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There are four main types of bonds, split into two groups. The strong primary bonds (Ionic, macromolecular covalent & metallic), and the weaker secondary bonds (molecular covalent and van der Waals): Ionic This is the electrostatic force of attraction between oppositely charged ions It is formed by electron transfer: metal lose electrons while non-metals gain them It is non-directional High melting/boiling points Easily soluble Poor conductivity when solid, but when molten, charged ions are free to move around Generally crystalline solids at room temperature & pressure Covalent This is formed from a shared pair of electrons Typically occurs if an atom’s outer shell is about half empty (gaining/losing ~four electrons requires too much energy) It is directional: the formation and orientation affects the overall molecular shape There are two types: Molecular Bonds (simple covalent): These have a low melting and boiling point, due to the weak intermolecular forces They have poor solubility in water Conductivity is also poor, as there are no ions, and all electrons are fixed Generally gaseous or liquidous at room temperature and pressure Macromolecular Bonds (giant covalent): These have very high melting and boiling points, as the bonds themselves are very strong and there are very many of them, so a vast amount of energy is required to break these Insoluble in water Mostly do not conduct, but graphite does Generally solid at room temperature and pressure Metallic This is the electrostatic force between positive metal ions (cations) and a sea of delocalised electrons (which are negatively charged) The cations are in regular rows, with the electrons free to move (hence conduct) around them Melting/boiling points are high, as there are strong electrostatic forces between cations and electrons Insoluble in water Very good conduction Generally shiny, malleable solid at room temperature and pressure Van der Waals Dipoles form when an atom has a net charge, due to an imbalance in protons and electrons Dipoles can be temporary or permanent Once a dipole forms, it encourages the neighbouring atom to become an oppositely charged dipole This +/- force of attraction is the van der Waals bond A hydrogen bond is a particular type of van der Waals bond. Since the hydrogen atom is the smallest (with just 1 electron in one shell), when it combines with much larger atoms such as nitrogen, oxygen and fluorine, there is a major electromagnetic imbalance. This causes permanent dipoles to form. These give rise to far stronger bonds than temporary dipoles. A common example is water. It is the hydrogen bonds that cause a solvent to be polar. As you can see from the tetrahedron above, most materials have predominantly one type of bond throughout their structure, but some (like ceramics and polymers) have a mixture. Polymers typically consist of long chains of carbon atoms that are covalently bonded with other atoms (hydrogen, boron etc). These bonds are extremely strong and rigid, with deformation only occurring in extreme conditions. It is the weaker van der Waals bonds that also exist between molecules in polymers that make them flexible and easy to break. Interatomic Forces & Bond Energies When two oppositely charged atoms (or ions) are far apart, their attraction is negligible. As they get closer, however, the attractive force between them increases. Yet at a certain point, the atoms get so close that the electron clouds start to overlap. At this point, the negatively charged electrons repel each other, causing a repulsive force, between the oppositely charged atoms. The repulsive force only acts at a very short range The net force between the atoms, F, will therefore be given by: The potential energy in the bond is calculated as the integral of the resultant force with respect to separation: There is a certain separation, r₀, where the resultant force is equal to zero and the bond energy is at a minimum Potential energy of a bond is often also defined using power laws. A common example is: The first term is from the attractive force, the second from the repulsive force A, B, m & n are system constants Typical values for m and n are: m = 1 for ionic bonds m = 2, n = 9 for covalent bonds 1 ≤ m ≤ 4 for metallic bonds Differentiating this gives an expression for the force: This equation can be used to determine the equilibrium spacing, by setting F to equal zero. Potential Energy – Interatomic Separation Graphs At the equilibrium: the atoms are a distance of r₀ apart the force is zero the energy is at a minimum: E₀ As r < r₀, there is an immense amount of energy in the strongly repulsive bond. When r > r₀, there is never a positive energy (it tends to zero) Interatomic Force – Separation Graphs These are the exact opposite of the energy graph. Finding Metallic Young’s Modulus We can approximate the Young’s Modulus for metals very accurately from the force in the metallic bonds: Young’s Modulus, E, is directly proportional to the gradient of the straight-line segment immediately after r₀: This does not apply to non-metals, as their structure is fundamentally different. This must not be used to estimate UTS, as it will give a far higher value: Predictions of UTS using this method will be in the range of E/10 or E/15 The actual UTS of metals is more like E/100 or E/1000 Thermal Expansion We can use the energy-separation graph to explain thermal expansion: r₀ is given as the midpoint on the horizontal line connecting the two points on the curve that have the same energy As the temperature increases, so does the energy in the system Therefore, the horizontal line connecting the two points gets longer Since the gradient on the right of the minimum point is shallower, the horizontal line grows more on the right than it does on the left Therefore, the midpoint, r₀, moves to the right as the energy increases As the temperature increases, the equilibrium spacing between atoms increases. This is thermal expansion.
