EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 2- ALGEBRAIC TECHNIQUES TUTORIAL 2 - COMPLEX NUMBERS

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1 EDEXCEL NATIONAL CERTIFICATE UNIT 8 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME - ALGEBRAIC TECHNIQUES TUTORIAL - COMPLEX NUMBERS CONTENTS Be able to apply algebraic techniques Arithmetic progression (AP): first term (a), common difference (d), n th term e.g. n a +(n )d ; arithmetic series e.g. sum to n terms, S n a n ) d Geometric progression (GP): first term (a), common ratio (r), n th term e.g. ar n- ; n ar a geometric series e.g. sum to n terms, Sn, sum to infinity S n r r solution of practical problems e.g. compound interest, range of speeds on a drilling machine Complex numbers: addition, subtraction, multiplication of a complex number in Cartesian form, vector representation of complex numbers, modulus and argument, polar representation of complex numbers, multiplication and division of complex numbers in polar form, polar to Cartesian form and vice versa, use of calculator Statistical techniques: review of measure of central tendency, mean, standard deviation for ungrouped and grouped data (equal intervals only), variance It is assumed that the student has completed the module MATHEMATICS FOR TECHNICIANS. D.J.Dunn

2 . ROOTS OF NEGATIVE NUMBERS Ordinary numbers can be added, subtracted and multiplied and are good enough for every day use. In engineering, we come across problems that can not be solved with ordinary numbers and one of these problems is how to handle the square root of a negative number. You should already know that x = and that - x - = so it follows that is either or -. However there is no answer to the question what is the root of -? There is no number that can be multiplied by its self to give -. To get around this problem in the first instance, we simply designate - by the letter j and it follows that j = - and j = - Consider the equation x = -9. Using conventional numbers, there is no solution but using this new idea, the solution becomes j3 since (j3) = j x 3 = - x 9 = -9.. COMPLEX NUMBER Consider the number given as P A B If we use the j operator this becomes P A x B Putting j = -we get P = A + jb and this is the form of a complex number. WORKED EXAMPLE No. Find the solution of P 4 9 and express the answer as a complex number. SOLUTION P j3 SELF ASSESSMENT EXERCISE No.. Write down the solution to the following. x = -4 x = x = -5 x = x = -0 x = (Answer j, j5 and j 0). Express the following as complex numbers. P 3 6 P 8 P 5 Answers (3 + j4), ( j 9) and (-5 j ) D.J.Dunn

3 3. FURTHER PROPERTIES OF THE OPERATOR j Consider a point A on a Cartesian plane situated at coordinates 4, 0 as shown. Now suppose that multiplying this point by j has the affect of rotating the line O A 90 o anticlockwise. This will produce point B which should be designated as j4 to indicate it is on the vertical axis. Figure If we multiply point B by j we rotate again to get point C. This is located at -4 and was obtained by multiplying A by j. Since j = - then point C is at j 4 = -4 which is correct. If we multiply by j again and we get point D and this is j 3 4 = -j4 so point D is designated j4. This work was produced by a French mathematician called Argand. We may simplify matters by labelling the vertical axis j. Numbers on the horizontal axis are called REAL NUMBERS and on the vertical axis are called IMAGINARY NUMBERS. Point A is +4, point B is j4, point C is 4 and point C is j4. This is fine for handling negative numbers but does not explain what a complex number is. 4. ARGAND DIAGRAM A complex number A + jb could be considered to be two numbers A and B that may be placed on the previous graph with A on the real axis and B on the imaginary axis. Adding them together as though they were vectors would give a point P as shown and this is how we represent a complex number. The diagram is now called an Argand Diagram. Figure If we draw a line from the origin to the point P it forms a vector and in some applications it is called a phasor. The length of the line is called the MODULUS and the angle formed with the real axis is called the ARGUMENT. D.J.Dunn 3

4 The complex number can hence be expressed in polar form as OP θ Consider four such numbers on the Argand diagram, one in each quadrant as shown. Now consider the vector labelled No.. The horizontal component is 5 and the vertical component is 8 so the vector may be written as P = 5 + j8. The angle of the vector is tan - (8/5) = tan - (.6) = 58 o Figure 3 SELF ASSESSMENT EXERCISE No.. Write down the other three vectors in the form A + jb and calculate their angles. 3 4 D.J.Dunn 4

