 8 5 1. 430 This leads you to be able to evaluate any positive integer power of i using the following property: i 4n 1 k 5 i k, k 5 0, 1, 2, 3. So, for example i 2122 5 i 2120 1 2 5 i 2 5 21. ___ Example 2 Simplify ____ ____ ____ a) √236 1 √ 249 ____ b) √236 √249 Solution ____ ____ ___ ___ ___ ___ a) √236 1 √ 249 5 √ 36 √21 1 √ 49 √21 5 6i 1 7i 5 13i ____ ____ b) √236 √249 5 6i 7i 5 42i 2 5 42(21) 5 242 Gauss introduced the idea of complex numbers by giving them the following definition. A complex number is a number that can be written in the form a 1 bi where a and b are real numbers and i 2 5 21. a is called the real part of the number and b is the imaginary part. We do not define i 5 √ 21 for a reason. It is the convention in mathematics that when we __ write √ 9 then we mean the non-negative square root of 9, namely 3. We do not mean 23! i does not belong to this category since we cannot say that i is the positive square root of 21, i.e. i . 0. If we do, then 21 5 i i . 0, which is false, and if we say i , 0, then 2i . 0, and 21 5 2i 2i . 0, which is also false. Actually 2i is also a square root of 21 because 2i 2i 5 i 2 5 21. With this in mind, we can use a ‘convention’ which calls i the principal square root of 21 ___ and write i 5 √ 21 . Notation It is customary to denote complex numbers with the variable z. z 5 5 1 7i is the complex number with real part 5 and imaginary part 7 and z 5 2 2 3i has 2 as real part and 23 as imaginary. It is usual to write Re(z) for the real part of z and Im(z) for the imaginary part. So, Re(2 1 3i ) 5 2 and Im(2 1 3i ) 5 3. Note that both the real and imaginary parts are real numbers! Algebraic structure of complex numbers A GDC can be set up to do basic complex number operations. For example, if you have a TI-84 Plus, the set up is as follows. SCI ENG FLOAT 0 1 2 3 4 5 6 7 8 9 RADIAN DEGREE FUNC PAR POL SEQ CONNECTED DOT SEQUENTIAL SIMUL REAL a+bi re^θi FULL HORIZ G-T SET CLOCK12/01/08 6:39AM Gauss’ definition of the complex numbers triggers the following understanding of the set of complex numbers as an extension to our number sets in algebra. The set of complex numbers C is the set of ordered pairs of real numbers C 5 {z 5 (x, y): x, y }, with the following additional structure: Equality Two complex numbers z1 5 (x1, y1) and z2 5 (x2, y2) are equal if their corresponding components are equal: (x1, y1) 5 (x2, y2) if x1 5 x2 and y1 5 y2. That is, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. 431 10 Complex Numbers This is equivalent to saying: a 1 bi 5 c 1 di ⇔ a 5 c and b 5 d. For example, if 2 2 ( y 2 2)i 5 x 1 3 1 5i, then x must be 21 and y must be 23. Explain why. An interesting application of the way equality works is in finding the square roots of complex numbers without a need for the trigonometric forms developed later in the chapter. Find the square root(s) of z 5 5 1 12i. Let the square root of z be x 1 yi, then (x 1 yi )2 5 5 1 12i ⇒ x2 2 y2 1 2xyi 5 5 1 12i ⇒ x2 2 y2 5 5 and 6 , and when we substitute this value in x2 2 y2 5 5, 2xy 5 12 ⇒ xy 5 6 ⇒ y 5 __ ( ) x 6 2 5 5. This simplifies to x 4 2 5x2 2 36 5 0 which yields x2 5 24 x we have x2 2 __ or x2 5 9, ⇒ x 5 63. This leads to x 5 62i, that is, the two square roots of 5 1 12i are 3 1 2i or 23 2 2i. (3+2i)2 (-3–2i)2 5+12i 5+12i Addition and subtraction for complex numbers are defined as follows: Addition (x1, y1) 1 (x2, y2) 5 (x1 1 x2, y2 1 y2) This is equivalent to saying: (a 1 bi) 1 (c 1 di) 5 (a 1 c) 1 (b 1 d)i. Multiplication (x1, y1)(x2, y2) 5 (x1x2 2 y1y2, x1y2 1 x2y1) This is equivalent to using the binomial multiplication on (a 1 bi)(c 1 di): (a 1 bi) (c 1 di) 5 ac 1 bdi 2 1 adi 1 bci 5 ac 2 bd 1 (ad 1 bc)i Addition and multiplication of complex numbers inherit most of the properties of addition and multiplication of real numbers: z 1 w 5 w 1 z and zw 5 wz (Commutativity) z 1 (u 1 v) 5 (z 1 u) 1 v and z(uv) 5 (zu)v (Associativity) z (u 1 v) 5 zu 1 zv (Distributive property) A number of complex numbers take up unique positions. For example, the number (0, 0) has the properties of 0: (x, y) 1 (0, 0) 5 (x, y) and (x, y)(0, 0) 5 (0, 0). It is therefore normal to identify it with 0. The symbol is exactly the same symbol used to identify the ‘real’ 0. So, the real and complex zeros are the same number. Another complex number of significance is (1, 0). This number plays an important role in multiplication that stems from the following property: (x, y) (1, 0) 5 (x 1 2 y 0, x 0 1 y 1) 5 (x, y) 432 For complex numbers, (1, 0) behaves like the identity for multiplication for real numbers. Again, it is normal to write (1, 0) 5 1. The third number of significance is (0, 1). It has the notable characteristic of having a negative square, i.e. (0, 1)(0, 1) 5 (0 0 2 1 1, 0 1 1 1 0) 5 (21, 0) Using the definition above, (0, 1) 5 0 1 1i 5 i. So, the last result should be no surprise to us since we know that i i 5 21 5 (21, 0). Since (x, y) represents the complex number x 1 yi, then every real number x can be written as x 1 0i 5 (x, 0). The set of real numbers is therefore a subset of the set of complex numbers. They are the complex numbers whose imaginary part is 0. Similarly, pure imaginary numbers are of the form 0 1 yi 5 (0, y). They are the complex numbers whose real part is 0. Notation So far, we have learned how to represent a complex number in two forms: (x, y) and x 1 yi. Now, from the properties above (x, y) 5 (x, 0) 1 (0, y) 5 (x, 0) 1 (y, 0)(0, 1) (Check the truth of this equation.) This last equation justifies why we can write (x, y) 5 x 1 yi. Example 3 Simplify each expression. a) (4 2 5i) 1 (7 1 8i) b) (4 2 5i) 2 (7 1 8i) c) (4 2 5i)(7 1 8i) Solution a) (4 2 5i) 1 (7 1 8i) 5 (4 1 7) 1 (25 1 8)i 5 11 1 3i b) (4 2 5i) 2 (7 1 8i) 5 (4 2 7) 1 (25 2 8)i 5 23 2 13i (4–5i) (8i) -.625–.5i Ans Frac -5 8–1 2i (4–5i) (7+8i) 68–3i c) (4 2 5i)(7 1 8i) 5 (4 7 2 (25) 8) 1 (4 8 1 (25) 7)i 5 68 2 3i Division Multiplication can be used to perform division of complex numbers. a 1 bi , involves finding a complex The division of two complex numbers, ______ c 1 di number (x 1 yi) satisfying ______ a 1 bi 5 x 1 yi; hence, it is sufficient to find c 1 di the unknowns x and y. 433 10 Complex Numbers Example 4 . Find the quotient ______ 2 1 3i 1 1 2i Solution 2 1 3i 5 x 1 iy. Hence, using multiplication and the equality of Let ______ 1 1 2i complex numbers, 2 1 3i 5 (1 1 2i)(x 1 iy) ⇔ 2 1 3i 5 x 2 2y 1 i(2x 1y) ⇔ { 2 5 x 2 2y 8 , y 5 __ 1 ⇒ x 5 __ 5 5 3 5 2x 1 y 2 1 3i Thus, ______ 5 __ 8 2 __ 1 i. 1 1 2i 5 5 (2+3i) (1+2i) 1.6-.2i Ans Frac 8 5–1 5i a 1 bi 5 x 1 yi ⇔ a 1 bi 5 (x 1 yi)(c 1 di). Now, in general, ______ c 1 di With the