10 Assessment statements 1.5 Complex numbers: the number i 5 1 ; 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. 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, solving the quadratic equation 8
x 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 > 0 x 1 1 > 1) x 1 1 > 0 for any choice of a real number x. This means that x 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. Numbers such as 1 are not intuitive and many mathematicians in the past resisted their introduction, so they are called imaginary numbers. The situation with finding a solution to x 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 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 x 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 x 1 1 5 0. We extend our number system to include numbers such as 1 ; i.e. a number whose square is 1. Thanks to Euler s (1707 178) 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. 10.1 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 x 5 b a 6 b ac a 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 ac was negative. The use of the formula in cases where b ac is negative depends on two principles (in 9
10 addition to the other principles inherent in the set of real numbers, such as associativity and commutativity of multiplication). 1. 1 1 5 1. k 5 k 1 for any real number k. 0 Example 1 Multiply 6 9. First we simplify each square root using rule. 6 5 6 1 5 6 1 9 5 9 1 5 7 1 And hence using rule 1 with the other obvious rules: 6 9 5 6 1 7 1 5 1 1 5 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 1 where a and b are real numbers. For example, to add 5 1 7 1 to 1 we combine like terms as we do in polynomials: (5 1 7 1 ) 1 ( 1 ) 5 5 1 1 7 1 1 5 (5 1 ) 1 (7 ) 1 5 7 1 1 Similarly, to multiply these numbers we use the binomial multiplication procedures: (5 1 7 1 ) ( 1 ) 5 5 1 (7 1 ) ( 1 ) 1 5 ( 1 ) 1 (7 1 ) 5 10 1 ( 1 ) 15 1 1 1 1 5 10 1 (1) 1 (15 1 1) 1 5 1 1 Euler introduced the symbol i for 1. 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 5 1. 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 5 i i 5 1 i 5 i, and i 5 i i 5 (1) (1) 5 1, and so i 5 5 i i 5 1 i 5 i, and also i 6 5 i i 5 i 5 1; i 7 5 i, and finally i 8 5 1. 0
This leads you to be able to evaluate any positive integer power of i using the following property: i n 1 k 5 i k, k 5 0, 1,,. So, for example i 1 5 i 10 1 5 i 5 1. Example Simplify a) 6 1 9 b) 6 9 a) 6 1 9 5 6 1 1 9 1 5 6i 1 7i 5 1i b) 6 9 5 6i 7i 5 i 5 (1) 5 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 5 1. a is called the real part of the number and b is the imaginary part. We do not define i 5 1 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. We do not mean! i does not belong to this category since we cannot say that i is the positive square root of 1, i.e. i. 0. If we do, then 1 5 i i. 0, which is false, and if we say i, 0, then i. 0, and 1 5 i i. 0, which is also false. Actually i is also a square root of 1 because i i 5 i 5 1. With this in mind, we can use a convention which calls i the principal square root of 1 and write i 5 1. 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 i has as real part and as imaginary. It is usual to write Re(z) for the real part of z and Im(z) for the imaginary part. So, Re( 1 i ) 5 and Im( 1 i ) 5. Note that both the real and imaginary parts are real numbers! Algebraic structure of complex numbers 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: A GDC can be set up to do basic complex number operations. For example, if you have a TI-8 Plus, the set up is as follows. SCI ENG FLOAT 0 1 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 CLOCK1/01/08 6:9AM Equality Two complex numbers z 1 5 (x 1, y 1 ) and z 5 (x, y ) are equal if their corresponding components are equal: (x 1, y 1 ) 5 (x, y ) if x 1 5 x and y 1 5 y. That is, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. 1
10 This is equivalent to saying: a 1 bi 5 c 1 di a 5 c and b 5 d. For example, if ( y )i 5 x 1 1 5i, then x must be 1 and y must be. 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 1i. Let the square root of z be x 1 yi, then (x 1 yi ) 5 5 1 1i x y 1 xyi 5 5 1 1i x y 5 5 and xy 5 1 xy 5 6 y 5 6 x, and when we substitute this value in x y 5 5, we have x ( 6 x ) 5 5. This simplifies to x 5x 6 5 0 which yields x 5 or x 5 9, x 5 6. This leads to x 5 6i, that is, the two square roots of 5 1 1i are 1 i or i. (+i) (- i) 5+1i 5+1i Addition and subtraction for complex numbers are defined as follows: Addition (x 1, y 1 ) 1 (x, y ) 5 (x 1 1 x, y 1 y ) This is equivalent to saying: (a 1 bi) 1 (c 1 di) 5 (a 1 c) 1 (b 1 d)i. Multiplication (x 1, y 1 )(x, y ) 5 (x 1 x y 1 y, x 1 y 1 x y 1 ) 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 1 adi 1 bci 5 ac 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 y 0, x 0 1 y 1) 5 (x, y)
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 1 1, 0 1 1 1 0) 5 (1, 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 1 5 (1, 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 Simplify each expression. a) ( 5i) 1 (7 1 8i) b) ( 5i) (7 1 8i) c) ( 5i)(7 1 8i) a) ( 5i) 1 (7 1 8i) 5 ( 1 7) 1 (5 1 8)i 5 11 1 i b) ( 5i) (7 1 8i) 5 ( 7) 1 (5 8)i 5 1i c) ( 5i)(7 1 8i) 5 ( 7 (5) 8) 1 ( 8 1 (5) 7)i 5 68 i ( 5i) (8i) -.65.5i Ans Frac - 5 8 1 i ( 5i) (7+8i) 68 i Division Multiplication can be used to perform division of complex numbers. The division of two complex numbers, a 1 bi, involves finding a complex 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.
