Complex Numbers (L8th)

Complex numbers up until Argand Diagram and Loci

Showing 1-25 of 35 problems
2025 Paper 2 Q2
D: 1500.0 B: 1500.0

    1. Show that if the complex number \(z\) satisfies the equation \[z^2 + |z + b| = a,\] where \(a\) and \(b\) are real numbers, then \(z\) must be either purely real or purely imaginary.
    2. Show that the equation \[z^2 + \left|z + \frac{5}{2}\right| = \frac{7}{2}\] has no purely imaginary roots.
    3. Show that the equation \[z^2 + \left|z + \frac{7}{2}\right| = \frac{5}{2}\] has no purely real roots.
    4. Show that, when \(\frac{1}{2} < b < \frac{3}{4}\), the equation \[z^2 + |z + b| = \frac{1}{2}\] will have at least one purely imaginary root and at least one purely real root.
  2. Solve the equation \[z^3 + |z + 2|^2 = 4.\]

Show Solution
    1. Suppose \(z^2 + |z + b| = a\), then \(z^2 = a- |z + b| \in \mathbb{R}\), since \(a \in \mathbb{R}\). Since the square root of a real number is either purely real or purely imaginary, \(z\) is purely real or purely imaginary.
    2. Suppose \(z = it\) for some \(t \in \mathbb{R}\), then \begin{align*} && \frac72 &= -t^2 +\sqrt{t^2 + \frac{25}{4}} \\ \Rightarrow && \left ( \frac72 + t^2\right)^2 &= t^2 + \frac{25}{4} \\ \Rightarrow && t^4 + 7t^2 + \frac{49}{4} &= t^2 + \frac{25}{4} \\ \Rightarrow && t^4 + 6t^2 + 6 &= 0 \end{align*} but since \(\Delta = 6^2 - 4 \cdot 1 \cdot 6 < 0\) there are no real solutions.
    3. Suppose \(z = t\) for some \(t \in \mathbb{R}\), then either \(t^2 + t + \frac72 = \frac52 \Rightarrow t^2 + t + 1 = 0\) (no solutions) or \(t^2 - t - \frac72 = \frac52 \Rightarrow t^2 - t - 6 = (t-3)(t+2) = 0\). When \(t = 3\) then we must take the positive part for \(|z + \frac72|\) so this cannot work. When \(t = -2\) we also have \(\frac72-2 > 0\) so we are still taking the positive part. Hence no solutions
    4. Suppose \(\frac{1}{2} < b < \frac{3}{4}\), the equation then consider \(z^2 + |z + b| = \frac{1}{2}\). Case 1: \(z = t \in \mathbb{R}\), then we have two cases: Case 1a: \(z+b > 0\). \(z^2 + z + b = \frac12 \Rightarrow z = \frac{-1 \pm \sqrt{1-4b+2}}{2}\) which clearly is a valid real number an \(z + b > 0\). Case 1b: \(z+b < 0\) \(z^2 - z - b = \frac12 \Rightarrow z = \frac{1 \pm \sqrt{1+4b-2}}{2}\)
  2. Let \(\omega\) be a (primitive) cube root of unity. \begin{align*} && z^3 &= 4 - |z+2|^2 \\ \Rightarrow && z &\in \mathbb{R} \cup \omega \mathbb{R} \cup \omega^2 \mathbb{R} \end{align*} Case 1:
2023 Paper 2 Q7
D: 1500.0 B: 1500.0

  1. The complex numbers \(z\) and \(w\) have real and imaginary parts given by \(z = a + \mathrm{i}b\) and \(w = c + \mathrm{i}d\). Prove that \(|zw| = |z||w|\).
  2. By considering the complex numbers \(2 + \mathrm{i}\) and \(10 + 11\mathrm{i}\), find positive integers \(h\) and \(k\) such that \(h^2 + k^2 = 5 \times 221\).
  3. Find positive integers \(m\) and \(n\) such that \(m^2 + n^2 = 8045\).
  4. You are given that \(102^2 + 201^2 = 50805\). Find positive integers \(p\) and \(q\) such that \(p^2 + q^2 = 36 \times 50805\).
  5. Find three distinct pairs of positive integers \(r\) and \(s\) such that \(r^2 + s^2 = 25 \times 1002082\) and \(r < s\).
  6. You are given that \(109 \times 9193 = 1002037\). Find positive integers \(t\) and \(u\) such that \(t^2 + u^2 = 9193\).

2022 Paper 3 Q8
D: 1500.0 B: 1500.0

  1. Use De Moivre's theorem to prove that for any positive integer \(k > 1\), \[ \sin(k\theta) = \sin\theta\cos^{k-1}\theta \left( k - \binom{k}{3}(\sec^2\theta - 1) + \binom{k}{5}(\sec^2\theta - 1)^2 - \cdots \right) \] and find a similar expression for \(\cos(k\theta)\).
  2. Let \(\theta = \cos^{-1}(\frac{1}{a})\), where \(\theta\) is measured in degrees, and \(a\) is an odd integer greater than \(1\). Suppose that there is a positive integer \(k\) such that \(\sin(k\theta) = 0\) and \(\sin(m\theta) \neq 0\) for all integers \(m\) with \(0 < m < k\). Show that it would be necessary to have \(k\) even and \(\cos(\frac{1}{2}k\theta) = 0\). Deduce that \(\theta\) is irrational.
  3. Show that if \(\phi = \cot^{-1}(\frac{1}{b})\), where \(\phi\) is measured in degrees, and \(b\) is an even integer greater than \(1\), then \(\phi\) is irrational.

2019 Paper 3 Q6
D: 1500.0 B: 1500.0

The point \(P\) in the Argand diagram is represented by the the complex number \(z\), which satisfies $$zz^* - az^* - a^*z + aa^* - r^2 = 0.$$ Here, \(r\) is a positive real number and \(r^2 \neq a^*a\). By writing \(|z - a|^2\) as \((z - a)(z - a)^*\), show that the locus of \(P\) is a circle, \(C\), the radius and the centre of which you should give.

  1. The point \(Q\) is represented by \(\omega\), and is related to \(P\) by \(\omega = \frac{1}{z}\). Let \(C'\) be the locus of \(Q\). Show that \(C'\) is also a circle, and give its radius and centre. If \(C\) and \(C'\) are the same circle, show that $$(|a|^2 - r^2)^2 = 1$$ and that either \(a\) is real or \(a\) is imaginary. Give sketches to indicate the position of \(C\) in these two cases.
  2. Suppose instead that the point \(Q\) is represented by \(\omega\), where \(\omega = \frac{1}{z^*}\). If the locus of \(Q\) is \(C\), is it the case that either \(a\) is real or \(a\) is imaginary?