multiplication as described above: a 1 bi 5 (cx 2 dy) 1 (dx 1 cy)i Again by applying the equality of complex numbers property above we get a system of two equations that can be solved. { cx 2 dy 5 a bc 2 ad ⇒ x 5 _______ ac2 1 bd2 ; y 5 _______ c 1 d c 2 1 d 2 dx 1 cy 5 b The denominator c 2 1 d 2 resulted from multiplying c 1 di by c 2 di , which is its conjugate. Conjugate Although the conjugate notation z * will be used in the book, in your own work you can use any notation you feel comfortable with. You just need to understand that the IB questions use this one. With every complex number (a 1 bi) we associate another complex number (a 2 bi) which is called its conjugate. The conjugate of number z is most often denoted with a bar over it, sometimes with an asterisk to the right of it, occasionally with an apostrophe and even less often with the plain symbol Conj as in _ z 5 z * 5 z9 5 Conj(z). In this book, we will use z * for the conjugate. The importance of the conjugate stems from the following property (a 1 bi )(a 2 bi ) 5 a2 2 b 2i 2 5 a2 1 b 2 which is a non-negative real number. So the product of a complex number and its conjugate is always a real number. 434 Example 5 Find the conjugate of z and verify the property mentioned above. a) z 5 2 1 3i b) z 5 5i c) z 5 11 Solution a) z * 5 2 2 3i, and (2 1 3i )(2 2 3i) 5 4 2 9i 2 5 4 1 9 5 13. b) z * 5 25i, and (5i )(25i ) 5 25i2 5 (25)(21) 5 5. c) z * 5 11, and 11 11 5 121. So, the method used in dividing two complex numbers can be achieved by multiplying the quotient by a fraction whose numerator and denominator are the conjugate c 2 di. 2 di) ______ _____ a 1 bi 5 _____ a 1 bi _____ c 2 di 5 __________ (a 1 2bi)(c 5 ac2 1 bd2 1 ______ bc2 2 ad2 i c 1 di c 1 di c 2 di c 1 d 2 c 1 d c 1 d Example 6 Find each quotient and write your answer in standard form. 4 2 5i a) ______ 7 1 8i 4 2 5i b) ______ 8i 4 2 5i c) ______ 7 Solution 28 2 40 1 (232 2 35)i 7 2 8i 4 2 5i 12 67 i 5 ______ 4 2 5i ______ 5 ____________________ a) ______ 5 2 ___ 2 ___ 113 113 49 1 64 7 1 8i 7 1 8i 7 2 8i 4 2 5i 4 2 5i 28i 5 _________ 40 5 2 __ b) ______ 1 i 5 ______ ____ 232i 2 5 2 __ 8 2 64 8i 8i 28i 4 2 5i 4 5 i c) ______ 5 __ 2 __ 7 7 7 (4–5i) (7+8i) -.1061946903–.5… Ans Frac -12 113–67 113i (4–5i) (8i) -.625–.5i Ans Frac -5 8–1 2i Example 7 Solve the system of equations and express your answer in Cartesian form. (1 1 i )z1 2 iz2 5 23 2z1 1 (1 2 i )z2 5 3 2 3i 435 10 Complex Numbers Solution Multiply the first equation by 2, and the second equation by (1 1 i). 2(1 1 i )z1 2 2iz25 26 2(1 1 i )z1 1 (1 1 i)(1 2 i )z25 (1 1 i )(3 2 3i ) 2(1 1 i )z1 1 2z25 6 (1) (2) By subtracting (2) from (1), we get (22 2 2i )z25 212 5 3 2 3i z2 5 _______ 212 22 2 2i 23 1 i(3 2 3i) __ z15 _____________ 3 5 3 1 __ 2 2i 11i And hence Properties of conjugates Here is a theorem that lists some of the important properties of conjugates. In the next section, we will add a few more to the list. Theorem Let z, z1 and z2 be complex numbers, then (1)(z *) * 5 z (2)z * 5 z if and only if z is real. (3)(z1 1 z2) * 5 z1 * 1 z2 * The conjugate of the sum is the sum of conjugates. (4)(2z) * 5 2z * The product can be extended to powers of complex numbers, i.e. (z2)* 5 (z z)* 5 z* z* 5 (z*)2. This result can be generalized for any non-negative integer power n, i.e. (z n)* 5 (z *)n and can be proved by mathematical induction. The basis case, when n 5 0, is obviously true: (z 0)* 5 1 5 (z *)0. Now assume (z k)* 5 (z *) k. (z k 1 1)* 5 (z kz)* 5 (z k)*z* 5 (z *) kz * (using the product rule). Therefore, (z k 1 1)* 5 (z *) kz* 5 (z *)k 1 1. So, since if the statement is true for n 5 k, it is also true for n 5 k 1 1, then by the principle of mathematical induction it is true for all n 0. 436 (5)(z1 z2) * 5 z1 * z2 * The conjugate of the product is the product of conjugates. (6)(z ) * 5 (z *) , if z 0. 21 21 Proof (1) and (2) are obvious. For (1), ((a 1 bi) *) * 5 (a 2 bi) * 5 a 1 bi, and for (2), a 2 bi 5 a 1 bi ⇒ 2bi 5 0 ⇒ b 5 0. (3) is proved by straightforward calculation: Let z15 x1 1 iy1 and z2 5 x2 1 iy2, then (z1 1 z2) *5 ((x1 1 iy1) 1 (x2 1 iy2)) * 5 ((x1 1 x2) 1 i (y1 1 y2)) * 5 (x1 1 x2) 2 i (y1 1 y2) 5 (x1 2 iy1) 1 (x2 2 iy2) 5 z1 * 1 z2 *. (4) can now be proved using the above results: (z 1 (2z)) *5 0 * 5 0 but, (z 1 (2z)) *5 0 * 5 z * 1 (2z) *, so z * 1 (2z) *5 0, and (2z) * 5 2z *. Also (5) is proved by straightforward calculation: (z1 z2) *5 ((x1 1 iy1) (x2 1 iy2)) * 5 ((x1x2 2 y1y2) 1 i (y1x2 1 x1y2)) * 5 (x1x2 2 y1y2) 2 i (y1x2 1 x1y2) 5 (x1 2 iy1) (x2 2 iy2) 5 z1 * z2 * And finally, (6): (z(z21)) * 5 1 * 5 1 but, (z(z21)) * 5 z *(z21) *, so z *(z21) * 5 1, 1 5 (z *)21. and (z21) * 5 __ z * Conjugate zeros of polynomials In Chapter 3, you used the following result without proof. If c is a root of a polynomial equation with real coefficients, then c * is also a root. Theorem: If c is a root of a polynomial equation with real coefficients, then c * is also a root of the equation. We give the proof for n 5 3, but the method is general. P(x) 5 ax 3 1 bx 2 1 dx 1 e Since c is a root of P(x) 5 0, we have ac 3 1 bc 2 1 dc 1 e 5 0 ⇒ (ac 3 1 bc 2 1 dc 1 e) *5 0 Since 0* 5 0. ⇒ (ac ) * 1 (bc ) * 1 (dc) * 1 e * 5 0 Sum of conjugates theorem. ⇒ a(c *) 1 b(c *) 1 d(c *) 1 e 5 0 Result of product conjugate. 3 2 3 2 ⇒ (c *) is a root of P(x) 5 0. Example 8 1 1 2i is a zero of the polynomial P(x) 5 x 3 2 5x 2 1 11x 2 15. Find all other zeros. Solution Since the polynomial has real coefficients, then 1 2 2i is also a zero. Hence, using the factor theorem, P(x) 5 (x 2 (1 1 2i))(x 2 (1 2 2i))(x 2 c), where c is a real number to be found. Now, P(x) 5 (x2 2 2x 1 5)(x 2 c). c can either be found by division or by factoring by trial and error. In either case, c 5 3. Example 91 1 1 2i is a zero of the polynomial P(x) 5 x 3 1 (i 2 2)x 2 1 (2i 1 5)x 1 8 1 i. Find all other zeros. 1 Not included in present IB syllabus. 