10 Example Find the quotient 1 i 1 1 i. Let 1 i 5 x 1 iy. Hence, using multiplication and the equality of 1 1 i complex numbers, 1 i 5 (1 1 i)(x 1 iy) 1 i 5 x y 1 i(x 1y) { 5 x y x 5 8 5, y 5 1 5 5 x 1 y Thus, 1 i 1 1 i 5 8 1 i. 5 5 (+i) (1+i) 1.6-.i Ans Frac 8 5 1 5i Now, in general, a 1 bi 5 x 1 yi a 1 bi 5 (x 1 yi)(c 1 di). c 1 di With the multiplication as described above: a 1 bi 5 (cx 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 dy 5 a x 5 ac 1 bd c ; y 5 bc ad dx 1 cy 5 b 1 d c 1 d The denominator c 1 d resulted from multiplying c 1 di by c di, which is its 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. Conjugate With every complex number (a 1 bi) we associate another complex number (a 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 bi ) 5 a b i 5 a 1 b which is a non-negative real number. So the product of a complex number and its conjugate is always a real number.
Example 5 Find the conjugate of z and verify the property mentioned above. a) z 5 1 i b) z 5 5i c) z 5 11 a) z * 5 i, and ( 1 i )( i) 5 9i 5 1 9 5 1. b) z * 5 5i, and (5i)(5i) 5 5i 5 (5)(1) 5 5. c) z * 5 11, and 11 11 5 11. 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 di. a 1 bi c 1 di 5 a 1 bi c 1 di c di 5 (a 1 bi)(c di) 5 ac 1 bd 1 bc ad c di c 1 d c 1 d c 1 d i Example 6 Find each quotient and write your answer in standard form. a) 5i 7 1 8i b) 5i 8i c) 5i 7 a) 5i 7 1 8i 5 5i 7 1 8i 7 8i 7 8i 5 8 0 1 ( 5)i 5 1 9 1 6 11 67 11 i b) 5i 5 5i 8i 8i 8i 8i 5 i 0 5 5 1 i 6 8 c) 5i 5 7 5 i 7 7 ( 5i) (7+8i) -.10619690.5 Ans Frac -1 11 67 11i ( 5i) (8i) -.65.5i Ans Frac -5 8 1 i Example 7 Solve the system of equations and express your answer in Cartesian form. (1 1 i )z 1 iz 5 z 1 1 (1 i )z 5 i 5
10 Multiply the first equation by, and the second equation by (1 1 i). (1 1 i )z 1 iz 5 6 (1) (1 1 i )z 1 1 (1 1 i)(1 i )z 5 (1 1 i )( i ) By subtracting () from (1), we get And hence (1 1 i )z 1 1 z 5 6 () ( i )z 5 1 z 5 1 i 5 i z 1 5 1 i( i) 5 1 1 1 i i 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. The product can be extended to powers of complex numbers, i.e. (z )* 5 (z z)* 5 z* z* 5 (z*). 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 k z)* 5 (z k )*z* 5 (z *) k z * (using the product rule). Therefore, (z k 1 1 )* 5 (z *) k z* 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. Theorem Let z, z 1 and z be complex numbers, then (1) (z*)* 5 z () z * 5 z if and only if z is real. () (z 1 1 z )* 5 z 1 * 1 z * The conjugate of the sum is the sum of conjugates. () (z)* 5 z * (5) (z 1 z )* 5 z 1 * z * The conjugate of the product is the product of conjugates. (6) (z 1 )* 5 (z*) 1, if z 0. Proof (1) and () are obvious. For (1), ((a 1 bi) *)* 5 (a bi) * 5 a 1 bi, and for (), a bi 5 a 1 bi bi 5 0 b 5 0. () is proved by straightforward calculation: Let z 1 5 x 1 1 iy 1 and z 5 x 1 iy, then (z 1 1 z )* 5 ((x 1 1 iy 1 ) 1 (x 1 iy ))* 5 ((x 1 1 x ) 1 i (y 1 1 y ))* 5 (x 1 1 x ) i (y 1 1 y ) 5 (x 1 iy 1 ) 1 (x iy ) 5 z 1 * 1 z *. () can now be proved using the above results: (z 1 (z))* 5 0 * 5 0 but, (z 1 (z))* 5 0 * 5 z * 1 (z)*, so z * 1 (z)* 5 0, and (z)* 5 z *. Also (5) is proved by straightforward calculation: (z 1 z )* 5 ((x 1 1 iy 1 ) (x 1 iy ))* 5 ((x 1 x y 1 y ) 1 i (y 1 x 1 x 1 y ))* 5 (x 1 x y 1 y ) i (y 1 x 1 x 1 y ) 5 (x 1 iy 1 ) (x iy ) 5 z 1 * z * 6
And finally, (6): (z(z 1 )) * 5 1 * 5 1 but, (z(z 1 )) * 5 z *(z 1 ) *, so z *(z 1 ) * 5 1, and (z 1 ) * 5 1 z * 5 (z *) 1. Conjugate zeros of polynomials In Chapter, 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, but the method is general. P(x) 5 ax 1 bx 1 dx 1 e Since c is a root of P(x) 5 0, we have ac 1 bc 1 dc 1 e 5 0 (ac 1 bc 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. (c*) is a root of P(x) 5 0. Example 8 1 1 i is a zero of the polynomial P(x) 5 x 5x 1 11x 15. Find all other zeros. Since the polynomial has real coefficients, then 1 i is also a zero. Hence, using the factor theorem, P(x) 5 (x (1 1 i))(x (1 i))(x c), where c is a real number to be found. Now, P(x) 5 (x x 1 5)(x c). c can either be found by division or by factoring by trial and error. In either case, c 5. Example 9 1 1 1 i is a zero of the polynomial P(x) 5 x 1 (i )x 1 (i 1 5)x 1 8 1 i. Find all other zeros. 1 Not included in present IB syllabus. 7
