Pure Further Mathematics Revision Notes June 20
2 FP JUNE 20 SDB
Further Pure Complex Numbers... 3 Definitions an arithmetical operations... 3 Complex conjugate... 3 Properties... 3 Complex number plane, or Argan iagram... 4 Complex numbers an vectors... 4 Multiplication by i... 5 Moulus of a complex number... 5 Argument of a complex number... 5 Equality of complex numbers... Square roots... Roots of equations... 2 Numerical solutions of equations... 8 Accuracy of solution... 8 Interval bisection... 8 Linear interpolation... 9 Newton-Raphson... 0 3 Coorinate systems... Parabolas... Parametric form... Focus an irectrix... Graient... Tangents an normals... 2 Rectangular hyperbolas... 2 Parametric form... 3 Tangents an normals... 3
4 Matrices... 4 Orer of a matrix...4 Ientity matrix...4 Determinant an inverse... 4 Singular an non-singular matrices... 5 Linear Transformations... 5 Basis vectors...5 Fining the geometric effect of a matrix transformation... Fining the matrix of a given transformation.... Rotation matrix...7 Determinant an area factor...7 5 Series... 8 Proof by inuction... 9 Summation... 9 Recurrence relations... 20 Divisibility problems... 20 Powers of matrices... 2 7 Appenix... 22 Complex roots of a real polynomial equation... 22 Formal efinition of a linear transformation... 22 Derivative of x n, for any integer... 23 8 Inex... 24 2 FP JUNE 20 SDB
Complex Numbers Definitions an arithmetical operations i =, so = 4i, = i, etc. These are calle imaginary numbers Complex numbers are written as z = a + bi, where a an b R. a is the real part an b is the imaginary part. +,, are efine in the sensible way; ivision is more complicate. (a + bi) + (c + i) = (a + c) + (b + )i (a + bi) (c + i) = (a c) + (b )i (a + bi) (c + i) = ac + bi 2 + ai + bci = (ac b) + (a +bc)i since i 2 = So (3 + 4i) (7 3i) = 4 + 7i an (4 + 3i) (2 5i) = 23 4i Division this is just rationalising the enominator. 3+4i 5+2i = 3+4i 5+2i 5 2i 5 2i multiply top an bottom by the complex conjugate = 23+4i 25+4 = 23 29 + 4 29 i Complex conjugate z = a + bi The complex conjugate of z is z* = z = a bi Properties If z = a + bi an w = c + i, then (i) {(a + bi) + (c + i)}* = {(a + c) + (b + )i}* (z + w)* = z* + w* = {(a + c) (b + )i} = (a bi) + (c i) FP JUNE 20 SDB 3
(ii) {(a + bi) (c + i)}* = {(ac b) + (a + bc)i}* = {(ac b) (a + bc)i} = (a bi) (c i) = (a + bi)*(c + i)* (zw)* = z* w* Complex number plane, or Argan iagram We can represent complex numbers as points on the complex number plane: 3 + 2i as the point A (3, 2), an 4 + 3i as the point ( 4, 3). 4 Imaginary B: (-4, 3) 2 A: (3, 2) Real 5 5 0 2 Complex numbers an vectors Complex numbers uner aition (or subtraction) behave just like vectors uner aition (or subtraction). We can show complex numbers on the Argan iagram as either points or vectors. (a + bi) + (c + i) = (a + c) + (b + ) i a b + c + c = a b + (a + bi) (c + i) = (a c) + (b ) i a b c c = a b or Im Im z 2 z + z 2 z 2 z z 2 z Re z Re 4 FP JUNE 20 SDB
Multiplication by i i(3 + 4i) = 4 + 3i on an Argan iagram this woul have the effect of a positive quarter turn about the origin. Im In general; i(a + bi) = b + ai x ( b, a) iz x (a, b) z Re Moulus of a complex number This is just like polar co-orinates. The moulus of z is z an is the length of the complex number z = a 2 + b 2. Im θ z x z = a + bi Re z z* = (a + bi)(a bi) = a 2 + b 2 z z* = z 2. Argument of a complex number The argument of z is arg z = the angle mae by the complex number with the positive x-axis. By convention, π < arg z π. N.B. Always raw a iagram when fining arg z. Example: Fin the moulus an argument of z = + 5i. Solution: First sketch a iagram (it is easy to get the argument wrong if you on t). z = 2 + 5 2 = an tan α = 5 α = 0 9473827 arg z = θ = π α = 2.45 to 3 S.F. x (, 5) z 5 α θ Im Re FP JUNE 20 SDB 5