437 10 Complex Numbers Solution Since the polynomial does not have real coefficients, then 1 2 2i is not necessarily also a zero. To find the other zeros, we can perform synthetic substitution 1 1 1 2i i22 1 1 2i 21 1 3i 1 2i 1 5 27 1 i 22 1 3i 81i 28 2 i 0 This shows that P(x) 5 (x 2 1 2 2i)(x 2 1 (21 1 3i)x 2 2 1 3i). The second factor can be factored into (x 1 1)(x 2 2 1 3i) giving us the other two zeros as 21 and 2 2 3i. Note: x2 1 (21 1 3i)x 2 2 1 3i 5 0 can be solved using the quadratic formula. _____________________ _______ 1 2 3i 6 √ (21 1 3i)2 2 4(22 1 3i) ______________________________ 2b 6 √ b2 2 4ac 5 x 5 ______________ 2a 2 ________________ 1 2 3i 6 √ 28 2 6i 1 8 2 12i _________________________ _____ 1 2 3i 6 √ 218i 5 ______________ 5 2 2 _____ √ we 218i let (a 1 bi)2 5 218i ⇒ a2 2 b2 1 2abi 5 218i, then To find equating the real parts and imaginary parts to each other: a2 2 b2 5 0 _____ and 2ab 5 218 will yield √ 218i 5 63 7 3i, and hence _____ 1 2 3i 6 (63 7 3i) 1 2 3i 6 √ 218i ________________ 2 x 5 ______________ 2 2 which will yield x 5 21 or x 5 2 2 3i. Exercise 10.1 Express each of the following numbers in the form a 1 bi. ___ ___ 1 5 1 √ 24 2 7 2 √ 27 ___ 3 26 ____ ____ 4 2 √ 49 √ 6 2 ____ 225 16 5 √281 Perform the following operations and express your answer in the form a 1 bi. 7 (23 1 4i ) 1 (2 2 5i ) 8 (23 1 4i ) 2 (2 2 5i ) 9 (23 1 4i )(2 2 5i ) 10 3i 2 (2 2 4i ) 11 (2 2 7i )(3 1 4i ) 12 (1 1 i )(2 2 3i ) 3 1 2i 13 ______ 2 1 5i 2 2 i 14 ______ 3 1 2i ( ( ) ( ) ( ) ) 15 __ 2 2 __ 1 1 __ 1 i 1 __ 1 i 3 2 3 2 2 2 __ 1 1 __ 17 __ 1 i 4 __ 1 i 3 2 3 2 1 (3 2 7i ) 19 __ i 13 21 _______ 5 2 12i 438 ( )( ) 16 __ 2 2 __ 2 1 __ 1 i __ 1 i 3 2 3 2 18 (2 1 i )(3 2 2i ) 20 (2 1 5i ) 2 (22 2 5i ) 12i 22 ______ 3 1 4i ( ) 23 3i 3 2 __ 2 i 3 39 2 52i 25 _______ 24 1 10i 24 (3 1 5i )(6 2 10i ) 26 (7 2 4i )21 3 1 ______ 28 ______ 2 3 2 4i 6 1 8i _____ 52√ 2144 ____ 30 __________ 31√ 216 27 (5 2 12i )21 (7 1 8i )(2 2 5i ) 29 _____________ 5 2 12i 31 Let z 5 a 1 bi. Find a and b if (2 1 3i )z 5 7 1 i. 32 (2 1 yi )(x 1 i ) 5 1 13i, where x and y are real numbers. Solve for x and y. __ 33 a) Evaluate (1 1 i √ 3 )3. __ b) Prove that (1 1 i √3 )6n 5 82n, where n Z1. __ c) Hence, find (1 1 i √ 3 )48. __ __ 34 a) Evaluate (2 √ 2 1 i √2 )2. __ __ b) Prove that (2 √2 1 i √2 )4k 5 (216)k, where k Z1. __ __ c) Hence, find (2 √2 1 i √ 2 )46. 35 If z is a complex number such that |z 1 4i | 5 2|z 1 i |, find the value of |z |. ______ ( |z| 5 √ x2 + y 2 where z = x + iy.) . 36 Find the complex number z and write it in the form a 1 bi if z 5 3 1 _______ 2i __ 2 2 i √2 37 Find the values of the two real numbers x and y such that (x 1 iy)(4 2 7i ) 5 3 1 2i. 38 Find the complex number z and write it in the form a 1 bi if i(z 1 1) 5 3z 2 2. 39 Find the complex number z and write it in the form a 1 bi if ______ 2 2 i √ z 5 2 2 3i. 1 1 2i _ 40 Find the values of the two real numbers x and y such that (x 1 iy)2 5 3 2 4i. 41 a) Find the values of the two real numbers x and y such that (x 1 iy)2 5 2 8 1 6i. b) Hence, solve the following equation z2 1 (1 2 i )z 1 2 2 2i 5 0. 42 If z C, find all solutions to the equation z3 2 27i 5 0. 43 Given that z 5 _12 1 2i is a zero of the polynomial f (x) 5 4x3 2 16x2 1 29x 2 51, find the other zeros. 44 Find a polynomial function with integer coefficients and lowest possible degree __ that has _12 , 21 and 3 1 i √ 2 as zeros. 45 Find a polynomial function __ with integer coefficients and lowest possible degree that has 22, 22 and 1 1 i √3 as zeros. 46 Given that z 5 5 1 2i is a zero of the polynomial f (x) 5 x3 2 7x2 2 x 1 87, find the other zeros. __ 47 Given that z 5 1 2 i √3 is a zero of the polynomial f (x) 5 3x3 2 4x2 1 8x 1 8, find the other zeros. z 5 a 1 bi, show that |a 1 bi | 5 1. 48 Let z C. If __ z* 439 10 Complex Numbers 49 Given that z 5 (k 1 i )4 where k is a real number, find all values of k such that a) z is a real number b) z is purely imaginary. 50 Solve the system of equations. 51 Solve the system of equations. iz1 1 2z2 5 3 2 i 2z1 1 (2 1 i )z2 5 7 1 2i 10.2 iz1 2 (1 1 i )z2 5 3 (2 1 i )z1 1 iz2 5 4 The complex plane Our definition of complex numbers as ordered pairs of real numbers enables us to look at them from a different perspective. Every ordered pair (x, y) determines a unique complex number x 1 yi, and vice versa. This correspondence is embodied in the geometric representation of complex numbers. Looking at complex numbers as points in the plane equipped with additional structure changes the plane into what we call complex plane, or Gauss plane, or Argand plane (diagram). The complex plane has two axes, the horizontal axis is called the real axis, and the vertical axis is the imaginary axis. Every complex number z 5 x 1 yi is represented by a point (x, y) in the plane. The real part is measured along the real axis and the imaginary part along the imaginary axis. imaginary axis 5i 3 4i 4i 3i 5 2i 2i i 5 4 3 2 1 0 i 1 2 3 4 5 real axis 2i 3i 4i The diagram above illustrates how the two complex numbers 3 1 4i and 25 1 2i are plotted in the complex plane. z x yi 0 440 Real part x Imaginary part y imaginary axis real axis Let us consider the sum of two complex numbers: z1 5 x1 1 y1i, and z2 5 x2 1 y2i As we have defined addition before: z1 1 z2 5 (x1 1 x2) 1 (y1 1 y2)i This suggests that we consider complex numbers as vectors; i.e. we regard the complex number z 5 x 1 iy as a vector in standard form whose terminal point is the complex number (x, y). Since we are representing the complex numbers by vectors, this results in some analogies between the two sets. So, adding two complex numbers or subtracting them, or multiplying by a scalar, are similar in both sets. Example 10 Consider the complex numbers z1 5 3 1 4i and z2 5 25 1 2i. Find z1 1 z2 and z1 2 z2. imaginary axis 6i z1 z2 2 6i 4i z1 z2 8 2i 2i z1 3 4i z2 5 2i 5 4 3 2 1 0 1 2 3 real axis 2i Note here that the vector representing the sum, 22 1 6i, is the diagonal of the parallelogram with sides representing 3 1 4i and 25 1 2i, while the vector representing the difference is the second diagonal of the parallelogram. The length, norm, of a vector also has a parallel in complex numbers. You recall that for a vector v 5 (x, y) the length of the vector is ______ |v | 5 √ x 2 1 y 2 . For complex numbers, the modulus or absolute value (or magnitude) of the complex number z 5 x 1 yi is ______ |z | 5 √ x 2 1 y 2 . Also of interest is the following result. z z* 5 (x 1 iy)(x 2 iy) 5 x 2 1 y 2, |z|2 5 x 2 1 y 2, and |z*|2 5 x 2 1 y 2 ⇒ z z* 5 |z |2 5 |z*|2 For example: It follows immediately that since _________ 2 2 z * 5 x 2 yi ⇒ |z *| 5 √ x 1 (2y) ______ 5 x 2 1 y 2 , then √ ______ 2 (3 1 4i )(3 2 4i ) 5 9 1 16 5 25 5 ( √32 1 42 ) |z *| 5 |z |. 