Show Solution
\begin{align*} && |z-a|^2 &= (z-a)(z-a)^* \\ &&&= (z-a)(z^*-a^*) \\ &&&= zz^*-az^*-a^*z+aa^* \\ &&&= r^2 \end{align*} Therefore the locus of \(P\) is a circle centre \(a\) radius \(r\).
  1. \begin{align*} && 0 &= zz^* - az^* - a^*z + aa^* - r^2 \\ &&&= \frac{1}{\omega \omega^{*}} - \frac{a}{\omega^*} - \frac{a^*}{\omega} + aa^*-r^2 \\ \Rightarrow && 0 &= 1-a\omega-a^*\omega^*+(|a|^2-r^2)\omega\omega^* \\ \Rightarrow && 0 &= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)-\left ( \frac{a^*}{|a|^2-r^2}\right)\left ( \frac{a}{|a|^2-r^2}\right)+ \frac{1}{|a|^2-r^2} \\ &&&= \omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}-\frac{|a|^2}{(|a|^2-r^2)^2}+ \frac{1}{|a|^2-r^2} \\ &&&=\omega\omega^* - \left ( \frac{a^*}{|a|^2-r^2}\right)^*\omega - \left ( \frac{a^*}{|a|^2-r^2}\right)\omega^*+\frac{|a|^2}{(|a|^2-r^2)^2}- \frac{r^2}{(|a|^2-r^2)^2} \end{align*} Therefore \(\displaystyle \left|\omega-\left ( \frac{a^*}{|a|^2-r^2}\right)\right|^2 = \frac{r^2}{(|a|^2-r^2)^2}\) ie \(\omega\) lies on a circle centre \(\frac{a^*}{|a|^2-r^2}\), radius \(\frac{r}{||a|^2-r^2|}\). If these are the same circle then \(r = \frac{r}{||a|^2-r^2|} \Rightarrow (|a|^2-r^2)^2 = 1\) and \(a = \frac{a^*}{|a|^2-r^2} \Rightarrow a = \pm a^*\), ie \(a\) is purely real or imaginary.
  2. This is the same story, except we end up with centre \(\frac{a}{|a|^2-r^2}\), so we do not end up with the same conditions
2018 Paper 3 Q6
D: 1700.0 B: 1516.0

  1. The distinct points \(A\), \(Q\) and \(C\) lie on a straight line in the Argand diagram, and represent the distinct complex numbers \(a\), \(q\) and \(c\), respectively. Show that \(\dfrac {q-a}{c-a}\) is real and hence that \((c-a)(q^*-a^*) = (c^*-a^*)(q-a)\,\). Given that \(aa^* = cc^* = 1\), show further that \[ q+ ac q^* = a+c \,. \]
  2. The distinct points \(A\), \(B\), \(C\) and \(D\) lie, in anticlockwise order, on the circle of unit radius with centre at the origin (so that, for example, \(aa^* =1\)). The lines \(AC\) and \(BD\) meet at \(Q\). Show that \[ (ac-bd)q^* = (a+c)-(b+d) \,, \] where \(b\) and \(d\) are complex numbers represented by the points \(B\) and \(D\) respectively, and show further that \[ (ac-bd) (q+q^*) = (a-b)(1+cd) +(c-d)(1+ab) \,. \]
  3. The lines \(AB\) and \(CD\) meet at \(P\), which represents the complex number \(p\). Given that \(p\) is real, show that \(p(1+ab)=a+b\,\). Given further that \(ac-bd \ne 0\,\), show that \[ p(q+q^*) = 2 \,. \]

Show Solution
  1. \(A\), \(Q\), \(C\) lie on a straight line if \(q = \lambda a + (1-\lambda)c\) for some \(\lambda \in \mathbb{R}\), \begin{align*} && q &= \lambda a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)a + (1-\lambda)c \\ \Leftrightarrow && q - a &= (\lambda - 1)(a-c) \\ \Leftrightarrow && \frac{q - a}{c-a} &= 1-\lambda \\ \end{align*} therefore \(\frac{q-a}{c-a} \in \mathbb{R}\) \begin{align*} && \frac{q-a}{c-a} & \in \mathbb{R} \\ \Leftrightarrow && \left (\frac{q-a}{c-a} \right)^* &= \frac{q-a}{c-a} \\ \Leftrightarrow && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \end{align*} Given \(aa^* = cc^* = 1\), \begin{align*} && (q^*-a^*)(c-a) &= (q-a)(c^*-a^*) \\ \Leftrightarrow && q^*(c-a) - \frac{c}{a}+1 &= q \frac{a-c}{ca} - \frac{a}{c}+1 \\ \Leftrightarrow && (c-a)\l q^* +\frac{q}{ca}\r &= \frac{c}{a} - \frac{a}{c} \\ &&&= \frac{c^2-a^2}{ac} \\ \Leftrightarrow && q^* +\frac{q}{ca} &= \frac{c+a}{ac} \\ \Leftrightarrow && q^*ac +q &= a+c \end{align*}
  2. Since \(Q\) lies on \(AC\) and \(BD\) we must have \begin{align*} &&& \begin{cases} q^*ac +q &= a+c \\ q^*bd +q &= b+d \\ \end{cases} \\ \Rightarrow && q^*(ac-bd) &= (a+c)-(b+d) \\ \Rightarrow && q(ac-bd) &= (b+d)ac-(a+c)bd \\ \Rightarrow && (q+q^*)(ac-bd) &= (a+c)(1-bd)+(b+d)(ac-1) \\ &&&=a-abd+c-bcd+abc-b+acd-d \\ &&&= a(1+cd)-b(1+cd)+c(1+ab)-d(1+ab) \\ &&&= (a-b)(1+cd)+(c-d)(1+ab) \end{align*}
  3. If \(AB\) and \(CD\) meet at \(p\) we must have \(p^*ab + p = a+b\), ie \(p(1+ab) = a+b\) amd \(p(1+cd) = c+d\), so \begin{align*} && (q+q^*)(ac-bd) &= (a-b) \frac{c+d}{p} + (c-d) \frac{a+b}{p} \\ \Leftrightarrow && p(q+q^*)(ac-bd) &= (a-b)(c+d)+(c-d)(a+b) \\ &&&= ac+ad-bc-bd+ac+bc-ad-bd \\ &&&= 2(ac-bd) \\ \Leftrightarrow && p(q+q^*) &= 2 \end{align*}
2014 Paper 3 Q5
D: 1700.0 B: 1500.0

A quadrilateral drawn in the complex plane has vertices \(A\), \(B\), \(C\) and \(D\), labelled anticlockwise. These vertices are represented, respectively, by the complex numbers \(a\), \(b\), \(c\) and \(d\). Show that \(ABCD\) is a parallelogram (defined as a quadrilateral in which opposite sides are parallel and equal in length) if and only if \(a+c =b+d\,\). Show further that, in this case, \(ABCD\) is a square if and only if \({\rm i}(a-c)=b-d\). Let \(PQRS\) be a quadrilateral in the complex plane, with vertices labelled anticlockwise, the internal angles of which are all less than \(180^\circ\). Squares with centres \(X\), \(Y\), \(Z\) and \(T\) are constructed externally to the quadrilateral on the sides \(PQ\), \(QR\), \(RS\) and \(SP\), respectively.