- Crystal Structures
In this notes sheet: The Hard Sphere Model Packing in Two Dimensions Packing in Three Dimensions Simple Cubic Body-Centred Cubic (BCC) Face-Centred Cubic (FCC) Hexagonal Close-Packed (HCP) Atomic Density Understanding the molecular structure of a material is crucial in understanding how that material behaves. Many of its physical properties are directly caused and explained by its crystalline structure (known as a lattice). We define this as: A crystalline lattice is a regular repeating pattern of atoms or molecules. An amorphous material is one that does not display a crystalline structure. Instead, the structure is random – examples include some polymers and ceramics. Some other polymers are known as semi-crystalline The Hard Sphere Model The Hard Sphere Model allows us to model atoms or molecules as, you guessed it, hard spheres. There are a number of conditions for this to apply: The outer electron shell must be full The remaining electrons must be distributed rotationally symmetrically Electron clouds of neighbouring atoms do not overlap much or at all For the outer shell of electrons to be full, metals must lose electrons, while non-metals must gain electrons. This means they are actually ions. Packing in Two Dimensions It is impossible to arrange circles in such a way that there are no unfilled spaces between them: Squares, rectangles, triangles, and hexagons, however, can be arranged into a complete lattice: The primitive unit cell is the smallest cell that can be regularly repeated throughout the structure – the hexagon can be split into six equilateral triangles. The unit cell is the easiest cell to repeat: the hexagonal is more practical, as it does not require rotating, whereas the triangle does. The two arrangements of circles above can, however, be modelled as square and rectangular patterns respectively, where the centre of an atom is at each vertex: Shown in red on the diagrams are the close-packed directions of each structure. These are the lines along which ions continuously touch (the line never passes through empty space). As you can see: The square lattice has two close-packed dimensions The hexagonal lattice has three close-packed directions Since the hexagonal lattice is more densely packed, it is known as the close-packed arrangement. Packing in Three Dimensions If we look at the same arrangements in 3D, is clear that, as well as close-packed directions, there are close-packed planes. These are planes in which all the ions are arranged in a hexagonal lattice. There are four typical unit cells in three dimensions: Simple Cubic This is simply a 2D square lattice in 3D: 1 atom per unit cell (⅛ on each vertex) 6 nearest neighbours (each ion touches six others - also known as the coordination number) Lattice parameter: a = 2r No close-packed planes No slip systems Atomic packing factor: 52.4% Body-Centred Cubic (BCC) In the simple cubic cell above, there is a gap in the centre. When the outer atom eights are moved apart enough to fit another atom in the middle, you have the BCC arrangement: 2 atoms per unit cell (⅛ at each vertex, one in the centre) 8 nearest neighbours Lattice parameter a = ⁴/₃ √3 r No close-packed planes 2 close-packed directions No slip systems Atomic packing factor: 68.2% Face-Centred Cubic (FCC) This is the most densely packed cubic arrangement. In this instance, there are three close-packed planes stacked on top of one another, in an ABC arrangement: Again, the unit cell is a cube, but the atoms on the vertices are further apart. This allows 4 half atoms to fit in the centre: 4 atoms per unit cell (⅛ on each vertex, ½ on each side) 12 nearest neighbours Lattice parameter a = √2 r 4 close-packed planes 3 close-packed directions 12 slip systems Atomic packing factor: 74% Hexagonal Close-Packed (HCP) This is where there are two close-packed planes stacked one on top of the other, this time in an ABAB arrangement: This leads to a hexagonal unit cell: 6 atoms per unit cell (⅙ on each vertex, ½ on top and bottom, 3 in centre) 12 nearest neighbours Lattice parameter: a = 2r 1 non-parallel close-packed plane 3 close-packed directions 3 slip systems Atomic packing factor: 74% The height of the hexagonal close-packed unit cell is given as: Note that HCP and FCC both have the same atomic packing factor, 74% Atomic Density If we know the atomic radius, mass and the type of unit cell, we can estimate the density of a material: N is the number of atoms per unit cell M is the atomic mass V is the volume of the unit cell N(A) is Avogadro’s Number, 9.02x10²³
- Defects & Deformation of Crystal Structures