10 Since the polynomial does not have real coefficients, then 1 i is not necessarily also a zero. To find the other zeros, we can perform synthetic substitution 1 1 i 1 i i 1 5 8 1 i 1 1 i 7 1 i 8 i 1 1 1 i 1 i 0 This shows that P(x) 5 (x 1 i)(x 1 (1 1 i)x 1 i). The second factor can be factored into (x 1 1)(x 1 i) giving us the other two zeros as 1 and i. Note: x 1 (1 1 i)x 1 i 5 0 can be solved using the quadratic formula. x 5 b 6 b ac 5 1 i 6 (1 1 i) ( 1 i) a 5 1 i 6 8 6i 1 8 1i 5 1 i 6 18i To find 18i we let (a 1 bi) 5 18i a b 1 abi 5 18i, then equating the real parts and imaginary parts to each other: a b 5 0 and ab 5 18 will yield 18i 5 6 7 i, and hence x 5 1 i 6 18i 1 i 6 (6 7 i) which will yield x 5 1 or x 5 i. Exercise 10.1 Express each of the following numbers in the form a 1 bi. 1 5 1 7 7 6 9 5 81 6 5 16 Perform the following operations and express your answer in the form a 1 bi. 7 ( 1 i ) 1 ( 5i ) 8 ( 1 i ) ( 5i ) 9 ( 1 i )( 5i ) 10 i ( i ) 11 ( 7i )( 1 i ) 1 (1 1 i )( i ) 1 1 i 1 5i 1 i 1 i 15 ( 1 i ) 1 ( 1 1 1 i ) 16 ( 1 i ) ( 1 1 i ) 17 ( 1 i ) ( 1 1 1 i ) 18 ( 1 i )( i ) 19 1 i ( 7i ) 0 ( 1 5i ) ( 5i ) 1 1 5 1i 1i 1 i 8
i ( i ) ( 1 5i )(6 10i ) 5 9 5i 6 (7 i )1 1 10i 7 (5 1i ) 1 8 i 1 6 1 8i (7 1 8i )( 5i ) 9 0 5 1 5 1i 1 16 1 Let z 5 a 1 bi. Find a and b if ( 1 i )z 5 7 1 i. ( 1 yi )(x 1 i ) 5 1 1i, where x and y are real numbers. Solve for x and y. a) Evaluate (1 1 i ). b) Prove that (1 1 i ) 6n 5 8 n, where n Z 1. c) Hence, find (1 1 i ) 8. a) Evaluate ( 1 i ). b) Prove that ( 1 i ) k 5 (16) k, where k Z 1. c) Hence, find ( 1 i ) 6. 5 If z is a complex number such that z 1 i 5 z 1 i, find the value of z. ( z 5 x + y where z = x + iy.) 6 Find the complex number z and write it in the form a 1 bi if z 5 1 i i. 7 Find the values of the two real numbers x and y such that (x 1 iy)( 7i ) 5 1 i. 8 Find the complex number z and write it in the form a 1 bi if i(z 1 1) 5 z. 9 Find the complex number z and write it in the form a 1 bi if i 1 1 i _ z 5 i. 0 Find the values of the two real numbers x and y such that (x 1 iy) 5 i. 1 a) Find the values of the two real numbers x and y such that (x 1 iy) 5 8 1 6i. b) Hence, solve the following equation z 1 (1 i )z 1 i 5 0. If z C, find all solutions to the equation z 7i 5 0. Given that z 5 1_ 1 i is a zero of the polynomial f (x) 5 x 16x 1 9x 51, find the other zeros. Find a polynomial function with integer coefficients and lowest possible degree that has 1_, 1 and 1 i as zeros. 5 Find a polynomial function with integer coefficients and lowest possible degree that has, and 1 1 i as zeros. 6 Given that z 5 5 1 i is a zero of the polynomial f (x) 5 x 7x x 1 87, find the other zeros. 7 Given that z 5 1 i is a zero of the polynomial f (x) 5 x x 1 8x 1 8, find the other zeros. 8 Let z C. If z z* 5 a 1 bi, show that a 1 bi 5 1. 9
10 9 Given that z 5 (k 1 i ) 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. iz 1 1 z 5 i iz 1 (1 1 i )z 5 z 1 1 ( 1 i )z 5 7 1 i ( 1 i )z 1 1 iz 5 10. 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. 5 i imaginary axis 5i i i i i i 5 1 0 i 1 i i i 5 real axis The diagram above illustrates how the two complex numbers 1 i and 5 1 i are plotted in the complex plane. imaginary axis z x yi Imaginary part y 0 Real part x real axis 0
Let us consider the sum of two complex numbers: z 1 5 x 1 1 y 1 i, and z 5 x 1 y i As we have defined addition before: z 1 1 z 5 (x 1 1 x ) 1 (y 1 1 y )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 z 1 5 1 i and z 5 5 1 i. Find z 1 1 z and z 1 z. imaginary axis z 1 z 6i 6i i z 1 z 8 i i z 5 i 5 1 0 1 i z 1 i real axis Note here that the vector representing the sum, 1 6i, is the diagonal of the parallelogram with sides representing 1 i and 5 1 i, 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 1 y. For complex numbers, the modulus or absolute value (or magnitude) of the complex number z 5 x 1 yi is z 5 x 1 y. It follows immediately that since z * 5 x yi z * 5 x 1 (y) 5 x 1 y, then z * 5 z. Also of interest is the following result. z z* 5 (x 1 iy)(x iy) 5 x 1 y, z 5 x 1 y, and z* 5 x 1 y z z* 5 z 5 z* For example: ( 1 i )( i ) 5 9 1 16 5 5 5 ( 1 ) 1
10 Example 11 Calculate the moduli of the following complex numbers a) z 1 5 5 6i b) z 5 1 1 5i a) z 1 5 5 6i 5 5 1 6 5 61 b) z 5 1 1 5i 5 1 1 5 5 169 = 1 Example 1 Graph each set of complex numbers. a) A 5{z z 5 } b) B 5{z z < } a) A is the set of complex numbers whose distance from the origin is units. So, the set is a circle with radius and centre (0, 0) as shown. A O z b) B is the set of complex numbers whose distance from the origin is less than or equal to. So, the set is a disk of radius and centre at the origin. B O z Another important property is the following result: Proof: But, z 1 z 5 z 1 z z 1 z 5 (x 1 x y 1 y ) 1 (x 1 y 1 x y 1 )i 5 (x 1 x y 1 y ) 1 (x 1 y 1 x y 1 ) 5 (x 1 x ) x 1 x y 1 y 1 (y 1 y ) 1 (x 1 y ) 1 x 1 y x y 1 1 (x y 1 ) 5 (x 1 x ) 1 (y 1 y ) 1 (x 1 y ) 1 (x y 1 ) z 1 z 5 x 1 1 y 1 x 1 y 5 (x 1 1 y 1 )(x 1 y ) 5 (x 1 x ) 1 (y 1 y ) 1 (x 1 y ) 1 (x y 1 ) And so the result follows.