441 10 Complex Numbers Example 11 Calculate the moduli of the following complex numbers a) z1 5 5 2 6i b) z2 5 12 1 5i Solution ______ ___ a) |z1| 5 |5 2 6i | 5 √ 52 1 62 5√ 61 _______ ____ b) |z2| 5 |12 1 5i | 5 √ 122 1 52 5√ 169 = 13 Example 12 Graph each set of complex numbers. a) A 5{z| |z | 5 3} b) B 5{z | |z | < 3} Solution a) A is the set of complex numbers whose distance from the origin is 3 units. So, the set is a circle with radius 3 and centre (0, 0) as shown. A |z| 3 3 3 O B b) B is the set of complex numbers whose distance from the origin is less than or equal to 3. So, the set is a disk of radius 3 and centre at the origin. |z| 3 3 O 3 Another important property is the following result: |z1z2| 5 |z1| |z2| Proof: _________________________ (x1x2 2 y1y2)2 1 (x1y2 1 x2y1)2 |z1z2| 5 |(x1x2 2 y1y2) 1 (x1y2 1 x2y1)i| 5 √ _________________________________________________ 2 2 2 5√ (x1x2)2 2 2x1x2y 1y2 1 (y1y2) 1 (x1y2) 1 2x1y2x2y1 1 (x2y1) ____________________________ 2 5√ (x1x2)2 1 (y1y2) 1 (x1y2)2 1 (x2y1)2 But, ________ ________ _________________ |z1| |z2| 5 √ x12 1 y12 •√ x22 1 y22 5√ (x12 1 y12)(x22 1 y22) ____________________________ 2 5√ (x1x2)2 1 (y1y2) 1 (x1y2)2 1 (x2y1)2 And so the result follows. 442 Example 13 Evaluate |(3 1 4i)(5 1 12i)|. Solution ______ ________ √ |(3 1 4i)(5 1 12i)| 5 |3 1 4i| |5 1 12i | 5 √ 9 1 16 25 1 144 5 5 3 13 5 65, ____________ _____ or |(3 1 4i)(5 1 12i)| 5 |233 1 56i | 5 √ (233)2 1 562 5 √ 4255 5 65 Trigonometric/polar form of a complex number imaginary axis r |z| Imaginary part y z x yi θ 0 Real part x real axis We know by now that every complex number z 5 x 1 yi can be considered as an ordered pair (x, y). Hence, using our knowledge of vectors, we can introduce a new form for representing complex numbers – the trigonometric form (also known as polar form). The trigonometric form uses the modulus of the complex number as its distance from the origin, r > 0, and u the angle the ‘vector’ makes with the real axis. ______ y x2 1 y 2 ; and tan u 5 __ x . Clearly x 5 r cos u and y 5 r sin u ; r 5 √ Therefore, z 5 x 1 yi 5 r cos u 1 (r sin u)i 5 r(cos u 1 i sin u). The angle u is called the argument of the complex number, arg(z). Arg(z) is not unique. However, all values differ by a multiple of 2p. Note: The trigonometric form is called ‘modulus-argument’ by the IB. Please keep that in mind. Also this trigonometric form is abbreviated, for ease of writing, as follows: z 5 x 1 yi 5 r(cos u 1 i sin u) 5 r cis u. (cis u stands for cos u 1 i sin u.) 443 10 Complex Numbers Example 14 Write the following numbers in trigonometric form. __ a) z 5 1 1 i b) z 5 √ 3 2 i c) z 5 25i d) z 5 17 Solution ______ __ a) r 5 √ 12 1 12 5√ 2 ; tan u 5 __ 1 5 1. 1 Hence, by observing the real and imaginary parts being positive, we can conclude that the argument must be u 5 __ p . 4 __ __ p 1 i sin __ p 5 √ p 2 cis __ z5√ 2 ( cos __ 4 4) 4 y θ 11π 6 0 i y i 0 z1i θ π4 x ____________ 3 x z 3i __ __ 2 __ b) r 5 √ ( √3 )2 1 (21) 5 √ 4 5 2; tan u 5 ___ 21 . The real part is positive, √ 3 the imaginary part is negative, and the point is therefore in the fourth quadrant, so u 5 ____ . 11p 6 11p 11p 11p 1 i sin ____ 5 2 cis ____ z 5 2 cos ____ 8 6 6 p . We can also use u 5 2 __ 6 ( ) c) r 5 5 and u 5 ___ 3p since it is on the negative side of the imaginary axis. 2 y 3p 3 p z 5 5 cos ___ 1 i sin ___ 2 2 p . We can also use u 5 2 __ 2 θ 3π 2 d) r 5 17 and u 5 0 0 ( ) x z 5 17 (cos 0 1 i sin 0) z 5 i Example 15 Convert each complex number into its rectangular form. 4p a) z 5 3 cos 150° 1 3i sin 150° b) z 5 12 cis ___ 3 p 1 i sin __ p c) z 5 6(cos 50° 1 i sin 50°) d) z 5 15( cos __ 2 2) Solution ( __ ) ( ) __ 2 √3 23 √3 __ 3i 1 a) z 5 3 _____ 1 3i __ 5 _____ 1 2 2 2 2 444 __ √ 3 4p 5 12 ___ 4p 1 12i sin ___ 21 1 12i 2 ___ 5 26 2 ___ 6i__ b) z 5 12 cos ___ 3 3 2 2 √ 3 c) z 5 6 cos 50° 1 6i sin 50° 5 6 0.643 1 6i 0.766 5 3.857 1 4.596i d) z 5 15(0 1 i ) 5 15i Multiplication The trigonometric form of the complex number offers a very interesting and efficient method for multiplying complex numbers. The analogy between complex numbers and vectors stops at multiplication. As you recall, multiplication of vectors is not ‘well defined’ in the sense that there are two products – the scalar product which is a scalar, not a vector, and the vector product (discussed later) which is a vector but is not in the plane! Complex number products are complex numbers! Let z1 5 r1(cos u1 1 i sin u1) and z2 5 r2(cos u2 1 i sin u2) be two complex numbers written in trigonometric form. Then z1z25 (r1(cos u1 1 i sin u1))(r2(cos u2 1 i sin u2) 5 r1r2[(cos u1 cos u2 2 sin u1 sin u2) 1 i (sin u1 cos u2 1 sin u2 cos u1)]. Now, using the addition formulae for sine and cosine, we have z1z2 5 r1r2[(cos(u1 1 u2)) 1 i (sin(u1 1 u2))] This formula says: To multiply two complex numbers written in trigonometric form, we multiply the moduli and add the arguments. Example 16 __ __ Let z1 5 2 1 2i √3 and z2 5 21 2 i √3 . a) Evaluate z1z2 by using their standard forms (rectangular or Cartesian). b) Evaluate z1z2 by using their trigonometric forms and verify that the two results are the same. Solution __ __ __ __ __ a) z1z2 5 (2 1 2i √3 )(21 2 i √ 3 ) 5 (22 1 6) 1 (22 √3 2 2 √3 )i 5 4 2 4i √ 3 b) Converting both to trigonometric form, we get 4p , then p and z 5 2 cis ___ z1 5 4 cis __ 2 3 3 ( ( )) ( ) ( ) 5p 5p 1 i sin ___ 4p 5 8 cis ___ z1z2 5 4 2 cis __ p 1 ___ 5p 5 8 cos ___ 3 3 3 3 3 ( ( )) __ 2 √3 _____ __ 5 8 __ 1 1 i 5 4 2 4i √ 3 . 2 2 Note: You may observe here that multiplying z1 by z2 resulted in a new number whose magnitude is twice that of z1 and is rotated by an angle 4p . Alternatively, you can see it as multiplying z by z which results in of ___ 2 1 3 a complex number whose magnitude is 4 times that of z2 and is rotated by an angle of __ p . 3 445 10 Complex Numbers Example 17 __ Let z1 5 22 1 2i and z2 5 3 √3 2 3i. Convert to trigonometric form and multiply. Solution __ 3p and z 5 6 cis ____ 11p , then z15 2 √2 cis ___ 2 4 6 ( ( ( ) )) 7p 7p 1 i sin ___ 5 12 √2 cis( ___ 7p )5 12 √2 ( cos ___ 12 12 12 ) ( ) __ __ __ 11p z1z25 12 √2 cis ___ 3p 1 ____ 5 12 √ 2 cis ____ 31p 5 12 √ 2 cis ___ 7p 1 2p 4 12 12 6 __ __ Note: You can simplify this answer further to get an exact rectangular form. ( ( ( ) ) ) __ 7p 5 12 √__ 7p 1 i sin ___ 3p 1 3p 1 4p 4p 2 cos ________ 1 i sin ________ z1z25 12 √2 cos ___ 12 12 12 12 ( __ p 1 i sin __ 5 12 √2 cos( __ p 1 __ ( p4 1 __p3 ) 4 3) ) ( ( 2 2 2 2 ) ( 2 2 2 2 )) 2 1 6 2 2 6 ________ 5 12 2 ( ________ 1 i )5 (6 2 6 3 ) 1 i (6 1 6 3 ) 4 4 5 __ 12 √2 __ √ √ 3 2 ___ ___ __ __ __ √ 2 ___ __ __ √ √ √ 3 2 ___ 2 __ 1 1 ___ ___ 1 2 1 i __ __ __ √ √ __ √ __ √ __ √ __ √ __ √ Note: By comparing the Cartesian form of the product to the polar form, __ __ __ __ __ __ √ √ 2 2 √ 6 2 2 √ 6 7 7 p p ___ ___ ________ ________ i.e. 12 √2 cos 1 i sin and 12 √2 1 i , we can 4 4 12 12 __ __ __ __ √ 2 2 √ 2 2 √ 6 6 7p √ 7p 5 ________ . and sin ___ 5 ________ conclude that cos ___ 4 4 12 12 ( ) ( ) This observation gives us a way of using complex number multiplication in order to find exact values of some trigonometric functions. You may have noticed that the conjugate of a complex number z 5 r (cos u 1 i sin u) is z* 5 r (cos u 2 i sin u) 5 r (cos(2u) 1 i sin(2u)). imaginary axis z x yi Also, z z*5 r (cos u 1 i sin u) r (cos u 2 i sin u) 5 r2(cos2 u 1 sin2 u) 5 r2. Graphically, a complex number and its conjugate are reflections of each other in the real axis. See the figure opposite. θ real axis θ z x yi 446 Division of complex numbers A similar approach gives us the rules for division of complex numbers. Let z1 5 r1(cos u1 1 i sin u1) and z2 5 r2(cos u2 1 i sin u2) be two complex numbers written in trigonometric form. Then cos u2 2 i sin u2 r1(cos u1 1 i sin u1) _____________ z1 ________________ __ z2 5 r2(cos u2 1 i sin u2) cos u2 2 i sin u2 ( ) ( ) r (cos u1 cos u2 1 sin u1 sin u2) 1 i(sin u1 cos u2 2 sin u2 cos u1) 5 __ r1 ________________________________________________ 2 cos2 u2 1 sin2 u2) r (cos u1 cos u2 1 sin u1 sin u2) 1 i(sin u1 cos u2 2 sin u2 cos u1) . 