Multiplication of complex numbers When multiplying complex numbers, we use the general algebraic method for multiplication. Example 1: Consider the complex numbers z and w, where z = 3 + 2i and w = 5–4i. Then zw = (3 + 2i)(5– 4i) = 3(5 – 4i) + 2i(5 – 4i)………[Use distributive property] = 15 – 12i + 10i – 8i2 ……[Expand each bracket] = 15 – 2i – 8(–1) ……..[Simplify. Note that i2 = –1] = 15 – 2i + 8………….[Simplify. (–8)(–1) = + 8]] = 23 –2i …………..[Simplify] Example 2: Simplify ( 5 – i)2 and give your answer in the form a + bi, where a and b are real numbers. Solution: (5 – i)2 = (5 – i)(5 – i) = 5 ( 5 – i) – i(5 – i)…….[Use the distributive property] = 25 – 5i – 5i + i2 ……..[Expand] = 25 – 10i + (–1) ……..[Add like terms. Note the value of i2 = –1] = 25 – 1- 10i = 24 - 10i …………..[Simplify] Challenge to the reader. Determine the product of: ( 3 + 4i)(2 – 5i)(1 – 2i) Explain your method to your colleague. Activity 5 Simplify each of the following expressions. Write your answer in the form a + bi, where a and b are real numbers. a) (3 + i)(2 + 3i) b) (4 – 2i)(5 + 3i) c) (8 – i)(9 + 2i) d) (–3i + 9)(5 – i) e) i(2 – 3i)(i + 4) f) (3 – 2i)(7–5i) Simplify each of the following expressions, giving your answer in the form a + bi, where a and b are real numbers.
= ………[Expand numerator and denominator] = ……..[Simplify. Note i2 = –1] = ………..[Simplify] = = +i = + i Example 3: If = – 3 –5i, determine the value of in the form a + bi, where a and b are real numbers. Solution: Step 1: Determine the of – 1: – 1 = –3 –5i – 1 – 1 = – (4 + 5i) Step 2: Determine the value of + 1: + 1 = – 3 – 5i + 1 + 1 = – (2 + 5i) Step 3: Now proceed to determine the value of: = = ….[Multiply by conjugate] = ….[Expand] = …..[i2 = –1] = = = – i Activity 6 Express each of the following fractions in the form a + bi, where a and b are real numbers. a) b) c) d)
e) (3 – i)–2 f) g) h) i) + j) – If = , express in the form a + bi, a and b are real numbers. The Complex Plane In some branches of mathematics, it is convenient to view complex numbers, x + yi , as ordered pairs (x ; y), of real numbers. In that case, the set of complex numbers is defined by: ={z: z = ( x ; y), where x , y } When the complex number x + yi is regarded as the ordered pair (x ; y) of real numbers, it can be represented by the point in the XY-Plane with coordinates (x ; y). Consider the general complex number represented by: z = x + yi. This number can be represented by a vector from O to the P(x ; y); i.e.: z = .
Example: Represent the following complex numbers on an Argand diagram, and state the modulus for each of the numbers. z1 = 3 z2 = 4 + 3i z3 = 5i z4 = –2 + 3i z5 = –3 – i z6 = – 2i Solution: = = = 3 = = = = 5 = = 5 = = = = 2 Activity 7 Represent the following complex numbers on an Argand diagram. a) 5 b) 5i c) –5 d) –5i e) –5 + 5i f) –5 – 5i g) 5 – 5i h) – 1 + 2i i) 2–i j) –6 – 2i k) – 6i l) 0+ 0i Copy and complete the following table Complex Number (z) Complex Number (z) a) 4 b) 3i c) – 3 d) – i e) 1 + i f) 2 – 2i g) 1 + i h) –1 – i i) – i j) 2 + i
k) 1–3i l) – 1 + 3i If z = 3 + 5i, represent each of the following on an Argand diagram z zi zi2 zi3 zi4 If z1 = 5 + 2i and z2 = 2 + 4i. Determine z1 + z2 algebraically. Hence, show z1 + z2 graphically. Given: z1 = 8 –4i and z2 = 4 – 6i. Determine z1 – z2 algebraically. Hence show z1 – z2 graphically. The argument of z Examples