Example 1 Evaluate ( 1 i)(5 1 1i). ( 1 i)(5 1 1i) 5 1 i 5 1 1i 5 9 1 16 5 1 1 5 5 1 5 65, or ( 1 i)(5 1 1i) 5 1 56i 5 () 1 56 5 55 5 65 Trigonometric/polar form of a complex number imaginary axis z x yi r z Imaginary part y 0 θ 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. Clearly x 5 r cos u and y 5 r sin u ; r 5 x 1 y ; and tan u 5 y x. 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 p. 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.)
10 Example 1 Write the following numbers in trigonometric form. a) z 5 1 1 i b) z 5 i c) z 5 5i d) z 5 17 a) r 5 1 1 1 5 ; tan u 5 1 1 5 1. Hence, by observing the real and imaginary parts being positive, we can conclude that the argument y i z 1 i must be u 5 p. z 5 ( cos p 1 i sin p ) 5 cis p 0 θ π x θ 11π 6 y i 0 x z i b) r 5 ( ) 1 (1) 5 5 ; tan u 5 1. The real part is positive, the imaginary part is negative, and the point is therefore in the fourth quadrant, so u 5 11p 6. z 5 ( cos 11p 1 i sin 11p 6 6 ) 5 cis 11p 8 We can also use u 5 p 6. c) r 5 5 and u 5 p since it is on the negative side of the imaginary axis. z 5 5 ( cos p 1 i sin p y ) We can also use u 5 p. d) r 5 17 and u 5 0 z 5 17 (cos 0 1 i sin 0) θ π 0 x z 5 i Example 15 Convert each complex number into its rectangular form. a) z 5 cos 150 1 i sin 150 b) z 5 1 cis p c) z 5 6(cos 50 1 i sin 50 ) d) z 5 15 ( cos p 1 i sin p ) a) z 5 ( ) 1 i ( 1 ) 5 1 i
b) z 5 1 cos p 1 1i sin p 5 1 1 1 1i 5 6 6i c) z 5 6 cos 50 1 6i sin 50 5 6 0.6 1 6i 0.766 5.857 1.596i 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. Let z 1 5 r 1 (cos u 1 1 i sin u 1 ) and z 5 r (cos u 1 i sin u ) be two complex numbers written in trigonometric form. Then z 1 z 5 (r 1 (cos u 1 1 i sin u 1 ))(r (cos u 1 i sin u ) 5 r 1 r [(cos u 1 cos u sin u 1 sin u ) 1 i (sin u 1 cos u 1 sin u cos u 1 )]. Now, using the addition formulae for sine and cosine, we have z 1 z 5 r 1 r [(cos(u 1 1 u )) 1 i (sin(u 1 1 u ))] This formula says: To multiply two complex numbers written in trigonometric form, we multiply the moduli and add the arguments. 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! Example 16 Let z 1 5 1 i and z 5 1 i. a) Evaluate z 1 z by using their standard forms (rectangular or Cartesian). b) Evaluate z 1 z by using their trigonometric forms and verify that the two results are the same. a) z 1 z 5 ( 1 i )(1 i ) 5 ( 1 6) 1 ( )i 5 i b) Converting both to trigonometric form, we get z 1 5 cis p and z 5 cis p, then z 1 z 5 ( cis ( p 1 p ) ) 5 8 cis ( 5p ) 5 8 ( cos 5p 1 i sin 5p ) 5 8 ( 1 1 i ( ) ) 5 i. Note: You may observe here that multiplying z 1 by z resulted in a new number whose magnitude is twice that of z 1 and is rotated by an angle of p. Alternatively, you can see it as multiplying z by z 1 which results in a complex number whose magnitude is times that of z and is rotated by an angle of p. 5
10 Example 17 Let z 1 5 1 i and z 5 i. Convert to trigonometric form and multiply. z 1 5 cis p and z 5 6 cis 11p, then 6 z 1 z 5 1 ( cis ( p 1 11p 6 ) ) 5 1 cis ( 1p 1 ) 5 1 cis ( 7p 1 1 p ) 5 1 cis ( 7p 1 ) 5 1 ( cos 7p 1 i sin 7p 1 1 ) Note: You can simplify this answer further to get an exact rectangular form. z 1 z 5 1 ( cos 7p 1 i sin 7p 1 1 ) 5 1 ( cos ( p 1 p 1 ) 1 i sin p 1 p 1 ) 5 1 ( cos ( p 1 p ) 1 i sin ( p 1 p ) ) 5 1 5 1 ( ( ( 1 ) 1 i ( 1 1 ) ) 6 1 i 1 6 ) 5 (6 6 ) 1 i (6 1 6 ) Note: By comparing the Cartesian form of the product to the polar form, i.e. 1 ( cos 7p 1 i sin 7p 1 1 ) and 1 ( 6 1 i 6 ), we can conclude that cos 7p 1 5 6 and sin 7p 1 5 6. 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 i sin u) 5 r (cos(u) 1 i sin(u)). Also, z z* 5 r (cos u 1 i sin u) r (cos u i sin u) 5 r (cos u 1 sin u) 5 r. Graphically, a complex number and its conjugate are reflections of each other in the real axis. See the figure opposite. imaginary axis θ θ z x yi real axis z x yi 6