In this notes sheet: Slip & Slip Systems Calculating Deformations Defects in Crystal Structures Strengthening There are two types of deformation of crystalline materials: elastic and plastic. The former is fully reversible, whereas in the latter, atomic planes slip over one another to form a new, different formation. This kind of deformation is irreversible. Note that the elastic deformation graph could be non-linear (e.g. rubber) Slip & Slip Systems The slipping of atomic planes (slip planes) happens in whatever direction requires the least energy. It is brought about by the application of an external force, resulting in shear stresses within the crystal structure. In a perfect sample, the slip would occur over the whole plane at once, however imperfections in the crystal structure prevent this: slip is a gradual process. Close-packed planes are the most susceptible to slip. This is because more energy is required for a non-close-packed plane to slip, as there is a greater distance for each displaced atom to move. The number of slip systems a crystal lattice has reflects how likely slip in the structure is. The more slip systems present, the more susceptible to slip. This is determined as the product of close-packed directions and non-parallel close-packed planes: Lattices with few slip systems are said to show brittle deformation, while those that have many undergo ductile deformation. HCP Slip Systems There are three close-packed directions per close-packed plane, however all close-packed planes are parallel. Therefore, the number of slip systems is 3, so the deformation is brittle: FCC Slip Systems The close-packed planes in an FCC lattice are along the diagonal-corner, and there are four: Each plane has three close-packed directions, so the number of slip systems is 12, so the deformation is ductile. BCC Slip Systems There are no close-packed planes, but there are 6 nearly-close-packed planes, each with two close-packed directions: Therefore, the number of slip systems is 12, so the deformation is ductile. Calculating Deformation The applied stress must be resolved in terms of the slip plane, φ, and direction, λ: σ is the normal stress, given by F/A Under the application of a shear stress, slip will occur in the most favourably orientated slip system. This is the system with the largest value of resolved shear stress: The value of shear stress required for this slip to occur in the most favourable direction is the critical resolved shear stress: For single crystals, slip occurs when φ = λ = 45°. Defects in Crystal Structures Crystals are never perfect – they are littered with defects that allow slip to occur at significantly lower values of applied stress than you would calculate from the equations above. There are three types of defects: Point Defects – defects with atomic dimensions in all three directions, e.g. a missing atom (a vacancy) Line Defects – defects in two atomic dimensions but are normal in the third direction. Area Defects – defects in only one dimension, e.g. a grain boundary All defects distort the lattice, increasing the energy stored inside it. Line Defects Line defects, also known as dislocations, is when a row of atoms is added or removed. These allow lattices to slip in steps, instead of all at once. Two types of dislocations are edge and screw dislocations: An edge dislocation is when a row of atoms is removed: A screw dislocation is when a set of rows slips a certain amount to one side: In both cases, the Burgers vector, b, measures the slip. As you can see from this schematic, edge dislocations happen in steps, until the edge of the crystal has been reached and no more slip can occur. This is because far less energy is required to move only a small amount every time. Area Defects & Grain Boundaries Grain boundaries occur between uniform regions of crystal structures. These could be crystals of the same atom, or different. Either way, they provide a blockage to slip, as slip cannot occur across such a boundary. A material with multiple crystal grains is known as a polycrystal. The smaller the grains, the higher the stress required to cause slip. This is expressed by the Hall-Petch equation: σ(y) is the yield strength σ(i) is the intrinsic lattice strength k is a material constant d is the average diameter of grains Strengthening It may seem intuitive that more dislocations = less strong material, but this is not the case. Yes, a perfect material with no defects at all would be very strong, but this is impossible to achieve. A small number of dislocations reduces the strength, but as you increase the number of dislocations, the strength actually increases. This is because the defects then interact and obstruct each other, reducing the likeliness of slip. Therefore, the four main ways of strengthening a material are: Reduce the grain size Generate more dislocations Adding smaller atoms into the lattice Adding larger atoms into the lattice The easiest methods of generating more dislocations to a structure are work hardening or melting and solidifying. It is important to note, however, that you cannot add infinite dislocations to a structure. This is because two opposite dislocations will meet and annihilate: This leads to a resolved shear stress-strain curve like this:
- A-Level Maths Cheat Sheet
This is absolutely not a cheat sheet... We do not endorse cheating at all! It's just a better name than "List of vital but forgettable equations for A Level Maths" The Formula Book for Edexcel A-Level Maths and Further Maths can be seen here and downloaded here: Trigonometry Cosine Rule a² = b² + c² - 2bc cosA Area of a Triangle Area = ½ ab sinC Radians 1° = π/180 1 rad = 180/π Small Angle Approximations sinθ ≈ θ tanθ ≈ θ cosθ ≈ 1 - θ²/2 Trig Identities sin² θ + cos² θ ≡ 1 tan θ ≡ sin θ / cos θ 1 + tan² x ≡ sec² x 1 + cot² x ≡ cosec² x Angle Addition Formulae sin(A ± B) ≡ sinA cosB ± cosA sinB cos(A ± B) ≡ cosA cos B ∓ sinA sinB tan(A ± B) ≡ (tanA ± tanB) / (1 ∓ tanA tanB) Double Angle Formulae sin2A ≡ 2 sinA cosA cos2A ≡ cos²A - sin²A ≡ 2cos²A - 1 ≡ 1 - 2sin²A tan2A ≡ (2 tanA) / (1 - tan²A) R-Addition Formula a sinx ± b sinx can be expressed as R sin(x ± α) a cosx ± b sinx can be expressed as R cos(x ∓ α) Where: R cos α = a R sin α = b R = √(a² + b²) Differentiation & Integration From First Principles: Product Rule: Quotient Rule: Integration by Parts Trigonometric Integration ∫ xⁿ dx = (xⁿ⁺¹) / (n+1) + c ∫ e ˣ dx = e ˣ + c ∫ 1/x dx = ln|x| + c ∫ cos(x) dx = sin(x) + c ∫ sin(x) dx = -cos(x) + c ∫ tan(x) dx = ln|sec(x)| + c ∫ sec²(x) dx = tan(x) + c ∫ cosec(x) cot(x) dx = -cosec(x) + c ∫ cosec²(x) dx = -cot(x) + c ∫ sec(x) tan(x) dx = sec(x) + c ∫ f'(ax+b) dx = f(ax+b) / a + c Series Expansions The Binomial Expansion: The Taylor Series: The Maclaurin Expansion: Common Expansions:
- Elasticity & Collisions
According to Hooke's Law, tension, T, is directly proportional to extension, x, in elastic strings and springs: T = kx The constant of proportionality, k, depends on two things: the modulus of elasticity, λ, of the material, and its unstretched, natural length, l: Note: natural length is noted with a lowercase 'L' - do not confuse this with an uppercase 'i'. The font of these note sheets is not massively helpful here, so sorry! Tension, T, is a force so is measured in Newtons. Extension and length are both distances, measured in metres. This gives the modulus of elasticity must also be measured in Newtons. Strings can be compressed as well as stretched. In this instance, Hooke's law still applies but the force in the spring, thrust (helpfully also written as T), acts the other way. Modelling For all questions, strings and springs are modelled as light. This means you do not take into account its weight. See the notes sheet on modelling in mechanics for single maths - the same rules apply. Elastic Potential Energy When an elastic string or spring is stretched or compressed, it stores elastic potential energy within it. This is equal to the work done in stretching/compressing the string/spring. The work done is represented by the area under a force-extension graph: Using the equation for the area of a right-angled triangle, A = ½ab, we can find the equation for the elastic potential energy stored in an elastic system: Worked Example Elastic Collisions in One Dimension Newton's law of restitution is used to solve problems where two particles collide directly: Direct impact means the the particles are moving along the same straight line when they collide. This means the collision is "head-on", so to speak. Newton's law of restitution depends on the coefficient of restitution, e, between the two colliding objects: 0 ≤ e ≤ 1 In a perfectly elastic collision, e = 1, so the two particles rebound at the same speeds In a perfectly inelastic collision, e = 0, so the particles coalesce (join together) When there are two unknowns in a problem, you can use the principle of conservation of linear momentum to solve simultaneously: Collisions with a Wall When a particle collides with a wall, the same principles apply. The only difference is, the wall does not move. Therefore, Newton's law of restitution is just taken as: Kinetic Energy The definition of a perfectly elastic collision (when e = 1) is a collision in which both momentum and kinetic energy are conserved. When e ≠ 1, the collision is not perfectly elastic, so some kinetic energy is lost. This can be calculated easily once you know the masses of each particle and their initial and final speeds: When an object collides with a wall, only the object experiences