1. Determine the argument of: z = 3-3i x = 3 ; y = – 3 2. Determine the argument of: z = – 1 + 2i x = – 1 ; y = 2 tan = – 2 = 116,57o = arg z = 0,6476 3. Determine the argument of : z = – 1 – i x = – 1 ; y = – 1 = = 1 = . But this angle lies in the third quadrant, so to calculate the arg z( i.e. the angle from the positive real axis or in a clockwise direction) is : arg z = – = If the angle is measure anti-clockwise, then the angle will be: = = 4. Determine the argument of: z = 2 + 2 i x = 2 ; y = 2 = = = arg z = Activity 8 Determine the argument of the following complex numbers Complex Number (z) arg z Complex Number (z) a) 4 b) 3i c) – 3 d) – i e) 1 + i f) 2 – 2i g) 1 + i h) –1 – i i) – i j) 2 + i k) 1–3i l) – 1 + 3i Given: z = Write z in the form a + bi Hence determine the modulus and argument of z
The complex numbers z1 and z2 are given by: z1 = 5 + i ; z2 = 2 – 3i Represent the points z1 and z2 on an Argand diagram Determine z1 – z2 algebraically Hence determine the modulus of z1 – z2 Determine a complex number in the form a + bi, where a and b are rational numbers. Hence, determine the argument of , giving your answer in radians correct to two decimal places. Determine the values of the real constants p and q such that: = 2i Trigonometric (polar) form of complex numbers From the above diagram: OP2 = r2 = a2 + b2 …………….[Pythagoras] r = And, tan = => = tan–1 From our earlier work of Trigonometry, we know that: sin = and cos = Thus, b = r sin and a = r cos z = a + bi z = r cos + r i(sin ) z = r[cos + i(sin )] Thus, if z has a modulus and an argument , the polar form of z is given by: = r[cos + i sin ] = cis
Solving equations with complex numbers with two variables In this section we will solve equations with complex numbers involving two variables. Example 1: If (x + yi)(2 – i) = –i, determine the values of x and y if x, y . Solution: Method 1: (x + yi)(2 – i) = –i, ( x + yi) = ………..[Divide both sides by (2 – i)] (x + yi) = ……..[Multiply RHS by conjugate of denominator] (x + yi) = (x + yi) = ……..[ i2 = –1] ( x + yi) = ………...[Simplify numerator and denominator] x + yi = + x + yi = – Now, comparing real part and imaginary parts of the equation: x = and y = – Method 2: (x + yi)(2 – i) = –i, 2x – xi + 2yi – yi2 = – i ……[Expand LHS} 2x – y(–1) – xi + 2yi = – i ……[i2 = –1] 2x + y – (x –2y)i = – i ……..[Group real terms and imaginary terms] 2x + y – (x – 2y)i = 0 – i…….[Write RHS in form a + bi] Comparing real parts: 2 x + y = 0 ………..(1) Comparing imaginary parts: x – 2y = 1……..(2) Solving the above equations simultaneously as follows: (2) 2: 2x –4y = 2 …….(3) (3) – (1): –5y = 2 y = – Substitute y = in (1): 2x – = 0
2x = x = Activity 10 for x and y in each of the following equations, if x, y a) 2x + 3yi = – x –6i b) (x + yi)(2 – i) = 8 + i c) (3 + 2i)(x + yi) = – i d) x2 + xi = 4 – 2i e) 2(x + yi) = x – yi f) (x + yi)(2 + i) = 2x – (y + 1)i g) x + yi = 4 – 2i h) x + iy + 3 –2i = 4(–2 +5i) i) (2 + i)(3–2i) = x + yi j) (3–5i)(4 + i) = x + yi k) x + yi = l) (2 –3i)2 = x + yi POLYNOMIALS Outcomes for this section: By the end of this section, you should be able to: Define a polynomial Revise methods of factorization studied in earlier grades Apply the remainder and factor theorems to cubic polynomials Use long division and synthetic division to factorise polynomials Factorise cubic polynomials Polynomials You have studied polynomials in depth in earlier grades. Recall that a quadratic is a special form of a polynomial. Below are some examples of polynomials: 3x3–2x2+1 3 4–2x x2 1 2x4 1–2x+3x5 x2 x17 x The number n is called the degree of the polynomial; The expressions axn ; bxn–1,…,k which make up the polynomial are called terms. The numbers a, b, c,…..k; are called coefficients; a is called the leading coefficient. The coefficient k is called the constant term.
