Complex numbers

• Read \(|z - a|\) as “the distance from \(a\) to \(z\)” and \(\arg(z - a)\) as “the direction from \(a\) to \(z\)”, and sketch the quadrant before you touch \(\tan^{-1}\).
• Sketch the quadrant before tan−1; divide by the conjugate of the denominator; real coefficients give conjugate pairs; moduli multiply and arguments add; |z − a| is a distance and arg(z − a) is a half-line from a.

• 3.9.1 I can understand — understand the idea of a complex number, recall the meaning of the terms real part, imaginary part, modulus, argument, conjugate, and use the fact that two complex numbers are equal if and only if both real and imaginary parts are equal
• 3.9.2 I can carry out — carry out operations of addition, subtraction, multiplication and division of two complex numbers expressed in Cartesian form x + iy
• 3.9.3 I can use — use the result that, for a polynomial equation with real coefficients, any non-real roots occur in conjugate pairs
• 3.9.4 I can represent — represent complex numbers geometrically by means of an Argand diagram
• 3.9.5 I can carry out — carry out operations of multiplication and division of two complex numbers expressed in polar form r(cos θ + i sin θ) ≡ reiθ
• 3.9.6 I can find — find the two square roots of a complex number
• 3.9.7 I can understand — understand in simple terms the geometrical effects of conjugating a complex number and of adding, subtracting, multiplying and dividing two complex numbers
• 3.9.8 I can illustrate — illustrate simple equations and inequalities involving complex numbers by means of loci in an Argand diagram

Accuracy for this chapter. Arguments are in radians: exact where exact (\(\tfrac{\pi}{3}\), \(-\tfrac{5\pi}{6}\), \(-\tfrac{11\pi}{12}\)), otherwise to 3 significant figures (\(0.927\), \(2.21\), \(1.11\)); never in degrees in a final answer. Keep \(\tan^{-1}\) values unrounded (\(0.927\,30\)) until after the subtraction from \(\pi\). Moduli are exact (\(2\sqrt2\), \(13\sqrt2\), \(3 - \sqrt5\)), with a 3 significant figure decimal only as a second form. Multiplication, division and square roots must show full working: the syllabus says so, and an answer copied from a calculator's complex mode earns nothing without it.