Division of complex numbers A similar approach gives us the rules for division of complex numbers. Let z 1 5 r 1 (cos u 1 1 i sin u 1 ) and z 5 r (cos u 1 i sin u ) be two complex numbers written in trigonometric form. Then z 1 z 5 r 1(cos u 1 1 i sin u 1 ) r (cos u 1 i sin u ) cos u i sin u cos u i sin u 5 r 1 r ( (cos u 1 cos u 1 sin u 1 sin u ) 1 i(sin u 1 cos u sin u cos u 1 ) 5 r 1 cos u 1 sin u ) r ( (cos u 1 cos u 1 sin u 1 sin u ) 1 i(sin u 1 cos u sin u cos u 1 ) 1 Now, using the subtraction formulas for sine and cosine, we have z 1 z 5 r 1 r [(cos(u 1 u )) 1 i(sin(u 1 u ))] 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 z 1 5 1 and z 5 z (i.e. u 1 5 0 and u 5 u), we will have the following result. ) ). If z 5 r(cos u 1 i sin u) then 1 z 5 1 r (cos(u) 1 i sin(u)) 5 1 r (cos(u) i sin (u)) Example 18 Let z 1 5 1 1 i and z 5 i. a) Convert into trigonometric form. b) Evaluate 1 z. c) Evaluate z 1 z. d) Use the results above to find the exact values of sin 5p 1 and cos 5p 1. a) z 1 1 cis p ; z 5 cis 11p 5 cis p 6 6 b) 1 z 5 1 cis ( p 6 ) 5 1 cis p 6 c) z 1 z can be found by either multiplying z 1 by 1 z, or by using division as shown above. z 1 z 5 z 1 1 z 5 ( cis p ) ( 1 cis p ) 5 6 z cis p 1 z 5 cis p 6 5 cis ( p p 6 ) 5 cis ( 5p 1 ) cis ( p 1 p 6 ) 5 cis ( 5p 1 ), or 7
10 d) z 1 z 5 1 1 i i 1 i 1 i 5 1 1 ( 1 1)i Comparing this to part c). 1 5 cos 5p cos 5p 1 1 5 1 5 6. Also, 1 1 5 sin 5p sin 5p 1 1 5 1 1 5 6 1. Exercise 10. In questions 1 1, write the complex number in polar form with argument u, such that 0 < u, p. 1 1 i 1 i i 6 i 5 i 6 1 i 7 i 8 i 9 i 1 1 10 15 11 ( 1 i ) 1 1 i( 1 i ) 1 p 1 ei In questions 15, find z 1 z and z 1 z. 15 z 1 5 cos p 1 i sin p, z 5 cos p 1 i sin p 16 z 1 5 cos 5p 1 i sin 5p 6 6, z 5 cos 7p 1 i sin 7p 6 6 17 z 1 5 cos p 1 i sin p, z 6 6 5 cos p 1 i sin p 18 z 1 5 cos 1p 1 i sin 1p 1 1, z 5 cos 5p 1 1 i sin 5p 1 19 z 1 5 ( cos p 1 i sin p ), z 5 ( cos p 1 i sin p ) 0 z 1 5 ( cos 5p 1 i sin 5p ), z 5 ( cos 5p 1 i sin 5p ) 1 z 1 5 cos 15 1 i sin 15, z 5 cos 90 1 i sin 90 z 1 5 (cos 10 1 i sin 10 ), z 5 (cos 0 1 i sin 0 ) z 1 5 5 (cos 5 1 i sin 5 ), z 8 5 (cos 0 1 i sin 0 ) z 1 5 (cos 15 1 i sin 15 ), z 5 (cos 00 1 i sin 00 ) In questions 5 0, write z 1 and z in polar form, and then find the reciprocals 1 z1, 1 z, the product z 1 z, and the quotient z 1 z (p, u, p). 5 z 1 5 1 i and z 5 i 6 z 1 5 6 1 i and z 5 6i 7 z 1 5 1 i and z 5 i 8 z 1 5 i and z 5 i 6 9 z 1 5 5 1 i 5 and z 5 i 8
0 z 1 5 1 1 i and z 5 1 Consider the complex number z where z i 5 z 1 i. a) Show that Im(z) 5 1_. b) Let z 1 and z be the two possible values of z, such that z 5 1. (i) Sketch a diagram to show the points which represent z 1 and z in the complex plane. (ii) Find arg(z 1 ) and arg(z ). Use the Argand diagram to show that z 1 1 z < z 1 1 z. If z 5 ( cos p 1 i sin p ), express each of the following complex numbers in Cartesian form. a) 1 z b) z c) z 1 z 1 z Find the modulus and argument (amplitude) of each of the complex numbers z 1 5 i, z 5 1 i and z 5 ( i )( 1 i ). 5 If the numbers in question represent the vertices of a triangle in the Argand diagram, find the area of that triangle. 6 Identify, in the complex plane, the set of points that correspond to the following equations. a) z 5 b) z* 5 z c) z 1 z* 5 8 d) z 5 e) z 1 1 z 5 7 Identify, in the complex plane, the set of points that correspond to the following inequations. a) z < b) z i > 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. 10. Let z 5 r (cos u 1 i sin u), now z 5 (r(cos u 1i sin u))(r (cos u 1 i sin u)) 5 r ((cos u cos u sin u sin u) 1 i (sin u cos u 1 cos u sin u)) 5 r ((cos u sin u) 1 i ( sin u cos u)) 5 r (cos u 1 i sin u). Similarly, z 5 z z 5 (r(cos u 1 i sin u))(r (cos u 1 i sin u)) 5 r (cos(u 1 u) 1 i sin(u 1 u)) 5 r (cos u 1 i sin u). In general, we obtain the following theorem, named after the French mathematician A. De Moivre (1667 175). 9
10 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 (): z 5 r (cos u 1 i sin u), 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 k r [(cos kucos u 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 5 r k 1 1 (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. We convert the number into polar form first. 50
(1 1 i) 5 ( cos p 1 i sin p ) Now we can apply De Moivre s theorem. (1 1 i) 6 5 [ ( cos p 1 i sin p ) ] 6 5 ( ) 6 ( cos ( 6 p ) 1 i sin ( 6 p ) ) 5 8 ( cos p 1 i sin p ) 5 8(i) 5 8i Imagine you wanted to use the binomial theorem to evaluate the power. (1 1 i) 6 5 1 1 6i 1 15i 1 0i 1 15i 1 6i 5 1 i 6 5 1 1 6i 15 0i 1 15 1 6i 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 1, the theorem gives the following result. z 1 5 r 1 (cos(u) 1 i sin(u)) 5 1 r (cos u i sin u) Also, z n 5 (z 1 ) n 5 (r 1 (cos(u) 1 i sin(u))) n 5 r n (cos(nu) 1 i sin(nu)). If we take the case when r 5 1, then z n 5 cos nu 1 i sin nu and z n 5 cos(nu) 1 i sin(nu) 5 cos nu i sin nu z n 1 z n 5 cos nu and z n z n 5 i 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 0 Find cos u in terms of first powers of the cosine function. Starting with ( z 1 1 z ) 5 ( cos u) and expanding the left-hand side, we get z 1 z 1 z 1 1 z 5 8 cos u z 1 1 z 1 ( z 1 1 z ) 5 8 cos u cos u 1 ( cos u) 5 8 cos u cos u 5 1_ 8 ( cos u 1 ( cos u)) 5 1_ (cos u 1 cos u) 51