a change in kinetic energy Oblique Collisions of Spheres with Surfaces An oblique impact is when the collision does not act normally to the surfaces: While the collision may not be perpendicular to the surface, the impulse of the particle acts perpendicular to the surface. This means that the component of the velocity of the particle that acts parallel to the surface remains unchanged. v cos(β) = u cos(α) The components perpendicular to the surface are parallel to the impulse. Therefore, we can apply Newton's law of restitution: v sin(β) = eu sin(α) e is still the coefficient of restitution between the particle and the wall Dividing the second equation by the first eliminates u and v, leaving only the angles and e: tan(β) = e tan(α) Using sin² + cos² = 1 we can eliminate the angles to leave only u and v: Angle of Deflection The angle of deflection is the total angle through which the path of the particle changes. It is given by α + β Vectors Sometimes, the velocities will be given as vectors. If the surface is parallel to one of the unit vectors, this makes the question easier. To determine the angle of deflection, θ, if the velocities before and after are given as vectors u and v respectively, use the scalar product, u . v = |u| |v| cos(θ) Oblique Collisions of Spheres with Spheres In addition to collisions between spheres and surfaces, you need to be able to solve problems where two spheres collide in two dimensions: The impulse on each sphere acts along the line of centres of the two spheres Like with oblique collisions between spheres and surfaces: the component of the velocities perpendicular to the impulse is unchanged through the impact Newton's Law of restitution applies to the parallel components of the velocities the principle of conservation of momentum applies parallel to the line of centres
- Momentum, Impulse, Work, Energy & Power
As you know, momentum, p, is defined as the product of an object's mass and velocity: p = mv The Impulse-Momentum Principle When an object collides with something else, it experiences a change in momentum. This is equal to the impulse experienced by that object in the collision: I = Δmv The impulse can also be calculated as the product of the force of the collision, F, and its duration, t: I = Δmv = Ft Therefore, the units of impulse are Newton-seconds, Ns. Often, the force will have to be calculated using Newton's second law (F=ma), or SUVAT equations need to be used to calculate the duration or change in momentum. The Principle of Conservation of Momentum According to Newton's third law, for every reaction there is an equal but opposite reaction. This means that when two particles collide, both experience the same magnitude of impulse, but in opposite directions. Since impulse equals the change in momentum, the two particles experience equal and opposite changes in momentum, Δp and -Δp. Overall, these two changes in momentum cancel out: the total momentum before impact equals the total momentum after the impact Mathematically, this can be expressed as: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Work & Energy Work done is defined as the the amount of energy transferred from one form to another when a force causes movement. This is known as the work-energy principle. W = Fx Work done = force x distance moved. The force and motion must be in the same direction Because of this, you will often need to use trigonometry to find the correct components of forces: The units of energy are joules, J - so when the force is given in Newtons, N, and the distance in metres, m, the work done has units joules, J. Work Done Against Gravity & Gravitational Potential Energy Work done against gravity is given by the equation: W = mgh This can easily be derived from the work-done equation above: W = max: substitute F = ma into W = Fx W = mah: for work done against gravity, we tend to refer to the distance, x, as height, h W = mgh: acceleration due to gravity is g, 9.8 m/s² According to the Work-Energy Principle, work done is equal to energy transferred. Therefore, gravitational potential (G.P.E.) is also defined with this equation: G.P.E = mgh Kinetic Energy Kinetic Energy, KE, is given by the equation: KE = ½mv² This can also be derived from the work-done equation: a = (v²-u²) / 2s: rearrange v² = u ²+ 2as to make a the subject F = m(v²-u²) / 2s: substitute this into F = ma Fs = ½m(v²-u²): multiply both sides by displacement, s W = ½m(v²-u²): Fs = Fx = Work done ΔKE = ½m(v²-u²): Work done = the change in energy KE = ½mv² Power Power is defined as the rate at which work is done: P = Fv The units of power are Watts, W. If the force is given in Newtons and the velocity in m/s, the units of power are watts, 1 joule per second.