  1. If \(P\) and \(Q\) are represented by the complex numbers \(p\) and \(q\), respectively, show that \(X\) can be represented by \[ \tfrac 12 \big( p(1+{\rm i} ) + q (1-{\rm i})\big) \,. \]
  2. Show that \(XY\!ZT\) is a square if and only if \(PQRS\) is a parallelogram.

Show Solution
The vector representing the side \(AB\) is \(b - a\) and the vector representing the side \(DC\) is \(c - d\). \(ABCD\) is a parallelogram if and only if these opposite sides are parallel and equal in length, which is given by \(b - a = c - d\), or equivalently \(a + c = b + d\). Similarly, if \(a + c = b + d\), then \(c - b = d - a\), so the side \(BC\) is parallel and equal in length to the side \(AD\). Thus, \(a + c = b + d\) is the necessary and sufficient condition for \(ABCD\) to be a parallelogram. In a parallelogram, the shape is a square if and only if the diagonals are equal in length and perpendicular to each other. The diagonals are represented by the vectors \(c - a\) and \(d - b\). For these to be equal in length and perpendicular, one must be a \(90^\circ\) rotation of the other. Since \(A, B, C, D\) are labeled anticlockwise, a \(90^\circ\) anticlockwise rotation of the vector \(\vec{AC}\) (which is \(c-a\)) would point in the direction of \(\vec{DB}\) (which is \(b-d\) if we consider the relative orientation). Specifically: \(i(c - a) = d - b \implies -i(a - c) = d - b \implies i(a - c) = b - d\). Thus, \(ABCD\) is a square if and only if \(i(a - c) = b - d\).
  1. The midpoint of the side \(PQ\) is \(\frac{1}{2}(p + q)\). To find the centre \(X\) of the square built externally on \(PQ\), we start at the midpoint and move a distance equal to half the side length in a direction perpendicular to \(PQ\). Since \(P, Q, R, S\) are anticlockwise, the outward direction is a \(90^\circ\) clockwise rotation of the vector \(\vec{PQ}\). A clockwise rotation of \(90^\circ\) corresponds to multiplication by \(-i\). \[ x = \frac{p+q}{2} + (-i)\left(\frac{q-p}{2}\right) = \frac{p + q - iq + ip}{2} = \frac{1}{2} \big( p(1+i) + q(1-i) \big) \]
  2. From part (i), we have the representations for the centres: \begin{align*} x &= \tfrac{1}{2}(p(1+i) + q(1-i)) \\ y &= \tfrac{1}{2}(q(1+i) + r(1-i)) \\ z &= \tfrac{1}{2}(r(1+i) + s(1-i)) \\ t &= \tfrac{1}{2}(s(1+i) + p(1-i)) \end{align*} As shown in the first part of the problem, \(XYZT\) is a square if and only if: (1) \(x+z = y+t\) (it is a parallelogram) (2) \(i(x-z) = y-t\) (it is a square) First, examine condition (1): \begin{align*} x+z - (y+t) &= \tfrac{1}{2} \big[ (p+r)(1+i) + (q+s)(1-i) - (q+s)(1+i) - (r+p)(1-i) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(1+i - (1-i)) - (q+s)(1+i - (1-i)) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(2i) - (q+s)(2i) \big] \\ &= i(p+r - (q+s)) \end{align*} Thus, \(x+z = y+t\) if and only if \(p+r = q+s\), which is the condition that \(PQRS\) is a parallelogram. Next, examine condition (2): \begin{align*} i(x-z) &= \tfrac{1}{2} i \big[ p(1+i) + q(1-i) - r(1+i) - s(1-i) \big] \\ &= \tfrac{1}{2} \big[ p(i-1) + q(i+1) - r(i-1) - s(i+1) \big] \\ y-t &= \tfrac{1}{2} \big[ q(1+i) + r(1-i) - s(1+i) - p(1-i) \big] \\ \text{So, } i(x-z) - (y-t) &= \tfrac{1}{2} \big[ p(i-1 + 1-i) + q(i+1 - 1-i) + r(-i+1 - 1+i) + s(-i-1 + 1+i) \big] \\ &= 0 \end{align*} Since \(i(x-z) = y-t\) is an identity (always true for any \(PQRS\)), \(XYZT\) is a square if and only if it is a parallelogram. As established above, this occurs if and only if \(PQRS\) is a parallelogram.
2013 Paper 3 Q6
D: 1700.0 B: 1500.0

Let \(z\) and \(w\) be complex numbers. Use a diagram to show that \(\vert z-w \vert \le \vert z\vert + \vert w \vert\,.\) For any complex numbers \(z\) and \(w\), \(E\) is defined by \[ E = zw^* + z^*w +2 \vert zw \vert\,. \]

  1. Show that \(\vert z-w\vert^2 = \left( \vert z \vert + \vert w\vert\right)^2 -E\,\), and deduce that \(E\) is real and non-negative.
  2. Show that \(\vert 1-zw^*\vert^2 = \left ( 1 +\vert zw \vert \right)^2 -E\,\).
Hence show that, if both \(\vert z \vert >1\) and \(\vert w \vert >1\), then \[ \frac {\vert z-w\vert} {\vert 1-zw^*\vert } \le \frac{\vert z \vert +\vert w\vert }{1+\vert z w \vert}\,. \] Does this inequality also hold if both \(\vert z \vert <1\) and \(\vert w \vert <1\)?

Show Solution
  1. \(\,\) \begin{align*} && |z-w|^2 &= (z-w)(z^*-w^*) \\ &&&= zz^* - wz^*-zw^* + ww^* \\ &&&= |z|^2+|w|^2 - E + 2|zw| \\ &&&= (|z|+|w|)^2 - E \\ \Rightarrow && E &= (|z|+|w|)^2 - |z-w|^2 &\in \mathbb{R} \end{align*} and by the triangle inequality \(|z|+|w| \geq |z-w|\), so \(E \geq 0\)
  2. \(\,\) \begin{align*} && |1-zw^*|^2 &= (1-zw^*)(1-z^*w) \\ &&&= 1 - zw^*-z^*w + |zw|^2 \\ &&&= 1 - E + 2|zw| + |zw|^2 \\ &&&= (1+|zw|)^2 - E \end{align*} \begin{align*} && \frac{|z-w|^2}{|1-zw^*|^2} &= \frac{(|z|+|w|)^2-E}{(1+|zw|)^2-E} \\ \Leftrightarrow && (1+|zw|^2)|z-w|^2 -E|z-w|^2 &= (|z|+|w|)^2|1-zw^*|^2-E|1-zw^*|^2\\ \Leftrightarrow && (1+|zw|^2)|z-w|^2-(|z|+|w|)^2|1-zw^*|^2 &= E(|z-w|^2-|1-zw^*|^2)\\ &&&= E(|z|^2-zw^*-z^*w+|w|^2-1+zw^*+z^*w-|z|^2|w|^2) \\ &&&= E(|z|^2+|w|^2-1-|z|^2|w|^2) \\ &&&= -E(1-|z|^2)(1-|w|^2) \\ &&&\leq 0 \\ \Leftrightarrow&& (1+|zw|^2)|z-w|^2& \leq (|z|+|w|)^2|1-zw^*|^2\\ \Leftrightarrow&& \frac{|z-w|^2}{|1-zw^*|^2} &\leq \frac{(|z|+|w|)^2}{(1+|zw|)^2}\\ \Leftrightarrow && \frac{|z-w|}{|1-zw^*|} &\leq \frac{(|z|+|w|)}{(1+|zw|)}\\ \end{align*} Yes, this inequality holds if \(|z|, |w|\) are the same side of \(1\) and is reversed otherwise.
2012 Paper 3 Q6
D: 1700.0 B: 1516.0