We will call this sentence a division statement. Activity 1 Perform the following operations. For each operation identify the dividend, the divisor, quotient and remainder. b) c) Using the method outline, lets extend our method outlined above to polynomials. Examples 1. Find the quotient and the remainder when the polynomial x2 + 4x – 5 is divided by x + 3 We begin by writing x2 + 4x – 5 in terms of a quotient and a remainder gives an expression in the form: x2 + 4x – 5 (x +3).Q(x) + Remainder We observe that the quotient must be of the form: ax + b. So, x2 + 4x – 5 (x +3)(ax + b) + R ax2 + bx + 3ax + 3b + R ax2 + (b + 3a)x +(3b +R) Comparing coefficients: x2 : a = 1………..(1) x: b + 3a = 4…….(2) b + 3(1) = 4 b = 1 x0 : 3b + R = –5 3(1) + R = – 5 R = – 8 Q(x) = (x + 1) ; R = – 8 2. (x +2).Q(x) + R We observe that the quotient must be a quadratic, i.e: Q(x) = ax2 + bx + c (x +2)( ax2 + bx + c) + R ax3 + (2a + b)x2 + (2b + c)x + (2c + R) Comparing coefficients: x3: a = 2……..(1)
Introduction Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. To a in the days of the empire a “calculus” was a pebble used in counting and gambling. Centuries later “calculare” came to mean “to compute”, “to reckon”, “to figure out”. To the mathematician, physical scientist, and social scientist of today, calculus is elementary mathematics (algebra, geometry, trigonometry) enhanced by the limit process. Calculus ideas from elementary mathematics and extends them to a more general situation. The table below contains some examples. On the left-hand side, you will find an idea from elementary mathematics; on the right, this same idea as extended by calculus. Elementary Mathematics Calculus Slope of a line y = mx + c of a curve y = f(x) Tangent line to a circle Tangent line to a more general curve Average velocity, Average acceleration Instantaneous velocity, Instantaneous acceleration Distance moved under a constant velocity. Distance moved under varying velocity. Area of a region bounded by line segments. Area of a region bounded by curves Length of a line segment Length of a curve Differentiation is used in calculations involving velocity and acceleration, of change and maximum and minimum values of curves. Before, we proceed; let us first revise some concepts which we encountered in earlier grades.
Worked Examples 1. Consider the function f(x) = 1 – x2 Solution 1.1 Determine the value of: Step 1: Find the value of f(2 + h), i.e: f (2 + h) = 1 – (2 + h)2 = 1 – ( 4 + 4h + h2) f (2 + h) = –3 –4h –h2 Step 2: Find the value of f(2), i.e: f(2) = 1 – (2)2 f(2) = –3 Step 3: Now using your results from Steps 1 and 2 above, evaluate the given expression, i.e.: = = = = ; h 0 = –4 –h 1.2 Interpret your answer This answer represents the average gradient of f(x) = 1 –x2 between x = 2 and x = 2 + h 2. Given: f(x) = x2 + 1. 2.1 Find the average gradient between the points R (a ; f(a)) and S(a + h ; f(a +h)) Step1 : Find the value of f(a) , i.e: f(a) = a2 + 1 Step 2: Find the value of f(a + h) , i.e.: f(a+h) = (a + h)2 + 1 = a2 + 2ah + h2 + 1 Step 3: Now determine the average gradient,i.e: = = = ; h 0 = 2a + h
2. Differentiate, y = 3x + 6, from first principles. Solution: Let f(x) = 3x + 6 Then; f(x + h) = 3(x + h) + 6 = 3x + 3h + 6 = = = = (x) = 3 3. Use first principles, to determine the derivative of the function defined by g(x) = –5. Solution: g(x) = –5 is a horizontal line with gradient 0; i.e. g(x) = –5x0 Then: g(x + h) = –5(x +h)0 = – 5 = = = = (x) = 0 Activity 5 Differentiate the following functions from principles. a) f(x) = x b) f(x) = 3x c) f(x) = x + 2 d) f(x) = 4 e) f(x) = 3 – 6x f) f(x) = a + 2px g)** f(x) = 3x2 h)** f(x) = x2 + 4 Differentiation rules To calculate the derivative of, say f(x) = xgoogle, from first principles this may be a daunting task to the ordinary human being. To assist in accomplishing this and other similar tasks, rules for differentiation have been developed and accepted. Suppose that y = xn , where n is a real number Then: = nxn–1 8.1 Notation The following are the commonly used symbols depicting differentiation of a function.