5 5. ADDING AND SUBTRACTING COMPLEX NUMBERS This is simply adding vectors together. If we have complex numbers: P = A + jb P = A + jb P 3 = A 3 + jb 3 Then adding we have P = (A + A + A 3 ) + j (B + B + B 3 ) Subtracting, simply put a minus instead of a +. WORKED EXAMPLE No. Add the complex numbers 3 + j, 6 - j4, and -4+j7 SOLUTION P = (3+6-4) + j(-4+7) P = 5 +j5 SELF ASSESSMENT EXERCISE No. 3. Two A.C. voltages are represented by the phasors V = 0 + j5 and V = 0 - j3 What is the resulting voltage phasor?. Two forces acting on a mass represented by the phasors F = 60 + j8 and F = 0 - j What is the resulting force? D.J.Dunn 5

6 6. MULTIPLYING COMPLEX NUMBERS USING POLAR CO-ORDINATES A complex number may be expressed in polar co-ordinates as follows. Let the Modulus be R and the argument. Consider the two shown. We have R and R Figure 4 We should not confuse the multiplication of vectors (see dot and cross products in the vector tutorials) with the multiplication of complex numbers. The real and imaginary co-ordinates are A = R cos B = R sin A = R cos B = R sin The complex number for each vector is: P R cosθ jr sinθ P R cosθ jr sinθ Multiplying them together and treating j as - we get the following. R {cosθ jsinθ } R {cosθ jsinθ } P P P P P x P x P x P x P x P R {cos θ jsin θ } x R {cos θ R R R R R R R R cos θ cos θ cos θ cos θ cos θ cos θ cos θ j(cos θ sin θ sin θ sin θ cos(θ θ ) jsin(θ jsin θ θ jsin θ j(cos θ sin θ ) jsin θ cos θ } jsin θ jsin θ sin θ cos θ ) j sin θ sin θ sin θ cos θ ) This is a vector with a length R R and angle +. The rule for multiplying is : The Modulus is the product of the other Modulii and the argument is the sum of the angles. This rule applies for any number of vectors multiplied together. D.J.Dunn 6

7 WORKED EXAMPLE No. 3 Find the result of (3 45 o ) x ( 30 o ) SOLUTION Modulus = 3 x 3 = 6 Argument is = 75 o The result is hence 675 o SELF ASSESSMENT EXERCISE No.3 Find the vector that result for each below o x 3 70 o o x 30 o USING COMPLEX FORM Consider the following problem. Multiply 3 45 o x 30 o. The result is 675 o. Figure 5 To do this as complex numbers is more difficult as we shall now see. In the form A + j B we have the following. P has coordinates A = 3 cos 45 =. and B = 3 sin 45 =. P has coordinates A = cos 30 =.73 and B = sin 30 =.0 P =. + j. and P =.73 + j D.J.Dunn 7

8 The result is a complex number P = P x P = (. + j.) x (.73 + j) P = P x P = (. x.73) + (. x j) + (j. x.73) + (j. x j) P = P x P = j. + j j P = P x V = j P = P x P =.5 + j5.794 This is shown on the diagram from which we deduce the following. Figure 6 R = ( ) = 6 = tan - (5.794/.5) = 75 o V = 675 o Hence we have arrived at the same solution but in a more difficult way. WORKED EXAMPLE No. 4 Find the result of multiplying the following complex numbers. P = (4 + j) x ( + j3) SOLUTION P = (4 + j) x ( + j3) = 8 + j +j4 + 6j = 8 + j6 6 = + j6 SELF ASSESSMENT EXERCISE No. 5 Find the result of multiplying the following complex numbers.. (3 + j3) x (5 j) Answer ( + 9j ). ( + j) x ( j3) Answer (30-3j ) D.J.Dunn 8

9 7. CONJUGATE NUMBERS The conjugate of a complex number has the opposite sign for the j part. The conjugate of A + jb is A jb. If a complex number is multiplied by its conjugate the result is a real number. WORKED EXAMPLE No. 5 Find the result of multiplying ( + j3) by its conjugate. SOLUTION The conjugate is ( - j3) ( + j3) x (- j3) = 4-6j +6j -9j = 4 (-9) = 3 SELF ASSESSMENT EXERCISE No. 6 Find the result of multiplying the following by their conjugate.. (5 j) (Answer 9). (-4 j4) (Answer 3) 3. (7 + j6) (Answer 85) D.J.Dunn 9