Let \(x+{\rm i} y\) be a root of the quadratic equation \(z^2 + pz +1=0\), where \(p\) is a real number. Show that \(x^2-y^2 +px+1=0\) and \((2x+p)y=0\). Show further that either \(p=-2x\) or \(p=-(x^2+1)/x\) with \(x\ne0\). Hence show that the set of points in the Argand diagram that can (as \(p\) varies) represent roots of the quadratic equation consists of the real axis with one point missing and a circle. This set of points is called the root locus of the quadratic equation. Obtain and sketch in the Argand diagram the root locus of the equation \[ pz^2 +z+1=0\, \] and the root locus of the equation \[ pz^2 + p^2z +2=0\,.\]

Show Solution
\begin{align*} && 0 &= z^2 + pz + 1\\ &&&= (x+iy)^2 + (x+iy)p + 1 \\ &&& = (x^2-y^2+px+1) + (2xy+py)i \\ \Rightarrow && 0 &= x^2 - y^2 + px + 1 \\ && 0 &= (2x+p)y \\ \Rightarrow && p &= -2x \\ \text{ or } && y &= 0 \\ \Rightarrow && p &= -(x^2+1)/x \end{align*} Therefore as \(p\) varies with either have \(y = 0\) and \(x\) taking any real value except \(0\) ie the real axis minus the origin. Or \(p = -2x\) and \(-y^2-x^2+1 = 0 \Rightarrow x^2 + y^2 = 1\) which is a circle. Suppose \(pz^2 + z + 1 = 0\) \begin{align*} && 0 &= pz^2 + z +1\\ &&&= p(x+iy)^2 + (x+iy) + 1\\ &&&= (px^2-py^2+x+1) + (2xyp + y) i \\ \Rightarrow && 0 &= (2xp+1)y \\ \Rightarrow && y & = 0, p = \frac{-(x+1)}{x^2}, x \neq 0 \\ \text{ or } && p &= -\frac{1}{2x}\\ \Rightarrow && 0 &= -\frac{1}{2}x + \frac{y^2}{2x} + x + 1 \\ &&&= \frac{y^2 - x^2 +2x^2 + 2x}{2x} \\ &&&= \frac{(x+1)^2+y^2-1}{2x} \end{align*} So we either have the real axis (except \(0\)) or a circle radius \(1\) centre \((-1, 0)\) (excluding \(x = 0\)).
TikZ diagram
Suppose \(pz^2 + p^2 z + 2 = 0\) then \begin{align*} && 0 &= p(x+iy)^2 + p^2(x+iy) + 2 \\ &&&= (p(x^2-y^2) + p^2x + 2) + (2xyp + p^2y)i \\ \Rightarrow && 0 &= py(2x+p) \\ \Rightarrow && y &= 0, \Delta = x^4-8x \\ \Rightarrow && x &\in (-\infty, 0) \cup [2, \infty) \\ \text{ or } && p &= -2x \\ && 0 &= (-2x)(x^2-y^2) + 4x^3+2 \\ &&&= 2x^3+2xy^2+2 \\ \Rightarrow && 0 &= x^3+xy^2+1 \end{align*}
TikZ diagram
2011 Paper 3 Q8
D: 1700.0 B: 1500.0

The complex numbers \(z\) and \(w\) are related by \[ w= \frac{1+\mathrm{i}z}{\mathrm{i}+z}\,. \] Let \(z=x+\mathrm{i}y\) and \(w=u+\mathrm{i}v\), where \(x\), \(y\), \(u\) and \(v\) are real. Express \(u\) and \(v\) in terms of \(x\) and \(y\).

  1. By setting \(x=\tan(\theta/2)\), or otherwise, show that if the locus of \(z\) is the real axis \(y=0\), \(-\infty < x < \infty\), then the locus of \(w\) is the circle \(u^2+v^2=1\) with one point omitted.
  2. Find the locus of \(w\) when the locus of \(z\) is the line segment \(y=0\), \(-1 < x < 1\,\).
  3. Find the locus of \(w\) when the locus of \(z\) is the line segment \(x=0\), \(-1 < y < 1\,\).
  4. Find the locus of \(w\) when the locus of \(z\) is the line \(y=1\), \(-\infty < x < \infty\,\).

Show Solution
\begin{align*} w &= \frac{1+iz}{i+z} \\ &= \frac{1-y+ix}{x+i(1+y)} \\ &= \frac{((1-y)+ix)(x-i(1+y))}{x^2+(1+y)^2} \\ &= \frac{x(1-y)+x(1+y)}{x^2+(1+y)^2}+\frac{x^2+y^2-1}{x^2+(1+y)^2}i \\ &= \frac{2x}{x^2+(y+1)^2}+\frac{x^2+y^2-1}{x^2+(1+y)^2}i \end{align*} Therefore \(u = \frac{2x}{x^2+(y+1)^2}, v = \frac{x^2+y^2-1}{x^2+(1+y)^2}\)
  1. Suppose \(z = \tan(\theta/2) = t\) then \(u = \frac{2t}{t^2+1} = \sin \theta, v = \frac{t^2-1}{t^2+1} = \cos \theta\), ie \(u+iv\) is the unit circle, where \(-\frac{\pi}{2} < \theta/2 < \frac{\pi}{2}\) or \(-\pi < \theta < \pi\), ie excluding the point \((\sin \pi, \cos \pi) = (0,1)\).
  2. When \(-1 < x < 1\) we have \(-\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\) ie \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), ie the lower half of the unit circle.
  3. When \(x = 0, -1 < y < 1\) we have \(u = 0, v = \frac{y^2-1}{(1+y)^2}\) which is the negative imaginary axis.
  4. We have \(u = \frac{2t}{t^2+4}, v = \frac{t^2}{t^2+4}\), ie \(u^2 + v^2 = v\), ie \(u^2+(v-\frac12)^2 = \frac12^2\), so a circle centre \(\frac12i\) radius \(\frac12\), missing out \((0,1)\)
2008 Paper 3 Q7
D: 1700.0 B: 1500.0

The points \(A\), \(B\) and \(C\) in the Argand diagram are the vertices of an equilateral triangle described anticlockwise. Show that the complex numbers \(a\), \(b\) and \(c\) representing \(A\), \(B\) and \(C\) satisfy \[2c= (a+b) +\mathrm{i}\sqrt3(b-a).\] Find a similar relation in the case that \(A\), \(B\) and \(C\) are the vertices of an equilateral triangle described clockwise.

