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Lecture 1 Complex Numbers 1: Introduction The aim of these notes is to expand our present number system so we can solve a larger class of problems. Motivating Problem Consider the equation x y 10 and xy 25 . The solution to this equation are easily found; it is x=5 and y=5. We now consider a very similar equation x y 10 and xy 26 Unlike the first equation, this seems more difficult to solve. Since x y 10 we must have x 5 i and y 5 i where i is a number we will determine later. We put the conditions x 5 i and y 5 i into xy 26 , this gives: (5 i)(5 i) 26 Expanding now gives: 25 i 2 26 . Therefore, i 2 1 and hence i 1 This means if we want to solve the second equation we need to introduce a number i such that i 2 1 . This might seem rather strange, but it is similar to introducing negative numbers to solve equations such as: x 5 3 , or fractions to solve an equation like 2x 1 . Of course, we must ensure that our new number makes sense and doesn’t violate any of the previous rules we have laid down for our number system! (Note you might be tempted to think that we can just invent numbers to get ourselves out of any sort of trouble like not being able to divide by zero. This certainly doesn’t work if we want out invented numbers to be consistent with our existing number system. Suppose we made the suggestion that 1/0=∞, where ∞ is our new symbol we have invented for use when we divide by zero! This means 1=0×∞. Now it is certainly true that 2×0=3×0. Multiplying both sides by ∞ to get (2×0) ×∞ = (3×0) ×∞. We now have 2×(0 × ∞)=3×(0 × ∞). Using the fact that 1=0×∞ we get 2×1=3×1 and so 2=3! Clearly, this is rubbish! So we are not free to add any number we want to our number system.) 1 Lecture 1 Properties and uses of i Consider the quadratic equation x2 2x 2 0 . (1) Using the quadratic formula with a 1, b 2 and c 2 x 2 22 4 1 2 2 2 1 4 2 . Since the discriminant b 2 4ac is negative 4 we cannot obtain roots to this equation that are real numbers. The roots are can be simplified using our new number i. Returning to equation (1) x 2 4 1 2 4 1 2 2i 1 i . 2 2 2 The roots of equation 1 are therefore the complex numbers 1 i and 1 i . We now check whether 1 i really does solve the original equation. We put 1 i into equation (1) (1 i) 2 2(1 i) 2 The expression (1+i)2 reminds us of expanding brackets, so we expand brackets in the usual way. (1 i) 2 2(1 i) 2 (1 2i i 2 ) 2 2i 2 0 since i 2 1 . General Properties of Complex Numbers We now explore the general properties of i. The number i is an example of a complex number. The general form of a complex number is z x yi where x and y are real numbers. We call x the real part of z, written Re(z) and y is called the imaginary part of z, written Im(z). 2 Lecture 1 Example 1 i) ii) iii) Given z 3 5i , then Rez 3, Imz 5 If z 2i 0 2i , then Rez 0, Imz 2 If z 7 7 0i , then Rez 7, Imz 0 . Equal Complex Numbers Two complex numbers are defined x yi and u vi where x, y, u and v are real. They are equal if and only if (iff) x u and y v. This might seem a rather trivial statement, but you should compare equality of complex number with equal fractions. Operations on Complex Numbers For the complex numbers x yi and u vi we have the following basic operations addition: subtraction: multiplication: x yi u vi x u y v i , x yi u vi x u y v i , x yi u vi xu xvi yui yvi2 xu yv xv yu i . Example 2 i) ii) iii) iv) 2 5i 3 i 2 3 5 1 i 5 6i 8 3i 6 4i 8 6 3 4 i 14 i 9 3i 6 2i 9 6 3 2 i 3 i 8 3i 6 2i 48 16i 18i 6i 2 54 2i . So far we have not said how to divide complex numbers, before we can do this we need an additional operation. Complex Conjugate Consider 2 3i 2 3i 4 6i 6i 9i 2 13 . The product is a real number. 2 3i and 2 3i are complex conjugate numbers and 2 3i is the conjugate of 2 3i . Any pair of complex numbers of the form x yi and x yi are said to be conjugate, each is the conjugate of the other. The conjugate of z x yi is z x yi . 