This leaflet describes how complex numbers are added, subtracted, multiplied and divided.

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1 7. Introduction. Complex arithmetic This leaflet describes how complex numbers are added, subtracted, multiplied and divided. 1. Addition and subtraction of complex numbers. Given two complex numbers we can find their sum and difference in an obvious way. If = a 1 + b 1 j and z = a + b j then + z =(a 1 + a ) + (b 1 + b )j z =(a 1 a ) + (b 1 b )j So, to add the complex numbers we simply add the real parts together and add the imaginary parts together. If = j and z =8 j find a) + z, b) z. a) + z = (13 + 5j) + (8 j) = 1 + 3j. b) z = (8 j) (13 + 5j) = 5 7j. Multiplication of complex numbers. To multiply two complex numbers we use the normal rules of algebra and also the fact that j = 1. If and z are the two complex numbers their product is written z. If =5 j and z = + 4j find z. Replacing j by 1 we obtain In general we have the following result: z = (5 j)( + 4j) = 1 + j 4j 8j z = j 8( 1) = j c Pearson Education Ltd

2 If = a 1 + b 1 j and z = a + b j then z =(a 1 + b 1 j)(a + b j) = a 1 a + a 1 b j + b 1 a j + b 1 b j = (a 1 a b 1 b )+j(a 1 b + a b 1 ) 3. Division of complex numbers. To divide complex numbers we need to make use of the complex conjugate. Given a complex number, z, its conjugate, written z, is found by changing the sign of the imaginary part. For example, the complex conjugate of z = 3 + j is z =3 j. Division is illustrated in the following example. Find z when = 3 + j and z =4 3j. We require = 3 + j z 4 3j Both numerator and denominator are multiplied by the complex conjugate of the denominator. Overall, this is equivalent to multiplying by 1 and so the fraction remains unaltered, but it will have the effect of making the denominator purely real, as you will see. 3 + j 4 3j = 3 + j 4 3j 4 + 3j 4 + 3j (3 + j)(4 + 3j) = (4 3j)(4 + 3j) 1 + 9j +8j +6j = j 1j 9j = j (the denominator is now seen to be real) 5 = j Exercises 1. If = 1 + j and z = 3 + j find a) z, b), c) z, d), e) z z. If = 1 + j and z = 3 + j find: a) z, b) z, c) /, d) z /z. 3. Find a) 7 6j j, b) 3+9j 1 j, c) 1 j. Answers 1. a) 1 + 5j, b) 1 j, c) 3 j, d), e) 13. a) 5 + j, b) 5 j 5, c) j, d) + 1j a) 3 7 j, b) 3 + 3j, c) j c Pearson Education Ltd

3 7.4 Introduction. The polar form From an Argand diagram the modulus and the argument of a complex number, can be defined. These provide an alternative way of describing complex numbers, known as the polar form. This leaflet explains how to find the modulus and argument. 1. The modulus and argument of a complex number. The Argand diagram below shows the complex number z = a + bj. The distance of the point (a, b) from the origin is called the modulus, or magnitude of the complex number and has the symbol r. Alternatively, r is written as z. The modulus is never negative. The modulus can be found using Pythagoras theorem, that is z = r = a + b The angle between the positive x axis and a line joining (a, b) to the origin is called the argument of the complex number. It is abbreviated to arg(z) and has been given the symbol. b r z = a + bj (a, b) a We usually measure so that it lies between π and π, (that is between 18 and 18 ). Angles measured anticlockwise from the positive x axis are conventionally positive, whereas angles measured clockwise are negative. Knowing values for a and b, trigonometry can be used to determine. Specifically, ) tan = b a so that = tan 1 ( b a but care must be taken when using a calculator to find an inverse tangent that the solution obtained is in the correct quadrant. Drawing an Argand diagram will always help to identify the correct quadrant. The position of a complex number is uniquely determined by giving its modulus and argument. This description is known as the polar form. When the modulus and argument of a complex number, z, are known we write the complex number as z = r. Polar form of a complex number with modulus r and argument : z = r c Pearson Education Ltd

4 Plot the following complex numbers on an Argand diagram and find their moduli. a) = 3 + 4j, b) z = +j, c) z 3 =3j The complex numbers are shown in the figure below. In each case we can use Pythagoras theorem to find the modulus. a) = 3 +4 = 5 = 5, b) z = ( ) +1 = 5 or.36, c) z 3 = 3 + = 3. -+j 4 3j 3+4j -3 O 3 Find the arguments of the complex numbers in the previous example. a) = 3 + 4j is in the first quadrant. Its argument is given by = tan 1 4. Using a calculator 3 we find =.97 radians, or b) z = +j is in the second quadrant. To find its argument we seek an angle,, in the second quadrant such that tan = 1. To calculate this correctly it may help to refer to the figure below in which α is an acute angle with tan α = 1. From a calculator α =.464 and so = π.464 =.678 radians. In degrees, α = 6.57 so that = = j 3 α 1 O 3 c) z 3 =3j is purely imaginary. Its argument is π, or 9. Exercises 1. Plot the following complex numbers on an Argand diagram and find their moduli and arguments. a) z = 9, b) z = 5, c) z = 1 + j, d) z = 1 j, e) z =8j, f) 5j. Answers 1. a) z = 9, arg(z)=, b) z = 5, arg(z) =π, or 18, c) z = 5, arg(z) =1.17 or 63.43, d) z =, arg(z) = 3π 4 or 135, e) z = 8, arg(z) = π or 9, f) z = 5, arg(z) = π or c Pearson Education Ltd