Equality of complex numbers a + bi = c + i a c = ( b)i (a c) 2 = ( b) 2 i 2 = ( b) 2 squaring both sies But (a c) 2 0 an ( b) 2 0 (a c) 2 = ( b) 2 = 0 a = c an b = Thus a + bi = c + i real parts are equal (a = c), an imaginary parts are equal (b = ). Square roots Example: Fin the square roots of 5 + 2i, in the form a + bi, a, b R. Solution: Let 5 + 2i = a + bi 5 + 2i = (a + bi) 2 = a 2 b 2 + 2abi Equating real parts a 2 b 2 = 5, I equating imaginary parts 2ab = 2 a = b Substitute in I b 2 b 2 = 5 3 b 4 = 5b 2 b 4 + 5b 2 3 = 0 (b 2 4)(b 2 + 9) = 0 b 2 = 4 b = ± 2, an a = ± 3 5 + 2i = 3 + 2i or 3 2i. Roots of equations (a) Any polynomial equation with complex coefficients has a complex solution. The is The Funamental Theorem of Algebra, an is too ifficult to prove at this stage. Corollary: Any complex polynomial can be factorise into linear factors over the complex numbers. FP JUNE 20 SDB
(b) If z = a + bi is a root of α n z n + α n z n + α n 2 z n 2 + + α 2 z 2 + α z + α 0 = 0, an if all the α i are real, then the conjugate, z* = a bi is also a root. The proof of this result is in the appenix. (c) For any polynomial with zeros a + bi, a bi, (z (a + bi))(z (a bi)) = z 2 2az + a 2 b 2 will be a quaratic factor in which the coefficients are all real. () Using (a), (b), (c) we can see that any polynomial with real coefficients can be factorise into a mixture of linear an quaratic factors, all of which have real coefficients. Example: Show that 3 2i is a root of the equation z 3 8z 2 + 25z 2 = 0. Fin the other two roots. Solution: Put z = 3 2i in z 3 8z 2 + 25z 2 = (3 2i) 3 8(3 2i) 2 + 25(3 2i) 2 = 27 54i + 3i 2 8i 3 8(9 2i + 4i 2 ) + 75 50i 2 = 27 54i 3 + 8i 72 + 9i + 32 + 75 50i 2 = 27 3 72 + 32 + 75 2 + ( 54 + 8 + 9 50)i = 0 + 0i 3 2i is a root the conjugate, 3 + 2i, is also a root since all coefficients are real (z (3 + 2i))(z (3 2i)) = z 2 z + 3 is a factor. Factorising, by inspection, z 3 8z 2 + 25z 2 = (z 2 z + 3)(z 2) = 0 roots are z = 3 ± 2i, or 2 FP JUNE 20 SDB 7
2 Numerical solutions of equations Accuracy of solution When aske to show that a solution is accurate to n D.P., you must look at the value of f (x) half below an half above, an conclue that there is a change of sign in the interval, an the function is continuous, therefore there is a solution in the interval correct to n D.P. Example: Show that α = 2 094 is a root of the equation f (x) = x 3 2x 5 = 0, accurate to 4 D.P. Solution: f (2.09455) = 0 00005, an f (2.0945) = +0 00997 There is a change of sign an f is continuous there is a root in [2 09455, 2 0945] root is α = 2 094 to 4 D.P. Interval bisection (i) Fin an interval [a, b] which contains the root of an equation f (x) = 0. (ii) x = a+b 2 is the mi-point of the interval [a, b] Fin f a+b 2 a+b to ecie whether the root lies in a, or a+b, b. 2 2 (iii) Continue fining the mi-point of each subsequent interval to narrow the interval which contains the root. Example: (i) (ii) Show that there is a root of the equation f (x) = x 3 2x 7 = 0 in the interval [2, 3]. Fin an interval of with 0 25 which contains the root. Solution: (i) f (2) = 8 4 7 = 3, an f (3) = 27 7 = 4 There is a change of sign an f is continuous there is a root in [2, 3]. (ii) Mi-point of [2, 3] is x = 2 5, an f (2 5) = 5 25 5 7 = 3 25 change of sign between x = 2 an x = 2 5 root in [2, 2 5] 8 FP JUNE 20 SDB