- Complex Numbers, Argand Diagrams & Loci
We know from single maths that if the discriminant of a quadratic equation is positive, it has two solutions, if the discriminant = 0, there is one repeated root, and if it is negative there are no roots. Wrong. If the discriminant is negative, the quadratic has two imaginary roots, given in terms of i, the square root of -1: i = √-1 A complex number is a sum of a real and an imaginary number, given in the form a + b i, where a is the real number and b the coefficient of i. Often, complex numbers are represented by the letter z. Adding and Subtracting Complex Numbers To add or subtract complex numbers, treat the real and imaginary terms separately: (a+b i) + (c+d i) = (a+c) + (b+d)i A few examples: 3 + (2+4i) = (3+2) + (4i) = 5+4i (1+2i) - 4i = (1) + (2-4)i = 1-2i (2-2i) + (6+i) = (2+6) + (-2+1)i = 8-i Multiplying Complex Numbers To multiply an imaginary number by a real constant, just expand the brackets: k(a+b i) = ka+kb i For example: 3(2+4i) = 3(2) + 3(4i) = 6+12i To multiply two imaginary numbers, expand like any quadratic, using the fact that: i² = -1 For example: (2+3i)(3+4i) = 6 + 8i +9i +12i² = 6 + 17i - 12 = -6+17i Complex Conjugation Just like conjugate pairs when rationalising a denominator, the conjugate of a complex number has the opposite sign: For any complex number z = a +bi, the conjugate is given as z* = a -bi The complex conjugate is given an asterisk. For any complex number z, the product of z and z* is a real number For example, if z = 4+5i: z* = 4-5i zz* = (4+5i)(4-5i) = 16 + 20i - 20i -25i² = 16 + 25 = 41 Solving Complex Equations Complex roots always come in conjugate pairs. Therefore, if the discriminant of a quadratic is negative, if z is one of the roots, z* will be the other root. For example, the roots of the equation x² - 6x + 13 = 0 are x = (3+2i) and x = (3-2i) Often, equations with complex roots will use z instead of x. The same rule applies to cubic and quartic graphs: If a cubic graph has just one real root, it will have two complex roots, z and z* If a quartic graph has two real roots or one real repeated root, then it will also have two complex roots, z and z* If a quartic graph has no real roots, it will have two complex conjugate pairs as roots: z, z*, w and w* All cubic graphs must have a total of three roots, real or complex All quartic graphs must have a total of four roots, real or complex Argand Diagrams Complex numbers can be represented graphically on an Argand diagram. The x-axis is the real axis, Re, and the y-axis is the imaginary axis, Im. To plot a complex number, treat it as a coordinate: The complex number z = (a + bi) has coordinates (a, b) on an Argand diagram This can be represented as a position vector, at an angle, θ, to the positive real axis (generally in radians): Like any vector, the modulus (the magnitude) can be found using Pythagoras' theorem: |z| = √(a²+b²) and tan of the angle is given as: tan(θ) = b/a Modulus-Argument Form The magnitude of the position vector for a complex number, r, is known as the modulus, and the angle it makes with the positive real axis, arg z, is called the argument. Therefore, complex numbers can also be expressed in modulus-argument form: z = r(cos(θ) + i sin(θ)) r = |z| arg z = θ Multiplying & Dividing Mod-Arg Loci A locus of points is a collection of points that can be anywhere along a particular line. These can be represented with complex numbers on an Argand diagram. The distance between two complex numbers, z₁ and z₂, is given as |z₂ - z₁| If, instead of defining both complex numbers, you just define one and give one general complex number, you get the locus of a circle: |z - z₁| = r is a circle with centre z₁ and radius r This can also be given as |z - (a+bi)| = r, where r is the radius and (a, b) the centre. You can find the locus of points for a perpendicular bisector of a line segment between two points: The perpendicular bisector of a line segment between two complex numbers, z₁ and z₂, is given as |z - z₁| = |z - z₂| Using the argument of a complex number, z₁, we can express the locus of points for a half line. This is a line from, but not including, the fixed point z₁ at an angle of arg(z-z₁): arg(z-z₁) = θ is a half line