• “\(\arg(-3 + 4i) = \tan^{-1}\!\left(\dfrac{4}{-3}\right) = -0.927\).” Correct Sketch the quadrant before taking \(\tan^{-1}\). \(-3 + 4i\) is in the second quadrant; the calculator's answer is a fourth-quadrant angle. Take the acute angle \(\alpha = \tan^{-1}\tfrac43 = 0.927\,30\) and adjust: \(\arg = \pi - \alpha = 2.21\).
• “\(\arg(z - 1 - i) = \tfrac{\pi}{4}\) is the line \(y = x\).” Correct It is a half-line from \(1 + i\), not a line. It starts at \(1 + i\), which is excluded (open circle) because \(z - 1 - i = 0\) has no argument, and it goes one way only: up and to the right. The angle is measured from a line through \(1 + i\) parallel to the real axis, not from \(O\).
• “\(\mathrm{Im}(3 - 4i) = -4i\).” Correct The imaginary part is the real number \(-4\). \(z = x + iy\) has \(\mathrm{Re}\,z = x\) and \(\mathrm{Im}\,z = y\), both real.
• “\(1 + 2i\) is a root, so \(1 - 2i\) is a root” — of an equation with a non-real coefficient. Correct Conjugate pairs are guaranteed only when every coefficient is real. \(z^2 - 3z + (3 - i) = 0\) has roots \(2 + i\) and \(1 - i\), which are not conjugates.
• “\(\arg(zw) = \tfrac{\pi}{3} + \tfrac{3\pi}{4} = \tfrac{13\pi}{12}\).” Correct Adding arguments can leave the principal range. \(\tfrac{13\pi}{12} > \pi\), so subtract \(2\pi\): \(-\tfrac{11\pi}{12}\).
• “\(a^4 - 5a^2 - 36 = 0\), so \(a = \pm3\) or \(a = \pm 2i\).” Correct In \((a + ib)^2 = p + iq\) the numbers \(a\) and \(b\) are real by definition, so \(a^2 = -4\) is rejected. There are exactly two square roots, \(\pm(3 + 2i)\).
• “\(|z + 2 - 3i| = 4\) is the circle with centre \(2 - 3i\).” Correct Write it as \(|z - (-2 + 3i)| = 4\): the centre is \(-2 + 3i\). The number subtracted from \(z\) is the centre.
• “\(\mathrm{Im}(3 - 4i) = -4i\).” Repair The imaginary part is the real number \(-4\).
• “\(|3 - 4i| = \sqrt{9 - 16}\).” Repair \(|z| = \sqrt{x^2 + y^2}\) with real \(x\) and \(y\): \(\sqrt{9 + 16} = 5\). \(i\) is never inside the square root.
• “\(\arg(-3 + 4i) = \tan^{-1}\!\left(\dfrac{4}{-3}\right) = -0.927\).” Repair That is a fourth-quadrant angle; \(-3 + 4i\) is in the second quadrant, so \(\arg = \pi - 0.927\,30 = 2.21\). Sketch first.
• “\(\arg(-4) = 0\), because \(\tan^{-1}\dfrac{0}{-4} = 0\).” Repair \(-4\) is on the negative real axis: \(\arg(-4) = \pi\).
• “\((2 + 3i)(4 - i) = 8 + 10i - 3 = 5 + 10i\).” Repair \(-3i^2 = +3\), so the product is \(11 + 10i\).
• “\(\dfrac{5 + i}{2 - 3i} = \dfrac{(5 + i)(2 + 3i)}{4 - 9} = -\dfrac{7 + 17i}{5}\).” Repair \((2 - 3i)(2 + 3i) = 4 - 9i^2 = 4 + 9 = 13\). The denominator \(zz^* = x^2 + y^2\) is always a sum of squares, never a difference.
• “\(1 + 2i\) is a root, so \(1 - 2i\) is a root” for an equation with a non-real coefficient. Repair The conjugate-pair result needs real coefficients; \(z^2 - 3z + 3 - i = 0\) has roots \(2 + i\) and \(1 - i\).
• “The quadratic factor from \(1 + 2i\) is \(z^2 + 2z + 5\).” Repair \((z - w)(z - w^*) = z^2 - 2\,\mathrm{Re}(w)\,z + |w|^2 = z^2 - 2z + 5\). The middle coefficient is \(-(w + w^*) = -2\,\mathrm{Re}(w) = -2\).
• “\(\arg(z_1z_2) = \dfrac{13\pi}{12}\).” Repair The principal argument lies in \((-\pi, \pi]\); subtract \(2\pi\) to get \(-\dfrac{11\pi}{12}\).
• “The square root of \(5 + 12i\) is \(3 + 2i\).” Repair There are two: \(\pm(3 + 2i)\).
• “\(a^4 - 5a^2 - 36 = 0 \Rightarrow a^2 = 9\) or \(-4 \Rightarrow a = \pm 3\) or \(\pm 2i\).” Repair \(a\) is real by definition; reject \(a^2 = -4\).
• “\(|z + 2 - 3i| = 4\) is a circle with centre \(2 - 3i\).” Repair \(|z - (-2 + 3i)| = 4\): the centre is \(-2 + 3i\).
• Drawing \(\arg(z - 1 - i) = \dfrac{\pi}{4}\) as the whole line \(y = x\), or measuring \(\dfrac{\pi}{4}\) from \(O\). Repair It is a half-line from \(1 + i\), with \(1 + i\) excluded (open circle), at \(\dfrac{\pi}{4}\) to a line through \(1 + i\) parallel to the real axis.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Section 3.9 is the last section of Paper 3, so the returns run backwards: when you revise chapter 9, re-solve a real cubic from one complex root; when you revise chapter 11, re-derive the polar product rule from the compound-angle formulae; when you revise chapter 3, redo a locus as a circle or bisector in both notations. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3. Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. This chapter's content is in Paper 3 only; none of it is in Paper 2. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A real polynomial with one complex root given: show that it is a root, then find the others with the conjugate pair and division. A square root found by equating parts, then used (hence) in a quadratic. A locus or region sketched, then a greatest or least value of |z| or arg z found from the diagram. Cartesian and polar arithmetic appear inside all three.
• MF19 has no complex-numbers entry. The definitions, the division method, the conjugate-pair factor, polar form and its product and quotient rules, the square-root method and every locus must be known. Only the compound-angle formulae behind the polar product rule are printed, under Trigonometry.
• The syllabus asks for full working in multiplication, division and square roots, so the expansion is written before the answer. Arguments are in radians, exact where exact, otherwise 3 significant figures, with tan−1 kept unrounded until the quadrant adjustment. A principal argument outside (−π, π] is not in its final form. In a sketch, a half-line without its open circle or a strict boundary drawn solid is a different locus.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.