Mi-point of [2, 2 5] is x = 2 25, an f (2 25) = 39025 4 5 7 = 0 09375 change of sign between x = 2 25 an x = 2 5 root in [2 25, 2 5], which is an interval of with 0 25 Linear interpolation To solve an equation f (x) using linear interpolation. First, fin an interval which contains a root, secon, assume that the curve is a straight line an use similar triangles to fin where the line crosses the x-axis, thir, repeat the process as often as necessary. Example: (i) Show that there is a root, α, of the equation f (x) = x 3 2x 9 = 0 in the interval [2, 3]. (ii) Use linear interpolation once to fin an approximate value of α. Give your answer to 3 D.P. Solution: (i) f (2) = 8 4 9 = 5, an f (3) = 27 9 = 2 There is a change of sign an f is continuous there is a root in [2, 3]. (ii) From (i), curve passes through (2, 5) an (3, 2), an we assume that the curve is a straight line between these two points. Let the line cross the x-axis at (α, 0) Using similar triangles 3 α = 2 α 2 5 (3, 2) 5 5α = 2α 24 2 α = 39 7 = 2 5 7 α = 2 294 to 3 D.P. 5 2 α 2 α 3 α 3 (2, 5) Repeating the process will improve accuracy. FP JUNE 20 SDB 9
Newton-Raphson Suppose that the equation f (x) = 0 has a root at x = α, f (α) = 0 To fin an approximation for this root, we first fin a value x = a near to x = α (ecimal search). y y = f (x) P In general, the point where the tangent at P, x = a, meets the x-axis, x = b, will give a better approximation. α N b M a x At P, x = a, the graient of the tangent is f (a), an the graient of the tangent is also PM NM. PM = y = f (a) an NM = a b f (a) = PM = f(a) NM a b b = a f(a) f (a). Further approximations can be foun by repeating the process, which woul follow the otte line converging to the point (α, 0). This formula can be written as the iteration x n + = x n f(x n) f (x n ) Example: (i) (ii) Show that there is a root, α, of the equation f (x) = x 3 2x 5 = 0 in the interval [2, 3]. Starting with x 0 = 2, use the Newton-Raphson formula to fin x, x 2 an x 3, giving your answers to 3 D.P. where appropriate. Solution: (i) f (2) = 8 4 5 =, an f (3) = 27 5 = There is a change of sign an f is continuous there is a root in [2, 3]. (ii) f (x) = x 3 2x 5 f (x) = 3x 2 2 x = x 0 f(x 0) f (x 0 ) = 2 8 4 5 2 2 = 2 x 2 = 2 094582 = 2 095 x 3 = 2 09455482 = 2 095 0 FP JUNE 20 SDB
3 Coorinate systems Parabolas y y 2 = 4ax is the equation of a parabola which passes through the origin an has the x-axis as an axis of symmetry. M X P (at 2, 2at) Parametric form x = at 2, y = 2at satisfy the equation for all values of t. t is a parameter, an these equations are the parametric equations of the parabola y 2 = 4ax. irectrix x= a X focus S (a, 0) x Focus an irectrix The point S (a, 0) is the focus, an the line x = a is the irectrix. Any point P of the curve is equiistant from the focus an the irectrix, PM = PS. Proof: PM = at 2 ( a) = at 2 + a PS 2 = (at 2 a) 2 + (2at) 2 = a 2 t 4 2a 2 t 2 + a 2 + 4a 2 t 2 = a 2 t 4 + 2a 2 t 2 + a 2 = (at 2 + a) 2 = PM 2 PM = PS. Graient For the parabola y 2 = 4ax, with general point P, (at 2, 2at), we can fin the graient in two ways:. y 2 = 4ax 2y y = 4a y x x 2. At P, x = at 2, y = 2at y = 2a, x = 2at t t y x y = t x t = 2a = 2at t = 2a y, which we can write as y x = 2a 2at = t FP JUNE 20 SDB