  1. The quadrilateral \(DEFG\) lies in the Argand diagram. Show that points \(P\), \(Q\), \(R\) and \(S\) can be chosen so that \(PDE\), \(QEF\), \(RFG\) and \(SGD\) are equilateral triangles and \(PQRS\) is a parallelogram.
  2. The triangle \(LMN\) lies in the Argand diagram. Show that the centroids \(U\), \(V\) and \(W\) of the equilateral triangles drawn externally on the sides of \(LMN\) are the vertices of an equilateral triangle. \noindent [{\bf Note:} The {\em centroid} of a triangle with vertices represented by the complex numbers \(x\),~\(y\) and~\(z\) is the point represented by \(\frac13(x+y+z)\,\).]

2007 Paper 3 Q6
D: 1700.0 B: 1472.0

The distinct points \(P\), \(Q\), \(R\) and \(S\) in the Argand diagram lie on a circle of radius \(a\) centred at the origin and are represented by the complex numbers \(p\), \(q\), \(r\) and \(s\), respectively. Show that \[ pq = -a^2 \frac {p-q}{p^*-q^*}\,. \] Deduce that, if the chords \(PQ\) and \(RS\) are perpendicular, then \(pq+rs=0\). The distinct points \(A_1\), \(A_2\), \(\ldots\), \(A_n\) (where \(n\ge3\)) lie on a circle. The points \hbox{\(B_1\), \(B_2\), \(\ldots\), \(B_{n}\)} lie on the same circle and are chosen so that the chords \(B_1B_2\), \(B_2B_3\), \(\ldots\), \(B_nB_{1}\) are perpendicular, respectively, to the chords \(A_1A_2\), \(A_2A_3\), \(\ldots\), \(A_nA_1\). Show that, for \(n=3\), there are only two choices of \(B_1\) for which this is possible. What is the corresponding result for \(n=4\)? State the corresponding results for values of \(n\) greater than 4.

2006 Paper 3 Q5
D: 1700.0 B: 1516.0

Show that the distinct complex numbers \(\alpha\), \(\beta\) and \(\gamma\) represent the vertices of an equilateral triangle (in clockwise or anti-clockwise order) if and only if \[ \alpha^2 + \beta^2 +\gamma^2 -\beta\gamma - \gamma \alpha -\alpha\beta =0\,. \] Show that the roots of the equation \begin{equation*} z^3 +az^2 +bz +c=0 \tag{\(*\)} \end{equation*} represent the vertices of an equilateral triangle if and only if \(a^2=3b\). Under the transformation \(z=pw+q\), where \(p\) and \(q\) are given complex numbers with \(p\ne0\), the equation (\(*\)) becomes \[ w^3 +Aw^2 +Bw +C=0\,. \tag{\(**\)} \] Show that if the roots of equation \((*)\) represent the vertices of an equilateral triangle, then the roots of equation \((**)\) also represent the vertices of an equilateral triangle.

Show Solution
The complex numbers represent an equilateral triangle iff \(\gamma\) is a \(\pm 60^\circ\) rotation of \(\beta\) around \(\alpha\), ie \begin{align*} && \gamma - \alpha &= \omega(\beta - \alpha) \\ \Leftrightarrow && \omega &= \frac{\gamma - \alpha}{\beta - \alpha} \\ \Leftrightarrow && -1 &= \left (\frac{\gamma - \alpha}{\beta - \alpha} \right)^3 \\ \Leftrightarrow && -(\beta - \alpha)^3 &=(\gamma - \alpha)^3 \\ \Leftrightarrow && 0 &= (\gamma-\alpha)^3+(\beta-\alpha)^3 \\ &&&= \gamma^3-3\gamma^2\alpha +3\gamma\alpha^2-\alpha^3 +\beta^3-3\beta^2\alpha+3\beta\alpha^2-\alpha^3 \\ &&&= (\beta + \gamma - 2\alpha)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta) \\ \Leftrightarrow && 0 &= \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta \end{align*} The roots of the equation \(z^3+az^2+bz+c = 0\) represents the vertices of an equilateral triangle iff \(a^2-3b = (\alpha+\beta+\gamma^2) - 3(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta = 0\) as erquired. Suppose \(a^2 = 3b\), then consider \(z = pw +q\), we must have \begin{align*} && 0 &= (pw+q)^3+a(pw+q)^2 + b(pw+q)+c \\ &&&= p^3w^3 +(3p^2q+ap^2)w^2+(3pq^2+2apq+bp)w+(q^3+aq^2+bq+c) \\ &&&= p^3w^3+p^2(3q+a)w^2+p(3q^2+2aq+b)w+(q^3+aq^2+bq+c) \\ \end{align*} We need to check if \(\left(\frac{3q+a}{p} \right)^2 = 3 \left (\frac{3q^2+2qa+b}{p^2} \right)\). Clearly the denominators match, so consider the numerators \begin{align*} && (3q+a)^2 &= 9q^2+6aq+a^2 \\ &&&= 9q^2+6aq+3b \\ &&&= 3(3q^2+2qa+b) \end{align*} as required
2005 Paper 3 Q8
D: 1700.0 B: 1484.0

In this question, \(a\) and \(c\) are distinct non-zero complex numbers. The complex conjugate of any complex number \(z\) is denoted by \(z^*\). Show that \[ |a - c|^2 = aa^* + cc^* -ac^* - ca^* \] and hence prove that the triangle \(OAC\) in the Argand diagram, whose vertices are represented by \(0\), \(a\) and \(c\) respectively, is right angled at \(A\) if and only if \(2aa^* = ac^*+ca^*\,\). Points \(P\) and \(P'\) in the Argand diagram are represented by the complex numbers \(ab\) and \(\ds \frac{a}{b^*}\,\), where \(b\) is a non-zero complex number. A circle in the Argand diagram has centre \(C\) and passes through the point \(A\), and is such that \(OA\) is a tangent to the circle. Show that the point \(P\) lies on the circle if and only if the point \(P'\) lies on the circle. Conversely, show that if the points represented by the complex numbers \(ab\) and \(\ds \frac{a}{b^*}\), for some non-zero complex number \(b\) with \(bb^* \ne 1\,\), both lie on a circle centre \(C\) in the Argand diagram which passes through \(A\), then \(OA\) is a tangent to the circle.