y – (–7) = 2(x – 1) y + 7 = 2x – 2 y = 2x – 9 3. Calculate the point on y = 3x2 – 4x – 6 where the gradient is 2. Solution: Step 1: Determine ; i.e: = 6x – 4 Step 2: You are the value of m = 2 i.e. 6x – 4 = 2 6x = 6 x = 1 Step 3: Use the x –value obtained in Step 2 above to obtain the corresponding i.e: y = 3(1)2 – 4(1) – 6 y = – 7 Step 4: down the coordinates ; i.e (1; – 7) Activity 7 1. Determine, the gradient of the curve at the given value of x Function of x a) f(x) = x2 – 2x x = 3 b) g(x) = x3 – 2x2 + x – 1 x = 2 c) xy = 3 x = 1 d) y = (3x –1)2 x = 0 e) y = x = 2 f) y = x = 1 2. Determine the equation of the tangents to the curve at the given value of x a) y = x3 – 2x2 + 4 , at x = 2 b) xy = 4 at x = – 2 c) y = x2 – 2x – 8, at x = 1 d) y = x3 at x = 1 e) y = – 3x2 + 2x + 1 , at x = – 1 f)** y = at x = 2 3. Determine the coordinates of the point P at which the gradient of the curve y = x2 is 6. 4. Determine the equation of the tangent to the curve y = 3x – x2 which is parallel to the line y – x – 4 = 0. 5** Q(x ; y) is any point on the curve defined by y = x2 – 5x. a) Determine the equation of the tangent at Q which is parallel to the straight- line y = 2x + 1 b) Draw a sketch to illustrate the given information c) Calculate the co-ordinates of the point Q. d) Hence, determine the equation of the straight-line PQR
Y-intercepts: Let x = 0, calculate f(0), so y-intercept is (0 ; d) X -intercepts: let f(x) = 0. Factorise f(x) by: grouping of terms OR using the factor theorem OR synthetic division Step 3: Stationary points Determine (x). Let (x). = 0 and solve for x. The solutions are the x-coordinates of the stationary points, say x = p or x = t Now find f(p) and f(t). These give you the y-coordinates of the stationary point. Step 4: Sketch the graph by: Plotting the intercepts (x and y-intercepts) Stationary points Join them to form the graph Worked example: Use the method outlined above, to sketch the graph of y = f(x) = –x3 + 3x + 2. Solution: Shape - a < 0 Intercepts: Y-intercept. Let x = 0, y = 2 ; (0; 2) X-intercepts: Let y= 0 : Then 0 = –x3 + 3x + 2. f(–1) = –1(–1)3 + 3(–1) + 2 = 0. Thus, (x + 1) is a factor of –x3 + 3x + 2. Find the other factors by inspection or division. f(x) = –x3 + 3x + 2 = (x +1)(–x2 + x + 2) 0 = –(x + 1)(x2 – x – 2) 0 = –(x +1)(x –2)(x + 1) So, x-intercepts are given by: (–1 ; 0) ; (–1 ; 0) and (2 ; 0) 3.Stationary points: f(x) = –x3 + 3x + 2. (x) = –3x2 + 3 (x) = 0 –3x2 + 3 = 0 –3(x2 –1) = 0 (x – 1)(x + 1) = 0 x = 1 OR x = – 1
In differentiation, the differential coefficient indicates that a function of x is differentiated with respect to x, the dx indicating that it is “with respect to x”. In integration, the variable of integration is shown by adding d(the variable) after the function to be integrated. When we want to integrate a function, we use a special notation: . Thus, to integrate 4x, we will write it as follows: = 2x2 + c , c . Note that along with the integral sign, there is a term of the form dx, which must always be written, and which indicates the variable involved, in our example x. We say that 4x is integrated with respect x. The function being integrated is called the integrand. Technically, integrals of this type are called indefinite integrals, to distinguish them from definite integrals, which we will deal with later. When you are required to evaluate an indefinite integral, your answer must always include a constant of integration. Formally, we define the anti-derivative as: If f(x) is a continuous function and F(x) is the function whose derivative is f(x), i.e.: (x) = f(x) , then: = F(x) + c; where c is any arbitrary constant. Activity 2 In each of the following determine the function f(x), if the derivative (x) is given: No. (x) f(x) 1. 4x 2. x4 3. 2x2 4. 0 5. 6. -2x3 7. Explain how the derivative (x) and the function f(x)are related to each other