Tangents an normals Example: Fin the equations of the tangents to y 2 = 8x at the points where x = 8, an show that the tangents meet on the x-axis. Solution: x = 8 y 2 = 8 8 y = ±2 2y y x = 8 y x = ± 3 since y = ±2 tangents are y 2 = (x 8) x 3y + 8 = 0 at (8, 2) 3 an y + 2 = (x 8) x + 3y + 8 = 0. at (8, 2) 3 To fin the intersection, a the equations to give 2x + 3 = 0 x = 8 y = 0 tangents meet at ( 8, 0) on the x-axis. Example: Fin the equation of the normal to the parabola given by x = 3t 2, y = t. Solution: x = 3t 2, y = t x t = t, y t =, y x y = t x t = = t t graient of the normal is t = t equation of the normal is y t = t(x 3t 2 ). Notice that this general equation gives the equation of the normal for any particular value of t: when t = 3 the normal is y + 8 = 3(x 27) y = 3x 99. Rectangular hyperbolas y A rectangular hyperbola is a hyperbola in which the asymptotes meet at 90 o. xy = c 2 xy = c 2 is the equation of a rectangular hyperbola in which the x-axis an y-axis are perpenicular asymptotes. x 2 FP JUNE 20 SDB
Parametric form x = ct, y = c t are parametric equations of the hyperbola xy = c 2. Tangents an normals Example: Fin the equation of the tangent to the hyperbola xy = 3 at the point where x = 3. Solution: x = 3 3y = 3 y = 2 y = 3 x y = 3 x x2 = 4 when x = 3 tangent is y 2 = 4(x 3) 4x + y 24 = 0. Example: Fin the equation of the normal to the hyperbola given by x = 3t, y = 3 t. Solution: x = 3t, y = 3 t x t = 3, y = 3 t t 2 y x y = t x t = 3 t 2 3 = t 2 graient of the normal is t 2 = t 2 equation of the normal is y 3 t = t 2 (x 3t) t 3 x ty = 3t 4 3. FP JUNE 20 SDB 3
4 Matrices You must be able to a, subtract an multiply matrices. Orer of a matrix An r c matrix has r rows an c columns; the first number is the number of Rows the secon number is the number of Columns. Ientity matrix The ientity matrix is I = 0 0. Note that MI = IM = M for any matrix M. Determinant an inverse Let M = a b then the eterminant of M is c Det M = M = a bc. To fin the inverse of M = a c Note that M M = M M b = I (i) (ii) Fin the eterminant, a bc. If a bc = 0, there is no inverse. Interchange a an (the leaing iagonal) Change sign of b an c, (the other iagonal) Divie all elements by the eterminant, a bc. M = Check: a bc c b a. M M = a bc b b a c a c = a bc a bc 0 0 = = I 0 cb + a 0 Similarly we coul show that M M = I. 4 FP JUNE 20 SDB
Example: M = 4 2 2 an MN =. Fin N. 5 3 2 Solution: Notice that M (MN) = (M M)N = IN = N multiplying on the left by M But MNM IN we cannot multiply on the right by M First fin M Det M = 4 3 2 5 = 2 M = 2 3 2 5 4 Using M (MN) = IN = N N = 2 3 2 2 5 4 2 = 2 7 4 5 2 = 3 3 5 3. Singular an non-singular matrices If et A = 0, then A is a singular matrix, an A oes not exist. If et A 0, then A is a non-singular matrix, an A exists Linear Transformations A matrix can represent a transformation, but the point must be written as a column vector before multiplying by the matrix. Example: The image of (2, 3) uner T = 4 5 5 is given by 4 2 2 2 = 23 3 8 the image of (2, 3) is (23, 8). Note that the image of (0, 0) is always (0, 0) the origin never moves uner a matrix (linear) transformation Basis vectors The vectors i = 0 an j = 0 are calle basis vectors, an are particularly important in escribing the geometrical effect of a matrix, an in fining the matrix for a particular geometric transformation. a b c 0 = a b an a c c 0 = b i = 0 a c, the first column, an j = 0 b, the secon column This is a more important result than it seems! FP JUNE 20 SDB 5