10 Example 1 Simplify the following expression: (cos 6u 1 i sin 6u)6 (cos u 1 i sin u) cos u 1 i sin u (cos 6u 1 i sin 6u)6 (cos u 1 i sin u) cos u 1 i sin u 5 (cos u 1 i sin u)6 (cos u 1 i sin u) (cos u 1 i sin u) Using the laws of exponents, we have (cos u 1 i sin u)6 (cos u 1 i sin u) 5 (cos u 1 i sin u)5 (cos u 1 i sin u) 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 w n 5 z. For example, w 5 1 1 i is a 6th root of z 5 8i because, as you have seen above, (1 1 i ) 6 5 8i, or w 5 1 i is a 10th root of 51 1 51i. This is also because w 10 5 ( 1 i ) 10 5 51 1 51i. 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, s n 5 r s 5 n _ r 5 r 1_ 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 p each; hence, { cos na 5 cos u na 5 u 1 kp, k 5 0, 1,,... sin na 5 sin u 5
This leads to a 5 u 1 n kp 5 n u 1 kp n ; k 5 0, 1,,,..., n 1. Notice that we stop the values of k at n 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 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 z k 5 n _ r ( cos ( u n 1 kp n ) 1 i sin ( n u 1 kp n ) ) where k 5 1,,,, n 1. Note: Each of the n nth roots of z has the same modulus n _ r 5 r n 1. Thus all these roots lie on a circle in the complex plane whose radius is n _ r 5 r n 1. Also, since the arguments of consecutive roots differ by p n, then the roots are also equally spaced on this circle. Example Find the cube roots of z 5 8 1 8i. r 5 8 and u 5 p, so the roots are w 5 n _ r ( cos ( u n 1 kp n ) 1 i sin ( n u 1 kp n ) ) 5 ( 8 ) ( cos ( p 1 kp ) 1 i sin ( p 1 kp ) ) 5 ( 6 ) ( cos ( p 1 kp ) 1 i sin ( p 1 kp ) ) ; k 5 0, 1, w 1 5 ( 6 ) ( cos ( p ) 1 i sin ( p ) ) w 5 ( 6 ) ( cos ( p 1 p ) 1 i sin ( p 1 p ) ) 5 6 ( cos ( 11p 1 ) 1 i sin ( 11p 1 ) ) w 5 ( 6 ) ( cos ( p 1 p ) 1 i sin ( p 1 p ) ) 5 6 ( cos ( 19p 1 ) 1 i sin ( 19p 1 ) ) y w 1 w π π 0 π π x w Notice how the arguments are distributed equally around a circle with radius ( 6 ). The difference between any two arguments is p. 5
10 Notice that if you try to go beyond k 5, then you get back to w 1. w 5 6 ( cos ( p 1 6p ) 1 i sin ( p 1 6p ) ) 5 6 ( cos ( p 1 p ) 1 i sin ( p 1 p ) ) 5 6 ( cos ( p ) 1 i sin ( p ) ) 5 w 1 Also, if you raise any of the roots to the third power, you will eventually get z; for example, (w ) 5 [ 6 ( cos ( 11p 1 ) 1 i sin ( 11p 1 ) ) ] 5 8 ( cos ( p 1 ) 1 i sin ( p 1 ) ) 5 8 ( cos ( 11p ) 1 i sin ( 11p ) ) 5 8 ( cos ( p ) 1 i sin ( p ) ) 5 z Example Find the six sixth roots of z 5 6 and graph these roots in the complex plane. Here r 5 6 and u 5 p. So the roots are y w 5 s ( cos ( u n 1 kp n ) 1 i sin ( u n 1 kp n ) ) 5 6 6 ( cos ( p 1 kp 6 6 ) 1 i sin ( p 1 kp 6 6 ) ) 5 ( cos ( p 1 kp 6 ) 1 i sin ( p 1 kp 6 ) ) ; k 5 0, 1,,,, 5 w 1 5 ( cos ( p 6 ) 1 i sin ( p 6 ) ) w w 1 π π 6 1 0 1 1 w w 5 w 1 w 6 x w 5 ( cos ( p 6 1 p ) 1 i sin ( p 6 1 p ) ) 5 ( cos ( p ) 1 i sin ( p ) ) w 5 ( cos ( p 1 p 6 ) 1 i sin ( p 1 p 6 ) ) 5 ( cos ( 5p 6 ) 1 i sin ( 5p 6 ) ) w 5 ( cos ( p 1 p 6 ) 1 i sin ( p 1 p 6 ) ) 5 ( cos ( 7p 6 ) 1 i sin ( 7p 6 ) ) w 5 5 ( cos ( p 1 p 6 ) 1 i sin ( p 1 p 6 ) ) 5 ( cos ( p ) 1 i sin ( p ) ) w 6 5 ( cos ( p 1 5p 6 ) 1 i sin ( p 1 5p 6 ) ) 5 ( cos ( 11p 6 ) 1 i sin ( 11p 6 ) ) 5
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 Or in degrees, z k 5 n _ r ( cos ( u n 1 kp n ) 1 i sin ( n u 1 kp n ) ) 5 n 1 ( cos ( 0 n 1 kp n ) 1 i sin ( 0 n 1 kp n ) ) 5 cos ( kp n ) 1 i sin ( kp n ) ; k 5 0, 1,,..., n 1 z k 5 cos ( 60k n ) 1 i sin ( 60k n ) ; k 5 0, 1,,..., n 1 Example Find a) the square roots of unity b) the cube roots of unity. a) Here k 5, and therefore the two roots are z k 5 cos ( 60k ) 1 i sin ( 60k ) ; k 5 0, 1 z 0 5 cos ( 0 ) 1 i sin ( 0 ) 5 1 z 1 5 cos ( 60 ) 1 i sin ( 60 ) 5 cos 180 1 i sin 180 5 1 b) Here k 5, and the three roots are z k 5 cos ( kp ) 1 i sin ( kp ) ; k 5 0, 1,, z 0 5 cos ( 0 ) 1 i sin ( 0 ) 5 1 z 1 5 cos ( p ) 1 i sin ( p ) 5 1 1 i z 5 cos ( p ) 1 i sin ( p ) 5 1 i 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. 55