5 __ r1 ________________________________________________ 1 2 Now, using the subtraction formulas for sine and cosine, we have z1 __ r1 __ z2 5 r2 [(cos(u1 2 u2)) 1 i(sin(u1 2 u2))] This formula says: To divide two complex numbers written in trigonometric form, we divide the moduli and subtract the arguments. In particular, if we take z1 5 1 and z2 5 z (i.e. u1 5 0 and u2 5 u), we will have the following result. 1 1 5 __ 1 __ If z 5 r(cos u 1 i sin u) then __ z r (cos(2u) 1 i sin(2u)) 5 r (cos(u) 2 i sin (u)) Example 18 __ Let z1 5 1 1 i and z2 5 √ 3 2 i. a) Convert into trigonometric form. 1 b) Evaluate __ z2 . z1 c) Evaluate __ z2 . 5p . 5p and cos ___ d) Use the results above to find the exact values of sin ___ 12 12 Solution __ 2p 11p p ; z 5 2 cis ____ a) z1 1 √ 2 cis __ 5 2 cis ____ 4 2 6 6 2p p 1 1 __ 1 __ ____ __ b) __ z2 5 2 cis( 2 6 )5 2 cis 6 z1 1 __ c) __ z2 can be found by either multiplying z1 by z2 , or by using division as shown above. ( ) __ __ ( ) __ z1 √ √ 2 __ 2 ___ p __ p 5 ___ p 5 ___ 1 __ 2 cis __ 1 cis __ cis ( p 1 __ cis 5p , or __ z2 5 z1 z2 5 ( √ 4) 2 4 2 2 12 6 6) __ p __ __ __ √ 2 cis √ z1 _______ √ 2 ___ 2 __ 2p 4 )5 ___ __ ( p 2 ____ cis 5p 5 ___ cis 2p z2 5 2 cis ____ 4 2 2 12 6 6 ( ) 447 10 Complex Numbers __ __ __ √ z 3 2 1 1 ( √3 1 1)i √ __ 3 1 i _________________ d) __1 5 ______ 1__1 i ______ 5 z2 4 √ 3 2 i √ 3 1 i Comparing this to part c). __ __ __ __ __ √ √ √ √ 3 2 1 ___ 3 2 1 ___ 6 2 √ 2 2 ___ 5p 5 ______ 2 5 cos 5p ⇒ cos ___ __ 5 ________ . ______ 4 2 12 __ √ 3 1 1 ______ Also, 4 12 __ √ 2 ___ 5p ⇒ sin ___ 5p 5 5 sin ___ 4 2 12 4 √ 2 __ __ √ 3 1 1 ______ 12 __ √ 6 1 √ 2 2__ 5 ________ ___ . 4 4 √ 2 Exercise 10.2 In questions 1–14, write the complex number in polar form with argument u, such that 0 < u , 2p. __ 1 2 1 2i 2 √3 1 i 3 2 2 2i 4 √6 2 i √ 2 __ __ __ 5 2 2 2i √ 3 6 23 1 3i 7 4i 8 23 √3 2 3i 9 i 1 1 10 215 11 (4 1 3i )21 12 i(3 1 3i ) 13 p 14 ei __ z z1 . In questions 15–24, find z1z2 and __ 2 p 1 i sin __ p , z 5 cos __ p 1 i sin __ p 15 z1 5 cos __ 2 2 2 3 3 5p 1 i sin ___ 5p , z 5 cos ___ 7p 1 i sin ___ 7p 16 z1 5 cos ___ 2 6 6 6 6 p 1 i sin __ p , z 5 cos ___ 2p 1 i sin ___ 2p 17 z1 5 cos __ 6 6 2 3 3 13p 5p 1 i sin ___ 13p, z 5 cos ___ 5p 1 i sin ____ 18 z1 5 cos ____ 12 12 2 12 12 3p 1 i sin ___ 3p , z 5 __ 4p 1 i sin ___ 4p 2 cos ___ 19 z1 5 3 cos ___ 2 4 4 3 3 3 __ p p p 5 5 5 5p 20 z1 5 3 √ 2 cos ___ 1 i sin ___ , z2 5 2 cos ___ 1 i sin ___ 4 4 3 3 ( ( ) ( ) ( ) ) 21 z1 5 cos 135° 1 i sin 135°, z2 5 cos 90° 1 i sin 90° 22 z1 5 3(cos 120° 1 i sin 120°), z2 5 2(cos 240° 1 i sin 240°) __ √ 3 5 (cos 225° 1 i sin 225°), z2 5 ___ (cos 330° 1 i sin 330°) 23 z1 5 __ 8 2 __ 24 z1 5 3 √ 2 (cos 315° 1 i sin 315°), z2 5 2(cos 300° 1 i sin 300°) z1 , __ 1 , In questions 25–30, write z1 and z2 in polar form, and then find the reciprocals __ z1 1 z2 __ the product z1z2, and the quotient z (2p , u , p). __ __ 2 25 z1 5 √ 3 1 i and z2 5 2 2 2i √ 3 __ __ __ 26 z1 5 √ 6 1 i √ 2 and z2 5 2 √3 2 6i __ 27 z1 5 4 √ 3 1 4i and z2 5 23 2 3i __ __ __ 28 z1 5 i √ 3 and z2 5 2 √ 2 2 i √ 6 __ __ __ 29 z1 5 √ 5 1 i √ 5 and z2 5 2i √ 2 448 __ __ 30 z1 5 1 1 i √ 3 and z2 5 2 √3 31 Consider the complex number z where |z 2 i | 5 |z 1 2i |. a) Show that Im(z) 5 2 _12 . b) Let z1 and z2 be the two possible values of z, such that |z | 5 1. (i) Sketch a diagram to show the points which represent z1 and z2 in the complex plane. (ii) Find arg(z1) and arg(z2). 32 Use the Argand diagram to show that |z1 1 z2| < |z1| 1 |z2|. ( ) __ p , express each of the following complex numbers in 2p 1 i sin 2 __ 33 If z 5 √ 3 cos ___ 3 3 Cartesian form. 3 3 2 z2 2z __ a) ______ b) ______ c) ______ 2 √ 31z 3 1 z2 3 1 z 34 Find the modulus and argument (amplitude) of each of the complex numbers __ __ z1 5 2 √ 3 2 2i, z2 5 2 1 2i and z3 5 (2 √ 3 2 2i )(2 1 2i ). 35 If the numbers in question 34 represent the vertices of a triangle in the Argand diagram, find the area of that triangle. 36 Identify, in the complex plane, the set of points that correspond to the following equations. a) |z | 5 3 b) z* 5 2z c) z 1 z* 5 8 d) |z 2 3| 5 2 e) |z 2 1| 1 |z 2 3| 5 2 37 Identify, in the complex plane, the set of points that correspond to the following inequations. a) |z | < 3 b) |z 2 3i | > 2 10.3 Powers and roots of complex numbers The formula established for the product of two complex numbers can be applied to derive a special formula for the nth power of a complex number. Let z5 r (cos u 1 i sin u), now z 2 5 (r (cos u 1i sin u))(r (cos u 1 i sin u)) 5 r 2((cos u cos u 2 sin u sin u) 1 i (sin u cos u 1 cos u sin u)) 5 r 2((cos2 u 2 sin2 u) 1 i (2 sin u cos u)) 5 r 2(cos 2u 1 i sin 2u). Similarly, z 35 z z 2 5 (r (cos u 1 i sin u))(r 2(cos 2u 1 i sin 2u)) 5 r 3(cos(u 1 2u) 1 i sin(u 1 2u)) 5 r 3(cos 3u 1 i sin 3u). In general, we obtain the following theorem, named after the French mathematician A. De Moivre (1667–1754). 