2002 Paper 3 Q8
D: 1700.0 B: 1469.7

Four complex numbers \(u_1\), \(u_2\), \(u_3\) and \(u_4\) have unit modulus, and arguments \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\), respectively, with \(-\pi < \theta_1 < \theta_2 < \theta_3 < \theta_4 < \pi\). Show that \[ \arg \l u_1 - u_2 \r = \tfrac{1}{2} \l \theta_1 + \theta_2 -\pi \r + 2n\pi \] where \(n = 0 \hspace{4 pt} \mbox{or} \hspace{4 pt} 1\,\). Deduce that \[ \arg \l \l u_1 - u_2 \r \l u_4 - u_3 \r \r = \arg \l \l u_1 - u_4 \r \l u_3 - u_2 \r \r + 2n\pi \] for some integer \(n\). Prove that \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]

2001 Paper 2 Q7
D: 1600.0 B: 1470.8

In an Argand diagram, \(O\) is the origin and \(P\) is the point \(2+0\mathrm{i}\). The points \(Q\), \(R\) and \(S\) are such that the lengths \(OP\), \(PQ\), \(QR\) and \(RS\) are all equal, and the angles \(OPQ\), \(PQR\) and \(QRS\) are all equal to \({5{\pi}}/6\), so that the points \(O\), \(P\), \(Q\), \(R\) and \(S\) are five vertices of a regular 12-sided polygon lying in the upper half of the Argand diagram. Show that \(Q\) is the point \(2 + \sqrt 3 + \mathrm{i}\) and find \(S\). The point \(C\) is the centre of the circle that passes through the points \(O\), \(P\) and \(Q\). Show that, if the polygon is rotated anticlockwise about \(O\) until \(C\) first lies on the real axis, the new position of \(S\) is $$ - \tfrac{1}{2} (3\sqrt 2+ \sqrt6)(\sqrt3-\mathrm{i})\;. $$

2001 Paper 3 Q8
D: 1700.0 B: 1500.0

  1. Prove that the equations $$ \left|z - (1 + \mathrm{i}) \right|^2 = 2 \eqno(*) $$ and $$ \qquad \quad \ \left|z - (1 - \mathrm{i}) \right|^2 = 2 \left|z - 1 \right|^2 $$ describe the same locus in the complex \(z\)--plane. Sketch this locus.
  2. Prove that the equation $$ \arg \l {z - 2 \over z} \r = {\pi \over 4} \eqno(**) $$ describes part of this same locus, and show on your sketch which part.
  3. The complex number \(w\) is related to \(z\) by \[ w = {2 \over z}\;. \] Determine the locus produced in the complex \(w\)--plane if \(z\) satisfies \((*)\). Sketch this locus and indicate the part of this locus that corresponds to \((**)\).

1998 Paper 1 Q5
D: 1484.0 B: 1529.9

  1. In the Argand diagram, the points \(Q\) and \(A\) represent the complex numbers \(4+6i\) and \(10+2i\). If \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) are the vertices, taken in clockwise order, of a regular hexagon (regular six-sided polygon) with centre \(Q\), find the complex number which represents \(B\).
  2. Let \(a\), \(b\) and \(c\) be real numbers. Find a condition of the form \(Aa+Bb+Cc=0\), where \(A\), \(B\) and \(C\) are integers, which ensures that \[\frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i}\] is real.

Show Solution
  1. TikZ diagram
    We are looking for \((10+2i) - (4+6i) = 6 - 4i\) rotated by \(\frac{\pi}{3}\) and then added to \(4+6i\), which is \begin{align*} (6-4i)(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) &= (6-4i)\left(\tfrac12 +\tfrac{\sqrt{3}}2i\right) \\ &= 3+2\sqrt{3} + (3\sqrt{3}-2)i \end{align*}
  2. \begin{align*} &&& \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &\in \mathbb{R} \\ \Longleftrightarrow && \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &= \frac{a}{1-i}+\frac{b}{1-2i}+\frac{c}{1-3i} \\ && 0 &= a\left ( \frac{1}{1+i} - \frac{1}{1-i} \right)+ b\left ( \frac{1}{1+2i} - \frac{1}{1-2i} \right)+ c\left ( \frac{1}{1+3i} - \frac{1}{1-3i} \right) \\ &&&= a\left ( \frac{(1-i)-(1+i)}{1^2+1^2} \right) + b\left ( \frac{(1-2i)-(1+2i)}{1^2+2^2} \right) + c\left ( \frac{(1-3i)-(1+3i)}{1^2+3^2} \right) \\ &&&= -\frac{2i}{2}a-\frac{4i}{5}b-\frac{-6i}{10}c \\ \Longleftrightarrow && 0 &= a+\tfrac45b+\tfrac35c \end{align*}
1998 Paper 2 Q5
D: 1600.0 B: 1470.9

Define the modulus of a complex number \(z\) and give the geometric interpretation of \(\vert\,z_1-z_2\,\vert\) for two complex numbers \(z_1\) and \(z_2\). On the basis of this interpretation establish the inequality $$\vert\,z_1+z_2\,\vert\le \vert\,z_1\,\vert+\vert\,z_2\,\vert.$$ Use this result to prove, by induction, the corresponding inequality for \(\vert\,z_1+\cdots+z_n\,\vert\). The complex numbers \(a_1,\,a_2,\,\ldots,\,a_n\) satisfy \(|a_i|\le 3\) (\(i=1, 2, \ldots , n\)). Prove that the equation $$a_1z+a_2z^2\cdots +a_nz^n=1$$ has no solution \(z\) with \(\vert\,z\,\vert\le 1/4\).

Show Solution
Suppose \(z = a+ib\), where \(a,b \in \mathbb{R}\) then the modulus of \(z\), \(|z| = \sqrt{a^2+b^2}\). Noting the similarity to the Pythagorean theorem, we can say that \(|z_1 - z_2|\) is the distance between \(z_1\) and \(z_2\) in the Argand diagram. \begin{align*} |z_1 + z_2| &= |(z_1 - 0) + (0 -z_2)| \\ &\underbrace{\leq}_{\text{the direct distance is shorter than going via }0} |z_1 - 0| + |0 - z_2| \\ &= |z_1| + |-z_2| \\ &= |z_1| + |z_2| \end{align*} Claim: \(\displaystyle \vert\,z_1+\cdots+z_n\,\vert \leq \sum_{i=1}^n |z_i|\) Proof: (By Induction) Base Case: \(n = 1, 2\) have been proven. Inductive step, suppose it is true for \(n = k\), then consider \(n = k+1\), ie \begin{align*} \vert\,z_1+\cdots+z_k+z_{k+1}\,\vert &\leq \vert\,z_1+\cdots+z_k\vert + \vert z_{k+1}\,\vert \\ &\underbrace{\leq}_{\text{inductive hypothesis}} \sum_{i=1}^k |z_i| + |z_{k+1}| \\ &= \sum_{i=1}^{k+1} |z_i| \end{align*} Therefore if our hypothesis is true for \(n = k\) it is true for \(n = k+1\), and so since it is true for \(n = 1\) it is true by the principle of mathematical induction for all integers \(n \geq 1\). Suppose \(|z| \leq 1/4\), then consider: \begin{align*} \vert a_1z+a_2z^2+\cdots +a_nz^n \vert &\leq \vert a_1 z\vert + \vert a_2z^2\vert + \cdots + \vert a_n z_n\ \vert \\ &= \vert a_1\vert\vert z\vert + \vert a_2\vert\vert z^2\vert + \cdots + \vert a_n\vert\vert z^n\ \vert \\ &\leq 3\left ( |z| + |z|^2 + \cdots + |z|^n \right) \\ &\leq 3 \left ( \frac{1}{4} + \frac1{4^2} + \cdots + \frac{1}{4^n} \right) \\ &< 3 \frac{1/4}{1-1/4} \\ &= 1 \end{align*} Therefore we cannot have equality and there are no solutions.
1997 Paper 2 Q5
D: 1600.0 B: 1484.0