Fining the geometric effect of a matrix transformation We can easily write own the images of i an j, sketch them an fin the geometrical transformation. Example: Fin the transformation represente by the matrix T = 2 0 0 3 Solution: Fin images of i, j an, an show on a sketch. Make sure that you letter the points 4 y C' B' 2 0 0 0 2 = 2 0 3 0 0 3 3 2 C B From sketch we can see that the transformation is a two-way stretch, of factor 2 parallel to the x-axis an of factor 3 parallel to the y-axis. A A' x 3 Fining the matrix of a given transformation. Example: Fin the matrix for a shear with factor 2 an invariant line the x-axis. Solution: Each point is move in the x-irection by a istance of (2 its y-coorinate). i = 0 (oes not move as it 0 is on the invariant line). This will be the first column of the matrix 0 j = 0 2. This will be the secon column of the matrix 2 Matrix of the shear is 2 0. C y B C' B' x A, A' 2 3 y Example: Fin the matrix for a reflection in y = x. B (0, ) Solution: First fin the images of i an j. These will be the two columns of the matrix. B' (, 0) A (, 0) x A A i = 0 0. A' (0, ) This will be the first column of the matrix 0 FP JUNE 20 SDB
B B j = 0 0. This will be the secon column of the matrix Matrix of the reflection is 0 0. 0 Rotation matrix From the iagram we can see that i = θ cos 0 sin θ, j = 0 sin θ cos θ These will be the first an secon columns of the matrix B ( sinθ, cosθ) y B (0, ) sinθ θ cosθ θ cosθ A (cosθ, sinθ) sinθ x A (, 0) cos θ sin θ matrix is R θ = sin θ cos θ. Determinant an area factor For the matrix A = a c b y b a a c b 0 = a c c an a c b 0 = b (b, ) (a, c) the unit square is mappe on to the parallelogram as shown in the iagram. The area of the unit square =. a c b x The area of the parallelogram = (a + b)(c + ) 2 (bc + ac + b) 2 2 = ac + a + bc + b 2bc ac b = a bc = et A. All squares of the gri are mappe onto congruent parallelograms area factor of the transformation is et A = a bc. FP JUNE 20 SDB 7
5 Series You nee to know the following sums n r r= n r 2 r= n = + 2 + 3 + + n = n(n + ) 2 = 2 + 2 2 + 3 2 + + n 2 = n(n + )(2n + ) r 3 = 3 + 2 3 + 3 3 + + n 3 = 4 n2 (n + ) 2 r= Example: Fin r(r 2 3). Solution: = n r= = n(n + 2 ) 2 = r a fluke, but it helps to remember it n r= 4 n2 (n + ) 2 3 n(n + ) 2 n r= r(r 2 3) = r 3 3 r n r= n r= 2 = = 4 4 n(n + ){n(n + ) } n(n + )(n + 3)(n 2) Example: Fin S n = 2 2 + 4 2 + 2 + + (2n) 2. Solution: S n = 2 2 + 4 2 + 2 + + (2n) 2 = 2 2 ( 2 + 2 2 + 3 2 + + n 2 ) = 4 n(n + )(2n + ) = 2 n(n + )(2n + ). 3 Example: Fin n+2 r 2 r=5 n+2 n+2 4 Solution: r 2 = r 2 r 2 notice that the top limit is 4 not 5 r=5 r= r= = = (n + 2)(n + 2 + )(2(n + 2) + ) 4 5 9 (n + 2)(n + 3)(2n + 5) 30. 8 FP JUNE 20 SDB
Proof by inuction Summation. Show that the result/formula is true for n = (an sometimes n = 2, 3..). Conclue therefore the result/formula. is true for n =. 2. Make inuction assumption Assume that the result/formula. is true for n = k. Show that the result/formula must then be true for n = k + Conclue therefore the result/formula. is true for n = k +. 