= [ + + + 4 + + + + 3 + + + + 0] = square OR 8,59 square units Activity 4 1. Using rectangles, calculate the areas described in each of the following cases. Use a partition with a sensible number of rectangles. Make sure each rectangle has the same width. Make a rough sketch for each case. a) f(x) = 2x + 7 between the x-axis, and x = – 2 and x = 3. b) k(x) = x2 bounded by the x –axis and x = – 4 and x = – 1. c) j(x) = , with x > 0, bounded by the x-axis and x = 1 and x = 4. d) m(x) = x2 + 2, between the x-axis and x = – 5 and x = 3. Is this area an or overestimate of the true area? e) p(x) = between the x-axis and the two x-intercepts. Can you work out the exact area under this curve? How accurate was your partitioning method? f) h(x) = 2 sinx, between the x-axis and x = and x = . 2. Create an Excel spreadsheet to approximate the area bounded by the curve t(x) = –x2 + 4 on the interval [–1 ; 2] for 3 different rectangle widths. 3. Determine the area under the function g(x) = , for x [– 1 ; 6], by dividing the given interval into five trapeziums. Let the height of each trapezium be 1 unit. Use the following formula to assist you to calculate the area of each trapezium: Area = (sum of parallel sides) height 3.1 How does this method compare with method of dividing the given interval into rectangles? What areas that fall below the x-axis? Consider the following example: h(x) = x(x – 2)(x + 2)
= = – A = square units Find the area between the curve y = x2 + 4x and the x-axis from: x = – 2 to x = 0 x = – 2 to x = 2 Solution: Let A1 = = = 0 –[ = 0 – [ ] = – …..the minus sign tells us the area is below the x-axis Thus, A1 = For the interval [– 2 ; 2], we observe that the required area comprises of two parts – one part below the x-axis and the other part above the x-axis. Calculate A2 , as follows: A2 = = (The reader should verify the accuracy) So, the area over the interval [–2 ; 2] is obtained by adding A1 and A2, i.e: A = A1 + A2 = + = 16 square With your partner, attempt the following. The given sketch is a graphical representation of the function defined by: f(x) = x3 – 4x2 + 3x. Determine the area between the curve and the x-axis from x = 0 to x = 3. (A challenge) Find the area enclosed between the curve y = x2 – 2x –3 and the straight-line y = x + 1. Activity 9 1. Use the properties of definite integrals which you have studied thus far, to evaluate the following definite integrals. a) b) c) d) 2. Calculate the areas of the following: a) b) c) d)
O is the centre of the circle, with OM AB. OP is the radius of the circle. O is the centre of the circle, with OM AB. OP is the radius of the circle. O is the centre of the circle, with OM AB. OP is the radius of the circle. In the above figure AB is a secant to the circle. In the above AB does not intersect the circle. In the above figure AB is a tangent to the circle. Note that AB: Intersects the circle in two places (see Figure 1) if OM < r Does not intersect the circle (see Figure 2) if OM > r Intersects the circle in one and only one place (see Figure 3) if OM = r Worked examples 1. Show that 2x + y + 5 = 0 is a tangent to the circle defined by: x2 + y2 = 5 Proving that there are two equal roots the given straight-line equation in the form y = mx + c, i.e: y = – 2x – 5 Substitute y = –2x – 5 in x2 + y2 = 5 i.e: x2 + (–2x –5)2 = 5 x2 + 4x2 + 20x + 25 – 5 = 0 5x2 + 20x + 20 = 0 5(x2 + 4x + 4) = 0 (x + 2)2 = 0 x = – 2 is a repeated root, which implies that the straight line: 2x + y + 5 = 0 is a tangent to the given circle. Can you solve the given problem differently? Discuss with your partner. 3.2 The equation of the tangent through a point (x1 ; y1) on the circle x2 + y2 = r2