The complex numbers \(w=u+\mathrm{i}v\) and \(z=x+\mathrm{i}y\) are related by the equation $$z= (\cos v+\mathrm{i}\sin v)\mathrm{e}^u.$$ Find all \(w\) which correspond to \(z=\mathrm{i\,e}\). Find the loci in the \(x\)--\(y\) plane corresponding to the lines \(u=\) constant in the \(u\)--\(v\) plane. Find also the loci corresponding to the lines \(v=\) constant. Illustrate your answers with clearly labelled sketches. Identify two subsets \(W_1\) and \(W_2\) of the \(u\)--\(v\) plane each of which is in one-to-one correspondence with the first quadrant \(\{(x,\,y):\,x>0,\,y>0\}\) of the \(x\)--\(y\) plane. Identify also two subsets \(W_3\) and \(W_4\) each of which is in one-to-one correspondence with the set \(\{z\,:0<\,\vert z\vert\,<1\}\). \noindent[{\bf NB} `one-to-one' means here that to each value of \(w\) there is only one corresponding value of \(z\), and vice-versa.]

1996 Paper 1 Q5
D: 1484.0 B: 1500.0

  1. Find all rational numbers \(r\) and \(s\) which satisfy \[ (r+s\sqrt{3})^{2}=4-2\sqrt{3}. \]
  2. Find all real numbers \(p\) and \(q\) which satisfy \[ (p+q\mathrm{i})^{2}=(3-2\sqrt{3})+2(1-\sqrt{3})\mathrm{i}. \]
  3. Solve the equation \[ (1+\mathrm{i})z^{2}-2z+2\sqrt{3}-2=0, \] writing your solutions in as simple a form as possible.
{[}No credit will be given to answers involving use of calculators.{]}

Show Solution
  1. Suppose \begin{align*} && 4 - 2\sqrt{3} &= (r+s\sqrt{3})^2 \\ &&&= r^2+3s^2+2sr \sqrt{3} \\ \Rightarrow && rs &= -1 \\ && r^2+3s^2 &= 4 \\ \Rightarrow && (r,s) &= (1,-1), (-1,1) \end{align*}
  2. \begin{align*} && (3-2\sqrt{3})+2(1-\sqrt{3})i &= (p+qi)^2 \\ &&&= p^2-q^2 + 2pq i \\ \Rightarrow && pq &= (1-\sqrt{3}) \\ && p^2 - q^2 &= 3-2\sqrt{3} \\ \Rightarrow &&3-2\sqrt{3} &= p^2 - \frac{(1-\sqrt{3})^2}{p^2} \\ \Rightarrow && 0 &= p^4-(3-2\sqrt{3})p^2-(4-2\sqrt{3}) \\ &&&= (p^2-(4-2\sqrt{3}))(p^2+1) \\ \Rightarrow && p &= \pm (1-\sqrt{3}) \\ && q &=\mp \frac12(1+\sqrt{3}) \end{align*}
  3. \begin{align*} && 0 &= (1+i)z^2 - 2z + 2(\sqrt{3}-1) \\ \Rightarrow && z &= \frac{2 \pm \sqrt{4-4(1+i)2(\sqrt{3}-1)}}{2(1+i)} \\ &&&= \frac{1 \pm \sqrt{1-(1+i)2(\sqrt{3}-1)}}{1+i} \\ &&&= \frac{1 \pm \sqrt{(3-2\sqrt{3})+(2-2\sqrt{3})i}}{1+i} \\ &&&= \frac{1 \pm (1 - \sqrt{3}) \mp \frac12 (1+\sqrt{3})i}{1+i} \\ &&&= \frac{5-\sqrt{3}}{4} + \frac{3-3\sqrt{3}}{4}i, \\ &&& \frac{\sqrt{3}-1}{4} + \frac{1+3\sqrt{3}}{4}i \end{align*}
1996 Paper 2 Q5
D: 1600.0 B: 1500.0

If $$ z^{4}+z^{3}+z^{2}+z+1=0\tag{*} $$ and \(u=z+z^{-1}\), find the possible values of \(u\). Hence find the possible values of \(z\). [Do not try to simplify your answers.] Show that, if \(z\) satisfies \((*)\), then \[z^{5}-1=0.\] Hence write the solutions of \((*)\) in the form \(z=r(\cos\theta+i\sin\theta)\) for suitable real \(r\) and \(\theta\). Deduce that \[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4} \ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]

Show Solution
\begin{align*} && 0 &= z^4+z^3+z^2+z+1 \\ \Rightarrow && 0 &= z^2+z+1+z^{-1}+z^{-2} \tag{\(z \neq 0\)} \\ &&&= \left ( z+z^{-1} \right)^2-2 + z+z^{-1} + 1 \\ &&&= u^2+u-1 \\ \Rightarrow && u &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && z+z^{-1} &= \frac{-1 \pm \sqrt{5}}{2} \\ \Rightarrow && 0 &= z^2-\left ( \frac{-1 \pm \sqrt{5}}{2}\right)z+1 \\ \Rightarrow && z &= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\left ( \frac{-1 \pm \sqrt{5}}{2}\right)^2-4}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{1+5\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{\left ( \frac{-1 \pm \sqrt{5}}{2}\right) \pm \sqrt{\frac{-10\mp2\sqrt{5}-16}{4}}}{2} \\ &&&= \frac{-1\pm\sqrt{5}}{4} \pm i\frac{\sqrt{10\pm 2\sqrt{5}}}{4} \end{align*} Since \(z^4+z^3+z^2+z+1 = 0\) we can multiply both sides by \(z-1\) to obtain \(z^5-1 = 0\). Therefore if \(z = r(\cos \theta + i \sin \theta)\) we see that \(z^5 = 1 \Rightarrow r^5 (\cos 5 \theta + i \sin 5 \theta) = 1 \Rightarrow r = 1, 5 \theta = 2n \pi\) ie \(z = \cos \frac{2n\pi}{5} + i\sin \frac{2n \pi}{5}\). We are looking for a solution in the first quadrant, therefore \(\cos \frac{2\pi}{5} = \frac{-1 + \sqrt{5}}4\) and \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)
1995 Paper 3 Q6
D: 1700.0 B: 1501.9

The variable non-zero complex number \(z\) is such that \[ \left|z-\mathrm{i}\right|=1. \] Find the modulus of \(z\) when its argument is \(\theta.\) Find also the modulus and argument of \(1/z\) in terms of \(\theta\) and show in an Argand diagram the loci of points which represent \(z\) and \(1/z\). Find the locus \(C\) in the Argand diagram such that \(w\in C\) if, and only if, the real part of \((1/w)\) is \(-1\).