449 10 Complex Numbers Note: As a matter of fact, de Moivre stated ‘his’ formula only implicitly. Its standard form is due to Euler and was generalized by him to any real n. De Moivre’s theorem If z 5 r (cos u 1 i sin u) and n is a positive integer, then z n 5 (r (cos u 1 i sin u))n 5 r n(cos nu 1 i sin nu). The theorem: To find the nth power of any complex number written in trigonometric form, we take the nth power of the modulus and multiply the argument with n. Proof The proof of this theorem follows as an application of mathematical induction. Let P(n) be the statement z n 5 r n(cos nu 1 i sin nu). Basis step: To prove this formula the basis step must be P(1). P (1): is true since z 1 5 r 1(cos u 1 i sin u), which is given! [If you are not convinced, you can try P (2): z 2 5 r 2(cos 2u 1 i sin 2u), which we showed above.] Inductive step: Assume that P(k) is true, i.e. z k 5 r k(cos ku 1 i sin ku). We need to show that P(k 1 1) is also true. So we have to show that z k11 5 r k 1 1(cos(k 1 1)u 1 i sin(k 1 1)u). Now, z k 1 1 5 z k z 5 (cos ku 1 i sin ku)(r (cos u 1 i sin u)) by assumption 5 r kr [(cos kucos u 2 sin ku sin u) 1 i (sin ku cos u 1 cos ku sin u)] 5 r k 1 1[cos(ku 1 u) 1 i sin(ku 1 u)] by addition formulae for sine and cosine k11 5 r (cos(k 1 1)u 1 i sin(k 1 1)u) Therefore, by the principle of mathematical induction, since the theorem is true for n 5 1, and whenever it is true for n 5 k, it was proved true for n 5 k 1 1, then the theorem is true for positive integers n. Note: In fact the theorem is valid for all real numbers n. However, the proof is beyond the scope of this course and this book and therefore we will consider the theorem true for all real numbers without proof at the moment. Example 19 Find (1 1 i)6. Solution We convert the number into polar form first. 450 __ p 1 i sin __ p (1 1 i) 5 √ 2 ( cos __ 4 4) Now we can apply De Moivre’s theorem. [ ( __ p 1 i sin __ p 6 5 ( √__ p 1 i sin 6 __ (1 1 i)6 5 √ 2 ( cos __ 2 ) 6 cos( 6 __ ( p4 ) 4 4) 4) 3p 5 8(2i) 5 28i 3p 1 i sin ___ 5 8 cos ___ 2 2 ] ( ) ) Imagine you wanted to use the binomial theorem to evaluate the power. (1 1 i)6 5 1 1 6i 1 15i 2 1 20i 3 1 15i 4 1 6i 5 1 i 6 5 1 1 6i 2 15 2 20i 1 15 1 6i 2 1 5 8i When the powers get larger, we are sure you will appreciate De Moivre! Applications of De Moivre’s theorem Several applications of this theorem prove very helpful in dealing with trigonometric identities and expressions. For example, when n 5 21, the theorem gives the following result. z21 5 r21(cos(2u) 1 i sin(2u)) 5 __ 1r (cos u 2 i sin u) Also, z2n 5 (z21)n 5 (r21(cos(2u) 1 i sin(2u)))n 5 r2n(cos(2nu) 1 i sin(2nu)). If we take the case when r 5 1, then zn 5 cos nu 1 i sin nu and z2n 5 cos(2nu) 1 i sin(2nu) 5 cos nu 2i sin nu ⇒ zn 1 z2n 5 2 cos nu and zn 2 z2n 5 2i sin nu. These relationships are quite helpful in allowing us to write powers of cos u and sin u in terms of cosines and sines of multiples of u. Example 20 Find cos3 u in terms of first powers of the cosine function. Solution Starting with ( z 1 __ 1z ) 5 (2 cos u)3 3 and expanding the left-hand side, we get 13 5 8 cos3 u⇒ z3 1 __ 3z 1 __ 13 1 3( z 1 __ 1z )5 8 cos3 u z3 1 3z 1 __ z z ⇕ ⇕ ⇒ 2 cos 3u 1 3(2 cos u) 5 8 cos3 u ⇒ cos3 u 5 _18 (2 cos 3u 1 3(2 cos u)) 5 _ 14 (cos 3u 1 3 cos u) 451 10 Complex Numbers Example 21 Simplify the following expression: (cos 6u 1 i sin 6u)6(cos 3u 1 i sin 3u) _____________________________ cos 4u 1 i sin 4u Solution 6 (cos 6u 1 i sin 6u) (cos 3u 1 i sin 3u) ______________________________ cos 4u 1 i sin 4u (cos u 1 i sin u)6(cos u 1 i sin u)3 5 ___________________________ (cos u 1 i sin u)4 Using the laws of exponents, we have (cos u 1 i sin u)6(cos u 1 i sin u)3 5 (cos u 1 i sin u)5 ___________________________ (cos u 1 i sin u)4 5 cos 5u 1 i sin 5u. nth roots of a complex number De Moivre’s theorem is an essential tool for finding nth roots of complex numbers. An nth root of a given number z is a number w that satisfies the following relation wn 5 z. For example, w 5 1 1 i is a 6th root of z 5 28i because, as you have seen above, (1 1 i )6 5 28i, or __ __ w 5 2 √ 3 1 i is a 10th root of 512 1 512i √ 3 . __ __ This is also because w 10 5 (2 √3 1 i )10 5 512 1 512i √ 3 . How to find the nth roots: To find them, we apply the definition of an nth root as mentioned above. Let w 5 s (cos a 1 i sin a) be an nth root of z 5 r (cos u 1 i sin u). This means that w n 5 z, i.e. (s(cos a 1 i sin a))n 5 r (cos u 1 i sin u) ⇒ s n(cos na 1 i sin na) 5 r (cos u 1 i sin u) However, two complex numbers are equal if their moduli are equal, that is, _ _1 sn 5 r ⇔ s 5 n√ r 5 r n . Also, cos na 5 cos u and sin na 5 sin u. From your trigonometry chapters, you recall that both sine and cosine functions are periodic of period 2p each; hence, cos na 5 cos u ⇒ na 5 u 1 2kp, k 5 0, 1, 2, ... { sin n a 5 sin u 452 This leads to u 1 ____ p 5 __ n 2knp ; k 5 0, 1, 2, 3, ..., n 2 1. a 5 _______ u 1n2k Notice that we stop the values of k at n 2 1. This is so because for values larger than or equal to n, principal arguments for these roots will be identical to those for k 5 0 till n 2 1. nth roots of a complex number Let z 5 r (cos u 1 i sin u) and let n be a positive integer, then z has n distinct nth roots ( ( )) ( ) n _ u 1 ____ u 1 ____ n 1 i sin __ n zk 5 √ r cos __ 2knp 2knp where k 5 1, 2, 3, …, n 2 1. _ 1 __ Note: Each of the n nth roots of z has the same modulus √ r 5 r n . Thus all these roots lie n 1 __ _ on a circle in the complex plane whose radius is √ r 5 r n . Also, since the arguments of 2p , then the roots are also equally spaced on this circle. consecutive roots differ by ___ n n Example 22 Find the cube roots of z 5 28 1 8i. Solution __ r 5 8 √2 and u 5 ___ 3p , so the roots are 4 ( ( ) ( )) 3p ___ ______ 3p ___ __ 3 u u 2k 2kp 2k 2k p p p 4 w 5 r cos( __ n )1 i sin( __ n ) 5 √ ( 8 √ n 1 ____ n 1 ____ 2 ) cos ___ 1 ____ 1 i sin ___ 4 1 ____ 3 3 3 3 _ ( ) n √ ( ( ( ) )) 6 __ 2kp 2kp 5 2( √ 2 ) cos __ p 1 ____ 1 i sin __ p 1 ____ ; k 5 0, 1, 2 4 4 3 3 6 __ w1 5 2( √ 2 ) cos( __ p )1 i sin( __ p ) 4 4 __ 6 2p 1 i sin __ 2p 5 26√ __ w2 5 2( √ 2 ) cos __ p 1 ___ p 1 ___ 2 cos ____ 11p 1 i sin ____ 11p 4 4 3 3 12 12 6 __ 4p 1 i sin __ 4p 5 26√ __ w3 5 2( √ 2 ) cos __ p 1 ___ p 1 ___ 2 cos ____ 19p 1 i sin ____ 19p 4 4 3 3 12 12 ( ) ( ( ( ( ) ) ( ( ( ( ) ( ( ) )) )) ( )) ( )) y w1 w2 2π 3 2π 3 0 π 4 2π 3 x w3 Notice how__the arguments are distributed equally around a circle with 6 2p . radius 2( √ 2 ). The difference between any two arguments is ___ 3 453 10 Complex Numbers Notice that if you try to go beyond k 5 2, then you get back to w1. ( ( ( ) )) ( 6 __ 6p 1 i sin __ 6p 5 26√ __2 cos __ w4 5 2√ 2 cos __ p 1 ___ p 1 ___ ( p4 1 2p )1 i sin( __ p4 1 2p ) 4 4 3 3 6 __ 5 2√ 2 cos( __ p )1 i sin( __ p ) 5 w1 4 4 ( ) ) Also, if you raise any of the roots to the third power, you will eventually get z; for example, ( )) ( )) ] ( ( ) [ ( ( ) 5 8 √2 ( cos( ____ 11p 1 i sin( ____ 11p 5 8 √ 2 ( cos( ___ 3p )1 i sin( ___ 3p ) )5 z 4 ) 4 )) 4 4 3 __ 6 __ 11p 1 i sin ____ 11p 5 8 √ 2 cos ____ 33p 1 i sin ____ 33p (w2)35 2√ 2 cos ____ 12 12 12 12 __ __ Example 23 Find the six sixth roots of z 5 264 and graph these roots in the complex plane. Solution Here r 5 64 and u 5 p. So the roots are ( ( ) )) ( u 1 ____ u 1 ____ 2knp 1 i sin __ 2knp n n w 5 s cos __ ( ( ( ( ( ) )) __ 6 2k 2kp p p 1 ____ 1 i sin __ p 1 ____ 5√ 64 cos __ 6 6 6 6 kp 1 i sin __ kp ; k 5 0, 1, 2, 3, 4, 5 p 1 ___ p 1 ___ 5 2 cos __ 3 3 6 6 ( w3 w2 3 2 1 w4 1 ( ( 0 1 2 w 5 π 6 1 2 w6 3 x ) ) ) ( )) ( ( ( )) ( ( ) 3p 1 i sin __ 3p w 5 2( cos( __ p 1 ___ ( p6 1 ___ 3 ) 3 )) 6 7p 1 i sin ___ 5 2( cos( ___ ( 76p ) ) 6 ) 4p 1 i sin __ 4p w 5 2( cos( __ p 1 ___ ( p6 1 ___ 3 ) 3 )) 6 3p )1 i sin( ___ 3p ) ) 5 2( cos( ___ 2 2 5p 1 i sin __ 5p w 5 2( cos( __ p 1 ___ ( p6 1 ___ 3 ) 3 )) 6 11p 11p 1 i sin( ____ 5 2( cos( ____ 6 ) 6 )) 2p 1 i sin __ 2p w3 5 2 cos __ p 1 ___ p 1 ___ 3 3 6 6 5 5 p p ___ ___ 5 2 cos 1 i sin 6 6 4 5 6 454 ) p 1 i sin __ w2 5 2 cos( __ p 1 __ ( p6 1 __p3 ) 3) 6 p )1 i sin( __ p ) 5 2 cos( __ 2 2 w1 π 3 )) w1 5 2 cos( __ p )1 i sin( __ p ) 6 6 y 2 ( ) nth roots of unity The rules we established can be applied to finding the nth roots of 1 (unity). Since 1 is a real number, then in polar/trigonometric form it has a modulus of 1 and an argument of 0. We can write it as 1 5 1(cos 0 1 i sin 0). Now applying the rules above, 1 has n distinct nth roots given by ) ( )) ( ( 0 1 ____ 0 1 ____ 2knp )1 i sin( __ 2knp ) ) n n 5 √ 1 ( cos( __ )1 i sin( ____ ); k 5 0, 1, 2, ..., n 2 1 2knp 2knp 5 cos( ____ _ n u 1 ____ u 1 ____ 2knp 1 i sin __ 2knp n n zk 5 √ r cos __ n __ Or in degrees, ( ) ( ) 360k 1 i sin ____ 360k ; k 5 0, 1, 2, ..., n 2 1 zk 5 cos ____ n n Example 24 Find a) the square roots of unity b) the cube roots of unity. Solution a) Here k 5 2, and therefore the two roots are ( ) ( ) ( ) ( ) ( ) ( ) 360k 1 i sin ____ 360k ; k 5 0, 1 zk 5 cos ____ 2 2 0 1 i sin __ 0 5 1 z0 5 cos __ 2 2 360 1 i sin ___ 360 5 cos 180 1 i sin 180 5 21 z1 5 cos ___ 2 2 b) Here k 5 3, and the three roots are ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2kp 1 i sin ____ 2kp ; k 5 0, 1, 2, 3 zk 5 cos ____ 3 3 0 1 i sin __ 0 5 1 z0 5 cos __ 3 3 __ √ 3 2 2 p p 1 ___ ___ __ ___ z1 5 cos 1 i sin 5 2 1 i 3 3 2 2__ √ 3 1 2 i ___ 4p 1 i sin ___ 4p 5 2 __ z2 5 cos ___ 3 3 2 2 Euler’s formula The material in this part depends on work that you will do in the Analysis option. Otherwise, you will have to accept the result without proof. 455 10 Complex Numbers In the options section on infinite series, we have the following results. Taylor’s (Maclaurin’s) series expansion for sin x, cos x and e x are 3 5 7 3! 5! 7! x 2 __ x 1 ... 5 sin x 5 x 2 __ x 1 __ ∞ 2n 1 1 x ∑(21)n ________ (2n 1 1)! 0 ∞ 2 4 6 2n x 2 __ x 1 ... 5 ∑(21)n _____ x cos x 5 1 2 __ x 1 __ 2! 4! 6! (2n)! 0 2 3 4 2! 3! 4! x 1 __ x 1 ... 5 x 1 __ e 5 1 1 x 1 __ x ∞ n x ∑ __ n! 0 Now if you add 2 3 4 5 6 7 sin x 1 cos x 5 1 1 x 2 __ x 2 __ x 1 __ x 1 __ x 2 __ x 2 __ x 1 ... 2! 3! 4! 5! 6! 7! and compare the result to e x expansion, we notice a stark similarity in the terms, except for the ‘discrepancy’ in the signs! The signs in the sum alternate in a way where pairs of terms alternate! This property is typical of powers of i. Look at i, i 2, i 3, i 4, i 5, i 6, i 7, i 8, ... 5 i, 21, 2i , 1, i , 21, 2i , 1, … This suggests expanding e ix 2 2 3 3 4 4 5 6 6 i 5x 1 ___ i x 1 ____ i x i x 1 … e ix 5 1 1 ix 1 ___ i x 1 ___ 1 ____ 2! 3! 4! 5! 6! 2 2 4 4 6 6 3 3 5 5 5 1 1 ___ i x 1 ____ i x 1 ix 1 ___ i x ! 1 ____ i x 1 ___ i x 1 … 4 2! 6! 3! 5! ( ) x 1 __ 5 1 2 __ x 2 ___ x 1 … 1 i ( x 2 __ x 1 … ) x 1 __ 2! 4! 6! 3! 5! 2 4 6 2 3 4 5 5 1 2 __ x 1 __ x 2 __ x 1 … 1 i x 1 ___ i x 1 … i x 1 ____ 2! 4! 6! 3! 5! 2 4 6 3 5 5 cos x 1 i sin x Since, for any complex number z 5 x 1 iy 5 r (cos u 1 i sin u) and since e iu 5 cos u 1 i sin u, then z 5 r (cos u 1 i sin u) 5 re iu. This is known as Euler’s formula. Example 25 Evaluate each of the following a) e ip p __ b)e i 2 Solution a) e ip 5 cos p 1 i sin p 5 21 p __ p 1 i sin __ p 5 i b)e i 2 5 cos __ 2 2 456 Example 26 Use Euler’s formula to prove DeMoivre’s theorem. Solution (r (cos u 1 i sin u))n 5 (re iu)n 5 r ne inu 5 r n(cos nu 1 i sin nu) Example 27 Find the real and imaginary parts of the complex numbers: p __ a) z 5 3e i 6 b) z 5 7e 2i Solution __ 3 √ 3 p 5 ____ and a) Since |z | 5 3 and arg(z) 5 __ p , Re(z) 5 3 cos __ 2 6 6 3 p Im(z) 5 3 sin __ 5 __ . 6 2 b) Since |z | 5 7 and arg(z) 5 2, Re(z) 5 7 cos 2 and Im(z) 5 7 sin 2. Example 28 Express z 5 5 1 5i in exponential form. Solution __ __ __p |z | 5 5 √2 and tan u 5 __ 5 5 1 ⇒ u 5 __ p , therefore z 5 5 √ 2 e i 4 . 5 4 Example 29 Evaluate (5 1 5i)6 and express your answer in rectangular form. Solution __ p __ Let z 5 5 1 5i. From the example above, z 5 5√ 2 e i 4 ; hence, z 6 5 ( 5√ 2 e i 4 ) 5 (5 √2 )6e i 4 3 6 5 125 000 ei 2 5 2125 000i. __ p 6 __ __ p __ 3p __ Alternatively, ( ( ) __ 6p 5 2125 000i. 6p 1 i sin ___ p 1 i sin __ p 6 5 (5 √__ (5 1 5i)6 5 5√ 2 ( cos __ 2 )6 cos ___ 4 4) 4 4 ) Example 30 Simplify the following expression: (cos 6u 1 i sin 6u)(cos 3u 1 i sin 3u) _____________________________ cos 4u 1 i sin 4u Solution (cos 6u 1 i sin 6u)(cos 3u 1 i sin 3u) _______ _____________________________ 5 e e 5 e 5iu 5 cos 5u 1 i sin 5u 4iu 6iu cos 4u 1 i sin 4u 3iu e 457 10 Complex Numbers Example 31 Use Euler’s formula to find the cube roots of i. Solution __ ___ __ __ _ i 5 e i( 2 1 2kp )⇒ i 3 5 ( e i( 2 1 2kp ) )35 e i( 6 1 3 ); k 5 0, 1, 2 p 1 _1 p Therefore, 2kp p __ p __ √ 3 __i p 1 i sin __ p 5 ___ z0 5 e i( 6 )5 cos __ 1 2 2 6 6 __ 2p 5p __ √ 3 __i i ( __ p6 1 __ 5 5 p p i ( 6 ) ) ___ ___ ___ 3 5 e 5 cos 1 i sin 5 2 1 z1 5 e 2 2 6 6 p 4p 3p __ __ __ 3p 5 2i 3p 1 i sin ___ z2 5 e i ( 6 1 3 )5 e i ( 2 )5 cos ___ 2 2 As you notice here, Euler’s formula provides us with a very powerful tool to perform otherwise extremely laborious calculations. Exercise 10.3 In questions 1–6, write the complex number in Cartesian form. 2p __ 1 z 5 4e2i 3 2 z 5 3e2pi 3 z 5 3e0.5pi 4 z 5 4 cis ___ 7p (exact value) 12 11 __p3 i 6 z 5 3e ( ) __ pi 5 z 5 13e3 In questions 7–16, write each complex number in exponential form. __ 7 2 1 2i __ 8 √3 1i __ __ 9 √6 2 i √2 10 2 2 2i √3 11 23 1 3i 12 4i __ 13 23 √3 2 3i 14 i(3 1 3i ) 15 p 16 ei In questions 17–25, find each complex number. Express in exact rectangular form when possible. __ 18 ( √3 2 i)6 17 (1 1 i )10 __ 19 (3 1 3i √3 )9 __ __ __ __ 20 (2 2 2i )12 21 ( √ 3 2 i √3 )8 22 (23 1 3i )7 __ 23 ( √ 3 2 i √3 )28 24 (23 √3 2 3i )27 __ 25 2( √ 3 1 i )7 In questions 26–30, find each root and graph them in the complex plane. __ 26 The square roots of 4 1 4i √ 3 . __ 27 The cube roots of 4 1 4i √ 3 . 28 The fourth roots of 21. 29 The sixth roots of i. __ 30 The fifth roots of 29 2 9i √ 2 . 458 In questions 31–36, solve each equation. 31 z 5 2 32 5 0 32 z 8 1 i 5 0 __ 33 z 3 1 4 √3 2 4i 5 0 34 z 4 2 16 5 0 35 z 5 1 128 5 128i 36 z 6 2 64i 5 0 In questions 37–40, use De Moivre’s theorem to simplify each of the following expressions. 