Worked examples 1. The P(3; – 2) is a point on the circle defined by: x2 + y2 = r2 a) Determine the equation of the circle. The equation of the circle is of the form: x2 + y2 = r2. Hence; (3)2 + (–2)2 = r2 r2 = 13 Required equation is: x2 + y2 = 13 b) Determine the equation of the tangent to the circle at the point P(3 ; – 2) The equation of the tangent through P(3 ;–2) is of the type: xx1 + yy1 = r2 x(3) + y(–2) = 13 Required equation is: 3x –2y =13 2. Determine the equation of the tangent to the circle x2 + y2 = 10 at the point K(–3 ; 1) Method 1: (Using the formula developed above) The equation of the tangent through K is of the type: xx1 + yy1 = r2 i.e: x(–3) + y(1) = 10 –3x + y = 10 y = 3x + 10 is the required equation. Method 2: First determine the gradient of the radius, OK; MOK = Mtangent = 3 (Do you recall, why?) Equation of tangent is: y -y1 = m(x – x1) y – (1) = 3(x –(–3)) y – 1 = 3x + 9 y = 3x + 10 is the required equation. Activity 2 1. Determine the equation of the tangent to the circle at the given point. Equation of circle Given point
Compare (1) and (2): a2 = 16 a = 4 OR a = – 4 b2 = 9 b = 3 OR b = – 3 Step 2: To draw the X-intercepts: (4 ; 0) and (–4 ; 0) Y-intercepts: (0 ; 3) and (0 ; –3) Major axis: The x -axis. Major axis = 8 units axis: The y-axis. Minor axis = 6 units Foci: c2 = a2 – b2 = 16 – 9 c = F1 ( ; 0) ; F2 ( ; 0) b) 2x2 + y2 = 10 Begin by dividing both sides of the equation by 10 to obtain the standard form: i.e: + = + = 1 Since 10 > 5, it implies that the y-axis is the major axis. a2 = 10 ; b2 = 5 Intercepts: X-axis: ( ; 0) ; ( ; 0) Y-axis: (0 ; ) ; (0 ; ) Minor axis: 2 Major axis: 2 Foci: c2= a2 – b2 = 10 – 5 = 5 c = F1 (0 ; ) ; F2 (0 ; ) Activity 4 1. In each of the following, determine: The intercepts with the axes; The coordinates of the foci;
ad = bc (The product of the mean terms = the product of the extreme Worked examples 1. Find the fourth proportional to 5 ; 8 ; Let x be the fourth proportional i.e.: = Then: 5x = 8() 5x = 100 x = 20 2. Find the third proportional to 5 and 8. Let x be the third proportional i.e: = 5x = 8(8) 5x = 64 x = = 12 Activity 2 Calculate the fourth proportional to 3, 5 and 8. Calculate the mean proportional between 14 and 56 Calculate the third proportional to 8 and 14. Calculate the mean proportional between 30 and 48. Solve for x if 28 : x = 7 : 3 Solve for x if 18 : x = x : 8 Theorem 1: A line drawn parallel to one side of a triangle divides the other two sides proportionally (line one side of )
= ……(line one side of ) = x = x = 78 mm 3. In the given diagram BD FE and HF CB. AF = 27 mm ; AH = 36 mm ; HC = 16 mm ; FB = x mm ; AE = a mm and ED = b mm. Calculate: The value of x The value of a : b In ABC : HF CB (Given) = …(line one side of ) x = = 12 mm In ABD : BD FE (Given) = …(line one side of ) = mm Activity 3 Calculate a) The first proportional to 7 , 15 and 21 b) The fourth proportional to 3 , 8 and 15 c) The third proportional to 4 and 9 d) The mean proportional to 4 and 2. In the diagram, A, B and C are the vertices of ABC, with BA DE. AE = x units; EC = 55 units; DC = 65 units and BD = 26 units. Determine the of x. 3. In the diagram KLMN is a parallelogram. ML is produced to R such that RL = 20 mm. LM = x mm. = . Determine the value of x.