Show Solution
TikZ diagram
\(z\) is a point on the circle shown: Therefore using the cosine rule \(|z|^2 = 1^2 + 1^2 - 2\cdot 1 \cdot 1 \cdot \cos (2 \theta) = 2 -2\cos 2\theta = 2\sin^2 \theta \Rightarrow |z| = \sqrt{2}|\sin \theta|\) \(\frac{1}{z}\) has modulus \(\frac{1}{\sqrt{2}|\sin \theta|}\) and argument \(-\theta\). \(|\frac{1}{z} - i| = 1 \Rightarrow |1-iz| = |z| \Rightarrow |-i-z| = |z|\) ie we're looking for the points on the perpendicular bisector of \(0\) and \(-i\). \(\textrm{Re}\left (\frac{1}{w}\right) = -1 \Rightarrow -1 = \textrm{Re} \left (\frac{1}{a+ib} \right) = \frac{a-ib}{a^2+b^2} = \frac{a}{a^2+b^2} \Rightarrow a^2+b^2 = -a \Rightarrow (a+\tfrac12)^2+b^2 = \tfrac14\) so we are looking at a circle radius \(\tfrac12\) centre \(-\frac12\)
1994 Paper 1 Q6
D: 1500.0 B: 1516.0

The function \(\mathrm{f}\) is defined, for any complex number \(z\), by \[ \mathrm{f}(z)=\frac{\mathrm{i}z-1}{\mathrm{i}z+1}. \] Suppose throughout that \(x\) is a real number.

  1. Show that \[ \mathrm{Re}\,\mathrm{f}(x)=\frac{x^{2}-1}{x^{2}+1}\qquad\mbox{ and }\qquad\mathrm{Im}\,\mathrm{f}(x)=\frac{2x}{x^{2}+1}. \]
  2. Show that \(\mathrm{f}(x)\mathrm{f}(x)^{*}=1,\) where \(\mathrm{f}(x)^{*}\) is the complex conjugate of \(\mathrm{f}(x)\).
  3. Find expressions for \(\mathrm{Re}\,\mathrm{f}(\mathrm{f}(x))\) and \(\mathrm{Im}\,\mathrm{f}(\mathrm{f}(x)).\)
  4. Find \(\mathrm{f}(\mathrm{f}(\mathrm{f}(x))).\)

Show Solution
  1. \begin{align*} && f(x) &= \frac{ix-1}{ix+1} \\ &&&= \frac{ix-1}{ix+1} \frac{1-ix}{1-ix} \\ &&&= \frac{ix-1+x^2+ix}{1^2+x^2} \\ &&&= \frac{x^2-1}{x^2+1} + i \frac{2x}{x^2+1} \\ \Rightarrow && \textrm{Re}(f(x)) &= \frac{x^2-1}{x^2+1} \\ && \textrm{Im}(f(x)) &= \frac{2x}{x^2+1} \end{align*}
  2. \begin{align*} && f(x)f(x)^* &= \frac{ix-1}{ix+1} \frac{(ix-1)^*}{(ix+1)^*} \\ &&&= \frac{ix-1}{ix+1} \frac{-ix-1}{-ix+1} \\ &&&= \frac{ix-1}{ix+1} \frac{-(ix+1)}{-(ix-1)} \\ &&&= 1
  3. \begin{align*} && \begin{pmatrix} i & -1 \\ i & 1 \end{pmatrix}^2 &= \begin{pmatrix} -1-i & -i-1 \\ -1+i & -i+1 \end{pmatrix} \\ \Rightarrow && f(f(z)) &= \frac{-(1+i)(z+1)}{(-1+i)(z-1)} \\ &&&= \frac{2i}{2} \frac{z+1}{z-1} \\ &&&= i \frac{z+1}{z-1} \\ \Rightarrow && \textrm{Re}(f(f(x))) &= 0 \\ && \textrm{Im}(f(f(x))) &= \frac{x+1}{x-1} \end{align*}
  4. \begin{align*} && \begin{pmatrix} i & -1 \\ i & 1 \end{pmatrix} \begin{pmatrix} i & i \\ 1 & -1 \end{pmatrix} &= \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \\ \Rightarrow && f(f(f(z))) &= z \end{align*}
1994 Paper 3 Q6
D: 1700.0 B: 1501.5

The four points \(A,B,C,D\) in the Argand diagram (complex plane) correspond to the complex numbers \(a,b,c,d\) respectively. The point \(P_{1}\) is mapped to \(P_{2}\) by rotating about \(A\) through \(\pi/2\) radians. Then \(P_{2}\) is mapped to \(P_{3}\) by rotating about \(B\) through \(\pi/2\) radians, \(P_{3}\) is mapped to \(P_{4}\) by rotating about \(C\) through \(\pi/2\) radians and \(P_{4}\) is mapped to \(P_{5}\) by rotating about \(D\) through \(\pi/2\) radians, each rotation being in the positive sense. If \(z_{i}\) is the complex number corresponding to \(P_{i},\) find \(z_{5}\) in terms of \(a,b,c,d\) and \(z_{1}.\) Show that \(P_{5}\) will coincide with \(P_{1},\) irrespective of the choice of the latter if, and only if \[a-c=\mathrm{i}(b-d)\] and interpret this condition geometrically. The points \(A,B\) and \(C\) are now chosen to be distinct points on the unit circle and the angle of rotation is changed to \(\theta,\) where \(\theta\neq0,\) on each occasion. Find the necessary and sufficient condition on \(\theta\) and the points \(A,B\) and \(C\) for \(P_{4}\) always to coincide with \(P_{1}.\)

1994 Paper 3 Q8
D: 1700.0 B: 1516.0

Let \(a,b,c,d,p,q,r\) and \(s\) be real numbers. By considering the determinant of the matrix product \[ \begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix}, \] where \(z_{1},z_{2},z_{3}\) and \(z_{4}\) are suitably chosen complex numbers, find expressions \(L_{1},L_{2},L_{3}\) and \(L_{4},\) each of which is linear in \(a,b,c\) and \(d\) and also linear in \(p,q,r\) and \(s,\) such that \[ (a^{2}+b^{2}+c^{2}+d^{2})(p^{2}+q^{2}+r^{2}+s^{2})=L_{1}^{2}+L_{2}^{2}+L_{3}^{2}+L_{4}^{2}. \]

Show Solution
Supppose \(z_1 = a+ib, z_2 = c+id, z_3 = p+iq, z_4 = r+is\) then: \begin{align*} && \det \left (\begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix} \right) &= \det \begin{pmatrix}z_{1} & z_{2}\\ -z_{2}^{*} & z_{1}^{*} \end{pmatrix}\det\begin{pmatrix}z_{3} & z_{4}\\ -z_{4}^{*} & z_{3}^{*} \end{pmatrix} \\ && \det \begin{pmatrix}z_{1}z_3-z_2z_4^* & z_1z_4+z_2z_3^*\\ -z_2^*z_3-z_1^*z_4*& -z_2^*z_4+z_{1}^*z_3^* \end{pmatrix}&= (z_1z_1^*+z_2z_2^*)(z_3z_3^*+z_4z_4^*) \\ && |z_{1}z_3-z_2z_4^*|^2+|z_1z_4+z_2z_3^*|^2&= (a^2+b^2+c^2+d^2)(p^2+q^2+r^2+s^2) \\ && L_1^2 + L_2^2+L_3^2+L_4^2 &= \ldots \end{align*}