3 Lecture 1 zz x yi x yi x 2 xyi xyi y 2i 2 x 2 y 2 . Example 3 i) ii) 8 i 8 i 82 12 65 4 2i 4 2i 42 22 20 . Division of a Complex Number When dividing by a complex number we make the denominator real by multiplying by the conjugate of the divisor. Example 4 i) ii) iii) iv) 1 , conjugate of 3 i is 3 i . 3i 1 1 3i 3i 3 i 2 2 . 3 i 3 i 3 i 3 1 10 10 5i 5i 6 2i 10 30i 1 3i 2 . 2 6 2i 6 2i 6 2i 6 2 4 4 7 5i 7 5i 6 2i 42 14i 30i 10 4 11i . 6 2i 6 2i 6 2i 40 5 10 Square roots of complex numbers We know how to find the square root of numbers such as 100, 4, 6 10 and so on. How do we find the square root of a complex number? We make use of equality of complex numbers. We find the square root of 3 4i . The square root of 3 4i will be a complex number, so we write 3 4i a bi , where a and b are real and will be determined below. Then 3 4i a bi a 2 2abi b 2i 2 a 2 b 2 2abi . 2 Equating real and imaginary parts (recall two complex numbers are equal if and only if their real and imaginary parts are equal) a 2 b2 3 (2), 2ab 4 (3) . From (3) 4 Lecture 1 b 2 a Substitute into (2) 2 a 3 a 4 a2 2 3 a 2 2 Multiply both sides by a 2 a 4 4 3a 2 a 4 3a 2 4 0 Factorising a 2 4 a2 1 0 a 2 4, a 2 1 a 2, a i. Discard a i since a is real. Substituting into (3) a2 a 2 4b 1, b 1 4b 1, b 1. Hence the square root of 3 4i are the complex numbers 2 i . 5 Lecture 1 Tutorial Questions The tutorial questions consist of three sections: Warm-up questions, section A and section B. The warm-up questions are designed to allow you attempt some simple questions on the lecture material and test your own learning. They also help prepare you for the section A and B questions. Section A questions show the level of learning required to answer the section A questions on the test and exam. Section B questions show the level of learning required to answer the section B questions on the test and exam. Note that not every set of tutorial questions will contain section A and B questions. Warm-up Questions Q1 Given that i 1 and i 2 1 simplify the following a) e) Q2 i4 i 3 c) g) d) h) i5 i9 i7 i n , where n is a positive integer x2 4x 5 0 x2 9 0 b) d) x2 2x 5 0 x4 1 0 Find the real and imaginary parts of the following complex numbers a) c) e) Q4 b) f) Solve the following equations a) c) Q3 i3 i 1 z 2 6i z 10 2 i 3 i b) d) f) z 3 7i z 16i 1 i 3 Simplify the following complex numbers a) c) e) g) i) k) 2 7i 5 3i 3 5i 7 i 3i7 4i 2 i 2 i1 i 2 3i b) d) f) 2 1 i l) h) j) 2 7i 5 3i 3 3i 4 i 3 4i 3 4i 1 i 3 74 3i 3i 4 3i 6 Lecture 1 m) o) 4i 4i 3i i n) 1 i 1 i Section A Questions Q5 Solve the following equations for x and y a) b) c) Q6 x yi 3 i 2 3i 3 4i x yi 1 i x yi 5i 2i Find the square roots of the following complex numbers a) 15 8i b) 21 20i c) 2i Q7 For the complex numbers z x yi and w u vi solve the simultaneous equations 3z 2w 38 and 2z iw 7 Q8 For z x yi solve the following z 2 16 0 z 2 5iz 6 0 a) b) Q9 Let z=(3+4i)/(2-3i). Find the complex number w which satisfies the equation zw=1. Q10 Find the values of a and b (with a>0) which satisfy (a+bi)2=5+12i. 7 Lecture 1 Solutions Should you get stuck on a tutorial question then the scheduled weekly tutorials will offer you support and advice with them. Note you are expected to have tried the questions before the weekly tutorials. 1 a e h i b 1 c i d i f i g i If n 4 without remainder, i n 1 if n 4 gives remainder 1, i n i if n 4 gives remainder 2, i n 1 if n 4 gives remainder 3, i n i 2 a c x 2i x 3i 3 a c e Rez 2, Imz 6 Rez 10, Imz 0 Rez 7, Imz 1 a c e g i 7 10i 26 32i 12 21i 3 4i 5i b d f h j k 1 i l m 4 16i 17 17 1 3i n i b 5 2i 4 o c 9 7i 7 i 2 2 11 3i 6 a c 4 i 1 i 7 z 8 a 5 a b b d x 1 2i x 1, i Rez 3, Imz 7 Rez 0, Imz 16 Re z 2, Im z 2 b d f 3 4i 1 4i 25 2 2i 28 21i 9 13i 25 25 34 22i 44 33i , w 5 5 5 5 z 4i b i z 2i or z 3i 8 Lecture 1 The following are sample worked solutions to questions 9 and 10. 9 If zw=1 then w=1/z, so 2 3i (2 3i)(3 4i) 6 17i 12 6 17i 6 17 w i 3 4i (3 4i)(3 4i) 25 25 25 25 10 (a bi ) 2 5 12i a 2 2abi b 2 5 12i Equating real and imaginary parts gives : 2ab 12 b 6 a 36 5 a 4 5a 2 36 0 (a 2 9)( a 2 4) 0 2 a Since a is real and positive, a 3 and therefore b 2. a2 b2 5 a2 9