10 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 Now if you add sin x 5 x x 1 x 5 x 7 1... 5! 5! 7! cos x 5 1 x 1 x x 6 1... 5!! 6! e x 5 1 1 x 1 x 1 x 1 x!!! 1... 5 0 0 x n 1 1 (1) n (n 1 1)! 0 (1) n x n (n)! sin x 1 cos x 5 1 1 x x x 1 x 1 x 5 x 6 x 7 1...!!! 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. x n n! Look at i, i, i, i, i 5, i 6, i 7, i 8,... 5 i, 1, i, 1, i, 1, i, 1, This suggests expanding e ix e ix 5 1 1 ix 1 i x! 1 i x! 1 i x! 1 i 5 x 5 5! 1 i 6 x 6 6! 1 5 1 1 i x! 1 i x! 1 i 6 x 6 6! 1 ix 1 i x! 1 i 5 x 5 5! 1 5 1 x 1 x x 6!! 1 1 i 6! ( x 1 i x! 1 i x 5 5! 1 ) 5 1 x 1 x x 6!! 6! 1 1 i ( x x 1 x 5 1! 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 5 Evaluate each of the following a) e ip b) e i p a) e ip 5 cos p 1 i sin p 5 1 b) e i p 5 cos p 1 i sin p 5 i 56
Example 6 Use Euler s formula to prove DeMoivre s theorem. (r (cos u 1 i sin u)) n 5 (re iu ) n 5 r n e inu 5 r n (cos nu 1 i sin nu) Example 7 Find the real and imaginary parts of the complex numbers: a) z 5 e i p 6 b) z 5 7e i a) Since z 5 and arg(z) 5 p, Re(z) 5 cos p 5 6 6 and Im(z) 5 sin p 5. 6 b) Since z 5 7 and arg(z) 5, Re(z) 5 7 cos and Im(z) 5 7 sin. Example 8 Express z 5 5 1 5i in exponential form. z 5 5 and tan u 5 5 5 1 u 5 p, therefore z 5 5 e i p. 5 Example 9 Evaluate (5 1 5i) 6 and express your answer in rectangular form. Let z 5 5 1 5i. From the example above, z 5 5 e i p ; hence, z 6 5 ( 5 e i p ) 6 5 (5 ) 6 e i p 6 5 15 000 e i p Alternatively, (5 1 5i) 6 5 ( 5 ( cos p 1 i sin p ) ) 6 5 (5 5 15 000i. ) ( 6 cos 6p 1 i sin 6p ) 5 15 000i. Example 0 Simplify the following expression: (cos 6u 1 i sin 6u)(cos u 1 i sin u) cos u 1 i sin u (cos 6u 1 i sin 6u)(cos u 1 i sin u) 5 e 6iu e iu cos u 1 i sin u e 5 e 5iu 5 cos 5u 1 i sin 5u iu 57
10 Example 1 Use Euler s formula to find the cube roots of i. i 5 e i ( p 1 kp ) i 1_ 5 ( e i ( p 1 kp ) ) 1_ 5 e i ( p 6 1 kp ) ; k 5 0, 1, Therefore, z 0 5 e i ( p 6 ) 5 cos p 1 i sin p 5 6 6 1 i z 1 5 e i ( p 6 1 p ) 5p i ( 5 e 6 ) 5 cos 5p 1 i sin 5p 6 6 5 1 i z 5 e i ( p 6 1 p ) p i ( 5 e ) 5 cos p 1 i sin p 5 i As you notice here, Euler s formula provides us with a very powerful tool to perform otherwise extremely laborious calculations. Exercise 10. In questions 1 6, write the complex number in Cartesian form. i p 1 z 5 e z 5 e pi z 5 e 0.5pi z 5 cis ( 7p 1 ) (exact value) 5 z 5 1 e pi 6 z 5 e 11 p i In questions 7 16, write each complex number in exponential form. 7 1 i 8 1i 9 6 i 10 i 11 1 i 1 i 1 i 1 i( 1 i ) 15 p 16 ei In questions 17 5, find each complex number. Express in exact rectangular form when possible. 17 (1 1 i ) 10 18 ( i) 6 19 ( 1 i ) 9 0 ( i ) 1 1 ( i ) 8 ( 1 i ) 7 ( i ) 8 ( i ) 7 5 ( 1 i ) 7 In questions 6 0, find each root and graph them in the complex plane. 6 The square roots of 1 i. 7 The cube roots of 1 i. 8 The fourth roots of 1. 9 The sixth roots of i. 0 The fifth roots of 9 9i. 58
In questions 1 6, solve each equation. 1 z 5 5 0 z 8 1 i 5 0 z 1 i 5 0 z 16 5 0 5 z 5 1 18 5 18i 6 z 6 6i 5 0 In questions 7 0, use De Moivre s theorem to simplify each of the following expressions. 7 (cos(9b) 1 i sin(9b))(cos(5b) i sin(5b)) 8 (cos(6b) 1 i sin(6b))(cos(b) 1 i sin(b)) (cos(b)) 1 i sin(b)) 9 (cos(9b) 1 i sin(9b) ) 1_ 0 n (cos(nb) 1 i sin(nb)) 1 Use e iu to prove that cos(a 1 b) 5 cos a cos b sin a sin b. Use De Moivre s theorem to show that cos a 5 8 cos a 8 cos a 1 1. Use De Moivre s theorem to show that cos 5a 5 16 cos 5 a 0 cos a 1 5 cos a. Use De Moivre s theorem to show that cos a 5 1_ 8 (cos a 1 cos a 1 ). 5 Let z 5 cos a 1 i sin a. a) Show that z 1 1 z 5 cos a and that i sin a 5 z 1 z. b) Find an expression for cos na and sin na in terms of z. 6 Let the cubic roots of 1 be 1, v and v. Simplify (1 1 v)(1 1 v ). 7 a) Show that the fourth roots of unity can be written as 1, b, b, and b. b) Simplify (1 1 b)(1 1 b 1 b ). c) Show that b 1 b 1 b 5 1. 8 a) Show that the fifth roots of unity can be written as 1, a, a, a and a. b) Simplify (1 1 a)(1 1 a ). c) Show that 1 1 a 1 a 1 a 1 a 5 0. 9 Show that (1 1 i ) n 1 (1 i ) n is real and find its value for n 5 18. 50 Given that z 5 (a 1 i ), and a 1, find the values of a such that arg z 5 15. Practice questions 1 Let z 5 x 1 yi. Find the values of x and y if (1 i )z 5 1 i. Let x and y be real numbers, and v be one of the complex solutions of the equation z 5 1. Evaluate: a) 1 1 v 1 v b) (vx 1 v y)(vy 1 v x) a) Evaluate (1 1 i ). b) Prove, by mathematical induction, that (1 1 i ) n 5 () n, where n N +. c) Hence or otherwise, find (1 1 i ). 59