37 (cos(9b) 1 i sin(9b))(cos(5b) 2 i sin(5b)) (cos(6b) 1 i sin(6b))(cos(4b) 1 i sin(4b)) 38 ________________________________ (cos(3b)) 1 i sin(3b)) _1 39 (cos(9b) 1 i sin(9b)) 3 __________________ 40n√ (cos(2nb) 1 i sin(2n b)) 41 Use e iu to prove that cos(a 1 b) 5 cos a cos b 2 sin a sin b. 42 Use De Moivre’s theorem to show that cos 4a 5 8 cos4 a 2 8 cos2 a 1 1. 43 Use De Moivre’s theorem to show that cos 5a 5 16 cos5 a 2 20 cos3 a 1 5 cos a. 44 Use De Moivre’s theorem to show that cos4 a 5 _ 18 (cos 4a 1 4 cos 2a 1 3). 45 Let z 5 cos 2a 1 i sin 2a. 1 5 2 cos 2a and that 2i sin 2a 5 z 2 __ a) Show that z 1 __ 1z . z b) Find an expression for cos 2na and sin 2na in terms of z. 46 Let the cubic roots of 1 be 1, v and v2. Simplify (1 1 3v)(1 1 3v2). 47 a) Show that the fourth roots of unity can be written as 1, b, b2, and b3. b) Simplify (1 1 b)(1 1 b2 1 b3). c) Show that b 1 b2 1 b3 5 21. 48 a) Show that the fifth roots of unity can be written as 1, a, a2, a3 and a4. b) Simplify (1 1 a)(1 1 a4). c) Show that 1 1 a 1 a2 1 a3 1 a4 5 0. __ __ 49 Show that (1 1 i √ 3 )n 1 (1 2 i √ 3 )n is real and find its value for n 5 18. 50 Given that z 5 (2a 1 3i )3, and a 1, find the values of a such that arg z 5 135°. Practice questions 1 Let z 5 x 1 yi. Find the values of x and y if (1 2 i )z 5 1 2 3i. 2 Let x and y be real numbers, and v be one of the complex solutions of the equation z 3 5 1. Evaluate: a) 1 1 v 1 v2 b) (vx 1 v2y)(vy 1 v2x) 3 a) Evaluate (1 1 i )2. b) Prove, by mathematical induction, that (1 1 i )4n 5 (24)n, where n N+. c) Hence or otherwise, find (1 1 i )32. 459 10 Complex Numbers __ __ √ i √2 6 2 4 Let z1 5 ________ z2 5 1 2 i. and 2 p < p . a) Write z1 and z2 in the form r (cos u 1 i sin u), where r > 0 and 2 __ u < __ 2 2 z1 p 1 i sin ___ p . ___ 5 cos b) Show that __ z2 12 12 z1 i n the form a 1 bi, where a and b are to be determined exactly c) Find the value of __ z2 p and sin ___ p . in radical (surd) form. Hence or otherwise, find the exact values of cos ___ 12 12 ( ( ) p 1 p and z 5 b cos __ p 1 p . i sin __ i sin __ 5 Let z1 5 a cos __ 2 4 4) 3 3 z1 3 __ Express z in the form z 5 x 1 yi. 2 ( ) 6 If z is a complex number and |z 1 16| 5 4|z 1 1|, find the value of |z |. 7 Find the values of a and b, where a and b are real, given that (a 1 bi )(2 2 i ) 5 5 2 i. 8 Given that z 5 (b 1 i )2, where b is real and positive, find the exact value of b when arg z 5 60°. ___ 9 The complex number z satisfies i (z 1 2) 5 1 2 2z, where i 5 √21 . Write z in the form z 5 a 1 bi, where a and b are real numbers. 10 a) Express z 5 2 1 as a product of two factors, one of which is linear. b) Find the zeros of z 5 2 1, giving your answers in the form r (cos u 1 i sin u ) where r . 0 and 2p , u < p. 4 c) Express z 1 z 3 1 z 2 1 z 1 1 as a product of two real quadratic factors. 11 a) Express the complex number 8i in polar form. b) The cube root of 8i which lies in the first quadrant is denoted by z. Express z (i) in polar form (ii) in Cartesian form. ( ) ( ) p 2 p 2 cos __ p 1 p 3 i sin __ i sin __ cos __ 4 4 3 3 _____________________________ 12 Consider the complex number z 5 . 4 p p ___ ___ cos 2 i sin ( 24 24 ) a) (i) Find the modulus of z. (ii) Find the argument of z, giving your answer in radians. 3 __ b) Using De Moivre’s theorem, show that z is a cube root of one, i.e. z 5 √ 1 . c) Simplify (1 1 2z)(2 1 z 2), expressing your answer in the form a 1 bi, where a and b are exact real numbers. 2 + 1 – 4i. 13 The complex number z satisfies the equation √ z = ____ 1–i Express z in the form x 1 i y where x, y Z. _ 14 a) Prove, using mathematical induction, that for a positive integer n, (cos u 1 i sin u)n 5 cos nu 1 i sin nu where i 2 5 21. b) The complex number z is defined by z 5 cos u 1 i sin u. 1 5 cos(2u) 1 i sin(2u). (i) Show that __ z (ii) Deduce that z n 1 z2n 5 2 cos nu. c) (i) Find the binomial expansion of (z 1 z 21)5. 1 (a cos 5u 1 b cos 3u 1 c cos u), where a, b and (ii) Hence, show that cos5 u 5 __ 16 c are positive integers to be found. 460 15 Consider the equation 2(p 1 iq) 5 q 2 ip 2 2(1 2 i ), where p and q are both real numbers. Find p and q. 16 Consider z 5 2 32 5 0. ( ) 2p 1 2p is one of the complex roots of this equation. (i) Show that z1 5 2 cos ___ i sin ___ 5 5 (ii) Find z 12, z 13, z 14 and z 15 giving your answer in the modulus argument form. (iii) Plot the points that represent z1, z 12, z 13, z 14 and z 15 in the complex plane. (iv) The point z n1 is mapped to z n 1 11 by a composition of two linear transformations, where n 5 1, 2, 3, 4. Give a full geometric description of the two transformations. 17 A complex number z is such that |z | 5 |z 2 3i |. 3 . a) Show that the imaginary part of z is __ 2 b) Let z1 and z2 be the two possible values of z, such that |z | 5 3. (i) Sketch a diagram to show the points which represent z1 and z2 in the complex plane, where z1 is in the first quadrant. p . (ii) Show that arg(z1) 5 __ 6 (iii)Find arg(z2). ( ) z k z2 1 5 p, find a value for k. c) Given that arg ____ 2i 18 Given that (a 1 i )(2 2 bi ) 5 7 2 i, find the value of a and of b, where a, b Z. 19 Consider the complex number z 5 cos u 1 i sin u. a) Using De Moivre’s theorem show that 1 5 2 cos nu. z n 1 __ z n 1 4 show that b) By expanding z 1 __ z ( ) cos4 u 5 _18 (cos 4u 1 4 cos 2u 1 3). 20 Consider the complex geometric series e iu 1 _12 e2iu 1 _14 e3iu 1 … a) Find an expression for z, the common ratio of this series. b) Show that |z | , 1. c) Write down an expression for the sum to infinity of this series. d) (i) Express your answer to part c) in terms of sin u and cos u. (ii) Hence, show that 4 cos u 2 2 cos u 1 _12 cos 2u 1 _14 cos 3u 1 … 5 _________ . 5 2 4 cos u 21 Let P (z) 5 z 3 1 az 2 1 bz 1 c, where a, b and c . Two of the roots of P (z ) 5 0 are 22 and (23 1 2i ). Find the value of a, of b and of c. __ 22 Given that |z | 5 2 √ 5 , find the complex number z that satisfies the equation 15 5 1 2 8i. ___ 25 2 ___ *z* z* 23 Solve the simultaneous system of equations giving your answers in x 1 i y form: iz1 1 2z2 5 3 z1 1 (1 2 i )z2 5 4 461 10 Complex Numbers 24 a) Solve the equation x2 2 4x 1 8 5 0. Denote its two roots by z1 and z2 and express them in exponential form with z1 in the first quadrant. z 4 b) Find the value of __21 and write it in the form x 1 yi. z 2 c) Show that z 14 5 z 24. z1 __ z2 d) Find the value of __ z2 1 z1 . e) For what values of n is z 1n real? 2p 1 2p is i sin ___ a root of the equation x7 2 1 5 0. 25 a) Show that z 5 cos ___ 7 7 b) Show that z 7 2 1 5 (z 2 1)(z 6 1 z 5 1 z 4 1 z 3 1 z 2 1 z 1 1) and deduce that z 6 1 z 5 1 z 4 1 z 3 1 z 2 1 z 1 1 5 0. 2p 1 4p 1 6p 5 c) Show that cos ___ cos ___ cos ___ 2 _12 . 7 7 7 Questions 1–23 © International Baccalaureate Organization 462