3. Final conclusion therefore the result/formula is true for all positive integers, n, by mathematical inuction. Example: Use mathematical inuction to prove that S n = 2 + 2 2 + 3 2 + + n 2 = n(n + )(2n + ) Solution: When n =, S = 2 = an S = ( + )(2 + ) = 2 3 = S n = n(n + )(2n + ) is true for n =. Assume that the formula is true for n = k S k = 2 + 2 2 + 3 2 + + k 2 = k(k + )(2k + ) S k + = 2 + 2 2 + 3 2 + + k 2 + (k + ) 2 = k(k + )(2k + ) + (k + )2 = (k + ){k(2k + ) + (k + )} = (k + ){2k2 + 7k + } = (k + )(k + 2)(2k + 3) = (k + ){(k + ) + }{2(k + ) + } The formula is true for n = k + S n = n(n + )(2n + ) is true for all positive integers, n, by mathematical inuction. FP JUNE 20 SDB 9
Recurrence relations Example: A sequence, 4, 9, 9, 39, is efine by the recurrence relation u = 4, u n + = 2u n +. Prove that u n = 5 2 n. Solution: When n =, u = 4, an u = 5 2 = 5 = 4, formula true for n =. Assume that the formula is true for n = k, u k = 5 2 k. From the recurrence relation, u k + = 2u k + = 2(5 2 k ) + u k + = 5 2 k 2 + = 5 2 (k + ) the formula is true for n = k + the formula is true for all positive integers, n, by mathematical inuction. Divisibility problems Consiering f (k + ) f (k), will lea to a proof which sometimes has hien ifficulties, an a more reliable way is to consier f (k + ) m f (k), where m is chosen to eliminate the exponential term. Example: Prove that f (n) = 5 n 4n is ivisible by for all positive integers, n. Solution: When n =, f () = 5 4 = 0, which is ivisible by, an so f (n) is ivisible by when n =. Assume that the result is true for n = k, f (k) = 5 k 4k is ivisible by. Consiering f (k + ) 5 f (k) we will eliminate the 5 k term. f (k + ) 5 f (k) = (5 k + 4(k + ) ) 5 (5 k 4k ) f (k + ) = 5 f (k) + k = 5 k + 4k 4 5 k + + 20k + 5 = k Since f (k) is ivisible by (inuction assumption), an k is ivisible by, then f (k + ) must be ivisible by, f (n) = 5 n 4n is ivisible by for n = k + f (n) = 5 n 4n is ivisible by for all positive integers, n, by mathematical inuction. 20 FP JUNE 20 SDB
Example: Prove that f (n) = 2 2n + 3 + 3 2n is ivisible by 5 for all positive integers n. Solution: When n =, f () = 2 2 + 3 + 3 2 = 32 + 3 = 35 = 5 7, an so the result is true for n =. Assume that the result is true for n = k f (k) = 2 2k + 3 + 3 2k is ivisible by 5 We coul consier either (it oes not matter which) f (k + ) 2 2 f (k), which woul eliminate the 2 2k + 3 term I or f (k + ) 3 2 f (k), which woul eliminate the 3 2k term II I f (k + ) 2 2 f (k) = 2 2(k + ) + 3 + 3 2(k + ) 2 2 (2 2k + 3 + 3 2k ) = 2 2k + 5 + 3 2k + 2 2k + 5 2 2 3 2k f (k + ) 4 f (k) = 9 3 2k 4 3 2k = 5 3 2k f (k + ) = 4 f (k) 5 3 2k Since f (k) is ivisible by 5 (inuction assumption), an 5 3 2k is ivisible by 5, then f (k + ) must be ivisible by 5. f (n) = 2 2n + 3 + 3 2n is ivisible by 5 for all positive integers, n, by mathematical inuction. Powers of matrices Example: If M = 2 0, prove that Mn = 2n 2 n for all positive integers n. 0 Solution: When n =, M = 2 2 0 the formula is true for n =. = 2 = M 0 Assume the formula is true for n = k M k = 2k 2 k. 0 M k+ = MM k = 2 0 2 k 2k = 2 2k 2 2 2 k 0 0 M k+ = 2k+ 2 k+ The formula is true for n = k + 0 M n = 2n 2 n is true for all positive integers, n, by mathematical 0 inuction. FP JUNE 20 SDB 2