10 Let z 1 5 6 i and z 5 1 i. a) Write z 1 and z in the form r (cos u 1 i sin u), where r > 0 and p < u < p. b) Show that z 1 z 5 cos p 1 1 i sin p 1. c) Find the value of z 1 z in the form a 1 bi, where a and b are to be determined exactly in radical (surd) form. Hence or otherwise, find the exact values of cos p 1 and sin p 1. 5 Let z 1 5 a ( cos p 1 i sin p ) and z 5 b ( cos p 1 i sin p ). Express ( z 1 z ) in the form z 5 x 1 yi. 6 If z is a complex number and z 1 16 5 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 )( i ) 5 5 i. 8 Given that z 5 (b 1 i ), 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 ) 5 1 z, where i 5 1. Write z in the form z 5 a 1 bi, where a and b are real numbers. 10 a) Express z 5 1 as a product of two factors, one of which is linear. b) Find the zeros of z 5 1, giving your answers in the form r (cos u 1 i sin u ) where r. 0 and p, u < p. c) Express z 1 z 1 z 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. ( cos p i sin p ) ( cos p 1 i sin p ) 1 Consider the complex number z 5 ( cos p i sin p. ) a) (i) Find the modulus of z. (ii) Find the argument of z, giving your answer in radians. b) Using De Moivre s theorem, show that z is a cube root of one, i.e. z 5 1. c) Simplify (1 1 z)( 1 z ), expressing your answer in the form a 1 bi, where a and b are exact real numbers. 1 The complex number z satisfies the equation _ z = + 1 i. 1 i Express z in the form x 1 i y where x, y Z. 1 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 5 1. b) The complex number z is defined by z 5 cos u 1 i sin u. (i) Show that 1 z 5 cos(u) 1 i sin(u). (ii) Deduce that z n 1 z n 5 cos nu. c) (i) Find the binomial expansion of (z 1 z 1 ) 5. (ii) Hence, show that cos 5 u 5 1 (a cos 5u 1 b cos u 1 c cos u), where a, b and 16 c are positive integers to be found. 60
15 Consider the equation (p 1 iq) 5 q ip (1 i ), where p and q are both real numbers. Find p and q. 16 Consider z 5 5 0. (i) Show that z 1 5 ( cos p 1 i sin p 5 5 ) is one of the complex roots of this equation. (ii) Find z, z, z and z 5 giving your answer in the modulus argument form. 1 1 1 1 (iii) Plot the points that represent z 1, z, z, z and z 5 in the complex plane. 1 1 1 1 (iv) The point z n 1 is mapped to z n 1 1 1 by a composition of two linear transformations, where n 5 1,,,. Give a full geometric description of the two transformations. 17 A complex number z is such that z 5 z i. a) Show that the imaginary part of z is. b) Let z 1 and z be the two possible values of z, such that z 5. (i) Sketch a diagram to show the points which represent z 1 and z in the complex plane, where z 1 is in the first quadrant. (ii) Show that arg(z 1 ) 5 p 6. (iii) Find arg(z ). c) Given that arg ( z 1 k z ) 5 p, find a value for k. i 18 Given that (a 1 i )( bi ) 5 7 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 z n 1 1 z n 5 cos nu. b) By expanding ( z 1 1 z ) show that cos u 5 1_ (cos u 1 cos u 1 ). 8 0 Consider the complex geometric series e iu 1 1_ eiu 1 1_ eiu 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 cos u 1 1_ cos u 1 1_ cos u 1 5 cos u 5 cos u. 1 Let P (z) 5 z 1 az 1 bz 1 c, where a, b and c. Two of the roots of P (z ) 5 0 are and ( 1 i ). Find the value of a, of b and of c. Given that z 5 5, find the complex number z that satisfies the equation 5 *z* 15 5 1 8i. z* Solve the simultaneous system of equations giving your answers in x 1 i y form: iz 1 1 z 5 z 1 1 (1 i )z 5 61
10 a) Solve the equation x x 1 8 5 0. Denote its two roots by z 1 and z and express them in exponential form with z 1 in the first quadrant. b) Find the value of z 1 and write it in the form x 1 yi. z c) Show that z 1 5 z. d) Find the value of z 1 z 1 z z. 1 e) For what values of n is z 1 n real? 5 a) Show that z 5 cos p 1 i sin p 7 7 is a root of the equation x7 1 5 0. b) Show that z 7 1 5 (z 1)(z 6 1 z 5 1 z 1 z 1 z 1 z 1 1) and deduce that z 6 1 z 5 1 z 1 z 1 z 1 z 1 1 5 0. c) Show that cos p 1 cos p 1 cos 6p 7 7 7 5 1_. Questions 1 International Baccalaureate Organization 6