from but not including the point z₁ at an angle of θ to the horizontal. Regions Loci can be used to represent regions on an Argand diagram: Exponential Form Complex numbers can also be expressed in exponential form, z = r e^iθ, using Euler's relation: Euler's relation is derived from the Maclaurin series expansions of sin(θ), cos(θ) and eˣ. See Further Maths Notes Sheet on Series. This leads to the exponential form for a complex number, z: Where r = |z| and arg z = θ just like in modulus-argument form. Multiplying & Dividing The same rules as for modulus-argument can be applied to multiplying and dividing complex numbers in exponential form: De Moivre's Theorem Euler's relation can be used to find powers of complex numbers that are in the modulus-argument form. (r (cos(θ) + i sin(θ))ⁿ = rⁿ(cos(nθ) + i sin(nθ)) For any integer n This can be proven with Euler's relation: It can also be proven using proof by induction (See the Notes Sheet on this) Trigonometric Identities De Moivre's theorem, combined with the binomial expansion, can be used to derive trigonometric identities. To drive an equation for sin(nθ) or cos(nθ): To drive an equation for sinⁿ(θ) or cosⁿ(θ): There are four standard results you need to know: An example of how to apply these: nth Roots of a Complex Number Just as a real number, x, has two square roots, ±√x, any complex number has n distinct nth roots. If z and w are non-zero complex numbers and n is a positive integer, the equation zⁿ = w has n distinct roots. We can use de Moivre's theorem to find the solutions to zⁿ = w by taking into account the fact that its argument is repeated every 2π. For any complex number z = r(cos(θ) + i sin(θ)), z = r(cos(θ + 2kπ) + i sin(θ + 2kπ)) Where k is any integer When zⁿ = 1 In general, the solutions to zⁿ = 1 are z = cos(2kπ / n) + i sin(2kπ / n) for k = 1, 2, 3, ..., n These are known as the nth roots of unity. If n is a positive integer, then there is an nth root of unity ω such that: the nth roots of unity are 1, ω², ω³, ω⁴, ..., ωⁿ‾¹ 1, ω, ω², ω³, ω⁴, ..., ωⁿ‾¹ form the vertices of a regular n-gon 1 + ω² + ω³ + ω⁴ + ... + ωⁿ‾¹ = 0 Using nth Roots Geometrically The nth roots of any complex number, z, form the vertices of a regular n-gon with the origin at its centre The size and orientation of the polygon depend on the complex number z. If you know the coordinates of a single vertex, you can find the others by rotating that point about the origin by 2π/n. This is the same as multiplying by the nth roots of unity. If z₁ is one of the roots of the equation zⁿ = w, and 1, ω, ω², ω³, ω⁴, ..., ωⁿ‾¹ are the roots of unity, then the roots of zⁿ = w are z₁, z₁ω, z₁ω², z₁ω³, z₁ω⁴, ..., z₁ωⁿ‾¹
- Proof by Induction
Proof by induction is used to prove that a general statement is true for all positive integer values. All proofs by mathematical induction follow four basic steps: Prove that the general statement is true when n = 1 Assume the general statement is true for n = k Show that, if it is true for n = 1, the general statement is also true for n = k+1 Conclude that the general statement is true whenever n ∈ ℕ All four steps must be shown clearly in your workings: prove, assume, show, conclude. Proving Sums Often, questions will use the standard results for the sums of r, r² and r³. Regardless, the method for all sums is the same and follows the four steps above. Quick Tip For the 3rd step, it is generally best to write the last line out first using the general function - just substitute (k+1) into it. You know that this is the answer you want to reach, so use it as a target to help you. Proving Divisibility Results Again, follow the four standard steps for proof by induction. For divisibility results, make step 2 equal any general multiple of the divisor: Proof Using Matrices Exactly the same four steps apply:
Sigma notation is used to write series quickly. This describes each term of a series as an equation, where substituting in values for r gives that term in the series. There are standard sums of series formulae that are used all the time: The Sum of Natural Numbers To find the sum of a series of constant terms, use the formula: To find the sum of the first n natural numbers (positive integers): If the series does not start at r = 1, but at r = k, subtract one series from another, where the first goes from r = 1 to r = k, and the second from r = 1 to r = (k-1): Expressions in sigma notation can be rearranged to simplify complicated series: The Sum of Squares The first sum formula above, for constant terms, is linear. The second, for the sum of natural numbers, is quadratic. Therefore, the formula for the sum of square terms is cubic. To find the sum of the sum of the squares of the first n natural numbers: The Sum of Cubics Following through with this pattern, the formula to find the sum of the cubes of the first n natural numbers is quartic: The Method of Differences The method of differences is used when the expansion of a series leads to pairs of terms that cancel out, leaving only a few single terms behind. When the general term, u₁, of a series can be expressed in the form f(r) - f(r+1), the method of differences applies. This means that: To solve problems with the method of differences Write out the first few terms to see what cancels Write out the last few terms to see what cancels Add the left overs from the beginning and the end Often, you will need to use partial fractions Example The Maclaurin Series We often work with first and second order derivatives, but why stop there? Functions that can be written as an infinite sum of terms in the form axⁿ can be differentiated infinite times. We already know of a few such examples: the binomial expansion of 1 / (1-x) forms an infinite polynomial, 1 + x + x² + x³ + x⁴ + ... the binomial expansion of 1 / √(1+x) forms an infinite polynomial, 1 + ½ x - ⅟₈ x² + ⅟₁₆ x³ - ⅟₃₂ x⁴ + ... eˣ = 1 + x + ½ x² + ⅟₆ x³ + ⅟₂₄ x⁴ + ... This happens when x = 0 is substituted into increasing orders of derivatives, which, in turn, increases the power of the polynomial. This is only true if the function can be differentiated an infinite number of times, and if the series converges. When the above criteria are satisfied and each derivative has a finite value, the Maclaurin series expansion is valid: There are many standard versions of the Maclaurin series expansion that are useful to know, as they crop up a great deal:
- Roots of Polynomials
Won't lie, this topic isn't particularly fun. It's just a whole lot of algebra and equations that you need to know and get right. At least the topic is small. To be perfectly honest, the only challenge is writing everything out clearly, and not getting confused between the Greek and Latin alphabets: don't confuse a with α don't confuse b with β don't confuse g with γ don't confuse d with δ a, b, c, d & e are real constants α, β, γ, δ are roots of polynomials Quadratic Polynomials A quadratic equation in the form ax² + bx + c = 0 has two possible roots, α and β. These can be distinct and real, repeated (the same), or complex conjugates. The roots can be added or multiplied to give the following results: Squaring or cubing the roots gives different results: Cubic Polynomials A cubic equation in the form ax³ + bx² + cx + d = 0 has three possible roots, α, β and γ. The roots can be added or multiplied to give the following results: Squaring or cubing the roots gives different results: Quartic Polynomials A quartic equation in the form ax⁴ + bx³ + cx² + dx + e = 0 has four possible roots, α, β, γ and δ. The roots can be added or multiplied to give the following results: Squaring the roots gives different results: Linear Transformations of Roots If you know the sums and products of roots of a polynomial, you can find the equation of another polynomial whose roots are a linear transformation of the first. For example, if the equation x² - 2x + 3 = 0 has roots α and β, find the equation that has roots (α+2) and (β+2). This can be done in two ways: Use the roots of polynomial equations above Rearrange the roots and substitute into the first equation Method 1 (α+2) + (β+2) = α + β + 4 = -b/a + 4 = 6 (α+2)(β+2) = αβ + 2α + 2β + 4 = αβ + 2(α+β) + 4 = c/a - 2b/a + 4 = 11 So the equation is w² - 6w + 11 = 0 The coefficient of each term must be the same as the initial equation Method 2 Let w = x + 2 Rearrange for x: x = w - 2 Substitute into first equation: (w-2)² - 2(w-2) + 3 = 0 w² - 4w + 4 -2w + 4 +3 = 0 w² - 6w + 11 = 0