7 Appenix Complex roots of a real polynomial equation Preliminary results: I (z + z 2 + z 3 + z 4 + + z n )* = z * + z 2 * + z 3 * + z 4 * + + z n *, by repeate application of (z + w)* = z* + w* II (z n )* = (z*) n (zw)* = z*w* (z n )* = (z n- z)* = (z n- )*(z)* = (z n-2 z)*(z)* = (z n-2 )*(z)*(z)* = (z*) n Theorem: If z = a + bi is a root of α n z n + α n z n + α n 2 z n 2 + + α 2 z 2 + α z + α 0 = 0, an if all the α i are real, then the conjugate, z* = a bi is also a root. Proof: If z = a + bi is a root of the equation α n z n + α n z n + + α z + α 0 = 0 then α n z n + α n z n + + α 2 z 2 + α z + α 0 = 0 (α n z n + α n z n + + α 2 z 2 + α z + α 0)* = 0 since 0* = 0 (α n z n )* + (α n z n )* + + (α 2 z 2 )* + (α z)* + (α 0)* = 0 using I α n *( z n )* + α n *(z n )* + + α 2*( z 2 )* + α *( z)* + α 0* = 0 since (zw)* = z*w* α n ( z n )* + α n (z n )* + + α 2( z 2 )*+ α ( z)* + α 0 = 0 α i real α i * = α i α n ( z*) n + α n (z*) n + + α 2(z*) 2 + α (z*) + α 0 = 0 using II z* = a bi is also a root of the equation. Formal efinition of a linear transformation A linear transformation T has the following properties: (i) T kx ky = kt x y (ii) T x y + x 2 y 2 = T x y + T x 2 y 2 It can be shown that any matrix transformation is a linear transformation, an that any linear transformation can be represente by a matrix. 22 FP JUNE 20 SDB
Derivative of x n, for any integer We can use proof by inuction to show that x (xn ) = nx n, for any integer n. ) We know that the erivative of x 0 is 0 which equals 0x, since x 0 =, an the erivative of is 0 x (xn ) = nx n is true for n = 0. 2) We know that the erivative of x is which equals x x (xn ) = nx n is true for n = Assume that the result is true for n = k x (xk ) = kx k x (xk+ ) = x (x xk ) = x x (xk ) + x k x (xk+ ) = x kx k + x k = kx k + x k = (k + )x k x (xn ) = nx n is true for n = k + prouct rule x (xn ) = nx n is true for all positive integers, n, by mathematical inuction. 3) We know that the erivative of x is x 2 which equals x x (xn ) = nx n is true for n = Assume that the result is true for n = k x (xk ) = kx k x (xk ) = x xk x x = x xk x k x 2 quotient rule x (xk+ ) = x kxk x k = (k )xk = (k )x k 2 = (k )x (k ) x 2 x 2 x (xn ) = nx n is true for n = k We are going backwars (from n = k to n = k ), an, since we starte from n =, x (xn ) = nx n is true for all negative integers, n, by mathematical inuction. Putting ), 2) an 3), we have prove that x (xn ) = nx n, for any integer n. FP JUNE 20 SDB 23
8 Inex Complex numbers Argan iagram, 4 argument, 5 arithmetical operations, 3 complex conjugate, 3 complex number plane, 4 efinitions, 3 equality, moulus, 5 multiplication by i, 5 polynomial equations, similarity with vectors, 4 square roots, Complex roots of a real polynomial equation, 22 Derivative of x n, for any integer, 23 Determinant area factor, 7 Linear transformation formal efinition, 22 Matrices eterminant, 4 ientity matrix, 4 inverse matrix, 4 non-singular, 5 orer, 4 singular, 5 Matrix transformations area factor, 7 basis vectors, 5 fining matrix for, geometric effect, rotation matrix, 7 Numerical solutions of equations accuracy of solution, 8 interval bisection, 8 linear interpolation, 9 Newton Raphson metho, 0 Parabolas irectrix, focus, graient - parametric form, normal - parametric form, 2 parametric form, tangent - parametric form, 2 Proof by inuction ivisibility problems, 20 general principles, 9 powers of matrices, 2 recurrence relations, 20 summation, 9 Rectangular hyperbolas graient - parametric form, 3 parametric form, 3 tangent - parametric form, 3 Series stanar summation formulae, 8 Transformations area factor, 7 basis vectors, 5 fining matrix for, geometric effect, linear transformations, 5 matrices, rotation matrix, 7 24 FP JUNE 20 SDB