10 8. DIVISION OF COMPLEX NUMBERS Suppose V = 8 + j8 and I = 4 - j8 and we wish to find V/I. (This is Ohms Law for complex impedance). This is done by multiplying the top and bottom by the conjugate of the bottom number as follows. WORKED EXAMPLE No. 6 If A = 8 + j8 and B = 4 j8 find the result of dividing A by B. SOLUTION V 8 j8 I 4 j8 V 8 j8 4 j8 x I 4 j8 4 j8 V 3 j64 j3 j 64 I 6 j3 - j3 - j 64 V 3 j j j I SELF ASSESSMENT EXERCISE No. 7 Find the following results V j (Answer j) I j3 V 5 j (Answer j) I j4 x y j9 (Answer j) j6 The remainder of this tutorial is optional as it is not specifically mentioned in the syllabus. D.J.Dunn 0

11 9. PHASOR DIAGRAMS A phasor is used to represent harmonic quantities such as alternating electricity and oscillating mechanical systems. The phasor is a rotating vector with a constant length and the speed of rotation is the same as the angular frequency of the quantity (always anticlockwise). Projecting a rotating vector onto the vertical scale of a graph with angle plotted horizontally will generate a sinusoidal waveform. If the vector represents voltage or current it is called a PHASOR. The rotation is anti-clockwise. Figure 7 Suppose we wish to represent a sinusoidal voltage by a phasor. The maximum voltage is V and the voltage at any moment in time is v. The phasor is drawn with a length V and angle as shown. The angle is given by = t where t is the time and is the angular frequency in radian/s. The voltage at any moment in time is the vertical projection such that v = V sin (t) Since there is no necessity to start plotting the graph at the moment = 0 a more general equation is v = V sin (t+ ) where is the starting angle and is often referred to as the phase angle. A phasor may also be given in polar form as V(+) If the phasor is drawn on an Argand diagram, the vertical component is the imaginary part and the horizontal component is the real part. It follows that a harmonic quantity can be represented as a complex number. An Argand diagram may be used to show a phasor at a particular moment in time. It might show more than one phasor. For example when the voltage across an inductor is shown together with the current through it, the current is ¼ cycle behind the current so at a given moment in time the relationship might be like this. Figure 8 D.J.Dunn

12 WORKED EXAMPLE No. 7 A sinusoidal voltage has a peak value of 00 V and a phase angle of 0 o. Represent it as a polar vector and a complex number. Sketch the phasor when = 50 o. SOLUTION The polar form is v = o The vertical component is v = 00 sin(+0 o ) The horizontal component is 00 cos(+0 o ) The complex number is hence v = 00 cos(+0 o ) + {00 sin(+0 o )}j Putting = 50 o this becomes v = j The angle is tan-(87.9/68.4) = 70 o as expected. Figure 9 D.J.Dunn

13 0. REPRESENTING IMPEDANCE AS A COMPLEX NUMBER When an electric circuit with alternating current contains resistance, inductance and capacitance, the current and voltage will not vary in time together but one will lead the other. The above diagram shows one example of this. The impedance of an electric circuit is defined as Z = V/I and in order to divide V by I we must represent the phasor as a complex number. Suppose V = 8 + j8 and I = 4 -j8. V 8 j8 I 4 - j8 Multiply he t top and bottom4 line into a real number. This does V I j8. This is called the conjugate number and it turns the bottom The not change the equality. 8 j8 4 j8 3 j64 j3 j 64 3 j j96 x j8 4 j8 6 j3 j3 - j The Impedance is Z = (3/80) + j (96/80) On an Argand diagram the voltage and current are like this j 80 Impedance is also a phasor and looks like this. Figure 0 Figure The real part is called the Resistance part (R) and the imaginary part is called the reactive part X. X It follows that. Z X R and φ tan and is called the phase angle. R It follows that complex impedance may be written in the form Z = R + jx and impedances may be added or subtracted. D.J.Dunn 3

14 SELF ASSESSMENT EXERCISE No.8. An electric circuit has a complex impedance of Z = j40. What is the resistance of the circuit and what is the reactance? (300 and 40) What is the phase angle? (7.6 o ). Two electric circuits are connected in series. The complex impedance of the first is Z = 50 +j3 and the second is Z = -0 + j. What is the combined impedance? (Add them). (30 +5j) What is the resistance and reactance of the combined circuit? (30 and 5) What is the phase angle? (9.5 o ) 3. The current in an electric circuit is represented by I = 5 + j and the voltage is V = 00 + j5. Determine the impedance as a complex number. ( j) Determine the resistance and reactance of the circuit. (7.6 and 6) Determine the phase angle. (8.8 o ) D.J.Dunn 4