5 7.5 The form r(cos + j sin ) Introduction. Any complex number can be written in the form z = r(cos + j sin ) where r is its modulus and is its argument. This leaflet explains this form. 1. The form r(cos + j sin ) Consider the figure below which shows the complex number z = a + bj = r. b r z = a + bj (a, b) a Using trigonometry we can write cos = a r and sin = b r so that, by rearranging, a = r cos and b = r sin We can use these results to find the real and imaginary parts of a complex number given in polar form: if z = r, the real and imaginary parts of z are: a = r cos and b = r sin, respectively Using these results we can then write z = a + bj as z = a + bj = r cos + jr sin = r(cos + j sin ) This is an alternative way of expressing the complex number with modulus r and argument c Pearson Education Ltd

6 z = a + bj = r = r(cos + j sin ) State the modulus and argument of a) z = 9(cos 4 + j sin 4 ), b) z = 17(cos 3.+j sin 3.). a) Comparing the given complex number with the standard form r(cos + j sin ) we see that r = 9 and = 4. The modulus is 9 and the argument is 4. b) Comparing the given complex number with the standard form r(cos + j sin ) we see that r = 17 and =3. radians. The modulus is 17 and the argument is 3. radians. a) Find the modulus and argument of the complex number z =5j. b) Express 5j in the form r(cos + j sin ). a) On an Argand diagram the complex number 5j lies on the positive vertical axis a distance 5 from the origin. Thus 5j is a complex number with modulus 5 and argument π. b) Using degrees we would write z =5j = 5(cos π + j sin π ) z =5j = 5(cos 9 + j sin 9 ) a) State the modulus and argument of the complex number z =4 (π/3). b) Express z =4 (π/3) in the form r(cos + j sin ). a) Its modulus is 4 and its argument is π 3. b) z = 4(cos π 3 + j sin π 3 ). Noting cos π 3 = 1 and sin π 3 = 3 the complex number can be written + 3j. Exercises 1. By first finding the modulus and argument express z =3j in the form r(cos + j sin ).. By first finding the modulus and argument express z = 3 in the form r(cos + j sin ). 3. By first finding the modulus and argument express z = 1 j in the form r(cos + j sin ). Answers 1. 3(cos π + j sin π ),. 3(cos π + j sin π), 3. (cos( 135 )+j sin( 135 )) = (cos 135 j sin 135 ) c Pearson Education Ltd

7 7.6 Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. This is an advantage of using the polar form. 1. Multiplication and division of complex numbers in polar form. if = r 1 1 and z = r then z = r 1 r ( 1 + ), z = r 1 r ( 1 ) Note that to multiply the two numbers we multiply their moduli and add their arguments. To divide, we divide their moduli and subtract their arguments. If =5 (π/6), and z =4 ( π/4) find a) z, b) z, c) z a) To multiply the two complex numbers we multiply their moduli and add their arguments. Therefore ( ( π z = 6 + π )) ( = π ) 4 1 b) To divide the two complex numbers we divide their moduli and subtract their arguments. = 5 ( π ( z 4 6 π )) = π 1 c) Exercises z = 4 ( 5 π 4 π ) = 4 ( 6 5 5π ) 1 1. If =7 π 3 and z =6 π find a) z, b) z, c) z, d) z 1, e) z 3. Answers 1. a) 4 5π, 6 b) 7 π, 6 6 c) 6 π, 7 6 d) 49 π, 3 e) 16 3π c Pearson Education Ltd

8 The modulus and argument of a complex number sigma-complex9-9-1 In this unit you are going to learn about the modulus and argument of a complex number. These are quantities which can be recognised by looking at an Argand diagram. Recall that any complex number, z, can be represented by a point in the complex plane as shown in Figure 1. The complex number z is represented by the point P P imaginary axis length OP is the modulus of z this angle is the argument of z real axis Figure 1. The complex number z is represented by point P. Its modulus and argument are shown. We can join point P to the origin with a line segment, as shown. We associate with this line segment two important quantities. The length of the line segment, that is OP, is called the modulus of the complex number. The angle from the positive axis to the line segment is called the argument of the complex number, z. The modulus and argument are fairly simple to calculate using trigonometry.. Find the modulus and argument of z = 4 + 3i.. The complex number z = 4 + 3i is shown in Figure. It has been represented by the point Q which has coordinates (4, 3). The modulus of z is the length of the line OQ which we can find using Pythagoras theorem. (OQ) =4 +3 = = 5 and hence OQ =5. y Q(4,3) 1 N x Figure. The complex number z = 4 + 3i. Hence the modulus of z = 4 + 3i is 5. To find the argument we must calculate the angle between the x axis and the line segment OQ. We have labelled this in Figure. 1 c mathcentre 9

9 By referring to the right-angled triangle OQN in Figure we see that tan = 3 4 = tan = To summarise, the modulus of z = 4 + 3i is 5 and its argument is = There is a special symbol for the modulus of z; this is z. So, in this example, z =5. We also have an abbreviation for argument: we write arg(z) = When the complex number lies in the first quadrant, calculation of the modulus and argument is straightforward. For complex numbers outside the first quadrant we need to be a little bit more careful. Consider the following example.. Find the modulus and argument of z =3 i.. The Argand diagram is shown in Figure 3. The point P with coordinates (3, ) represents z =3 i. y N x P(3,-) α Figure 3. The complex number z =3 i. We use Pythagoras theorem in triangle ONP to find the modulus of z: (OP) =3 + = 13 OP = 13 Using the symbol for modulus, we see that in this example z = 13. We must be more careful with the argument. When the angle shown in Figure 3 is measured in a clockwise sense convention dictates that the angle is negative. We can find the size of the angle by referring to the right-angled triangle shown. In that triangle tan α = 3 so that α = tan 1 3 = This is not the argument of z. The argument of z is = We often write this as arg(z) = In the next unit we show how the modulus and argument are used to define the polar form of a complex number. c mathcentre 9

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