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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.
  1. Solve the equation \[z^3 + |z + 2|^2 = 4.\]


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: \(z+b < 0\) \(z^2 - z - b = \frac12 \Rightarrow z = \frac{1 \pm \sqrt{1+4b-2}}{2}\)
  1. 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:

2025 Paper 3 Q6
D: 1500.0 B: 1500.0

  1. Let \(a\), \(b\) and \(c\) be three non-zero complex numbers with the properties \(a + b + c = 0\) and \(a^2 + b^2 + c^2 = 0\). Show that \(a\), \(b\) and \(c\) cannot all be real. Show further that \(a\), \(b\) and \(c\) all have the same modulus.
  2. Show that it is not possible to find three non-zero complex numbers \(a\), \(b\) and \(c\) with the properties \(a + b + c = 0\) and \(a^3 + b^3 + c^3 = 0\).
  3. Show that if any four non-zero complex numbers \(a\), \(b\), \(c\) and \(d\) have the properties \(a + b + c + d = 0\) and \(a^3 + b^3 + c^3 + d^3 = 0\), then at least two of them must have the same modulus.
  4. Show, by taking \(c = 1\), \(d = -2\) and \(e = 3\) that it is possible to find five real numbers \(a\), \(b\), \(c\), \(d\) and \(e\) with distinct magnitudes and with the properties \(a + b + c + d + e = 0\) and \(a^3 + b^3 + c^3 + d^3 + e^3 = 0\).


Solution:

  1. If \(a,b,c\) were all real then \(a^2+b^2+c^2 = 0 \Rightarrow a,b,c = 0\) but they are non-zero. Therefore they cannot all be real. Since \((a+b+c)^2 = 0\) we must have \(ab+bc+ca = 0\). Therefore \(a,b,c\) must satisfy \(x^3 -abc = 0 \Rightarrow\) they all have the same modulus, since they are all cube roots of the same number.
  2. Notice that \(a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab-bc-ca) \Rightarrow abc = 0\) but therefore they cannot all be non-zero.
  3. Suppose \(a+b+c+d = 0\) then note that \(\displaystyle a^2+b^2+c^2+d^2 = (a+b+c+d)^2 - 2\sum_{sym} ab\) and \(\displaystyle a^3+b^3+c^3+d^3 = (a+b+c+d)^3 - 3(a+b+c+d)(ab+ac+ad+bc+bd+cd) + 3(abc+abd+acd+bcd) \Rightarrow abc+abd+acd+bcd = 0\). Therefore \(a,b,c,d\) are roots of a polynomial of the form \(x^4 -kx^2 + l = 0\), but this means they must come in pairs with the same modulus.
  4. Suppose \(c = 1, d = -2, e = 3\) so \(c+d+e = 2\) and \(c^3 + d^3 + e^3 = 1 - 8 + 27 = 20\), so we need to find \(a,b\) satisfying \(a+b = -2, a^2+b^2 = -20\), ie \(4 = (a+b)^2 = -20 + 2ab \Rightarrow ab = 12\), so we need the roots of \(x^2 +2x + 12= 0\) which clearly have different modulus.

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 2 Q7
D: 1500.0 B: 1500.0

Let \(\mathrm{h}(z) = nz^6 + z^5 + z + n\), where \(z\) is a complex number and \(n \geqslant 2\) is an integer.

  1. Let \(w\) be a root of the equation \(\mathrm{h}(z) = 0\).
    1. Show that \(|w^5| = \sqrt{\dfrac{\mathrm{f}(w)}{\mathrm{g}(w)}}\), where \[\mathrm{f}(z) = n^2 + 2n\operatorname{Re}(z) + |z|^2 \quad \text{and} \quad \mathrm{g}(z) = n^2|z|^2 + 2n\operatorname{Re}(z) + 1.\]
    2. By considering \(\mathrm{f}(w) - \mathrm{g}(w)\), prove by contradiction that \(|w| \geqslant 1\).
    3. Show that \(|w| = 1\).
  2. It is given that the equation \(\mathrm{h}(z) = 0\) has six distinct roots, none of which is purely real.
    1. Show that \(\mathrm{h}(z)\) can be written in the form \[\mathrm{h}(z) = n(z^2 - a_1 z + 1)(z^2 - a_2 z + 1)(z^2 - a_3 z + 1),\] where \(a_1\), \(a_2\) and \(a_3\) are real constants.
    2. Find \(a_1 + a_2 + a_3\) in terms of \(n\).
    3. By considering the coefficient of \(z^3\) in \(\mathrm{h}(z)\), find \(a_1 a_2 a_3\) in terms of \(n\).
    4. How many of the six roots of the equation \(\mathrm{h}(z) = 0\) have a negative real part? Justify your answer.

2020 Paper 2 Q7
D: 1500.0 B: 1500.0

In this question, \(w = \dfrac{2}{z-2}\).

  1. Let \(z\) be the complex number \(3 + t\mathrm{i}\), where \(t \in \mathbb{R}\). Show that \(|w - 1|\) is independent of \(t\). Hence show that, if \(z\) is a complex number on the line \(\operatorname{Re}(z) = 3\) in the Argand diagram, then \(w\) lies on a circle in the Argand diagram with centre \(1\). Let \(V\) be the line \(\operatorname{Re}(z) = p\), where \(p\) is a real constant not equal to \(2\). Show that, if \(z\) lies on \(V\), then \(w\) lies on a circle whose centre and radius you should give in terms of \(p\). For which \(z\) on \(V\) is \(\operatorname{Im}(w) > 0\)?
  2. Let \(H\) be the line \(\operatorname{Im}(z) = q\), where \(q\) is a non-zero real constant. Show that, if \(z\) lies on \(H\), then \(w\) lies on a circle whose centre and radius you should give in terms of \(q\). For which \(z\) on \(H\) is \(\operatorname{Re}(w) > 0\)?

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?


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

2016 Paper 3 Q7
D: 1700.0 B: 1516.0

Let \(\omega = \e^{2\pi {\rm i}/n}\), where \(n\) is a positive integer. Show that, for any complex number \(z\), \[ (z-1)(z-\omega) \cdots (z - \omega^{n-1}) = z^n -1\,. \] The points \(X_0, X_1, \ldots\, X_{n-1}\) lie on a circle with centre \(O\) and radius 1, and are the vertices of a regular polygon.

  1. The point \(P\) is equidistant from \(X_0\) and \(X_1\). Show that, if \(n\) is even, \[ |PX_0| \times |PX_1 |\times \,\cdots\, \times |PX_{n-1}| = |OP|^n +1\, ,\] where \(|PX_ k|\) denotes the distance from \(P\) to \(X_k\). Give the corresponding result when \(n\) is odd. (There are two cases to consider.)
  2. Show that \[ |X_0 X_1|\times |X_0 X_2|\times \,\cdots\, \times |X_0 X_{n-1}| =n\,. \]


Solution: Notice that \(\omega^n = e^{2\pi i} = 1\), so \(\omega\) is a root of \(z^n - 1\), notice also that \((\omega^k)^n =1\) so therefore the \(n\) roots are \(1, \omega, \omega^2, \cdots, \omega^{n-1}\) and so \((z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)\). By considering the coefficient of \(z^n\) we can see that \(C = 1\).

  1. \(P\) lies on the perpendicular bisect of \(1\) and \(\omega\), so \(p = re^{\pi i/n}\), where \(r\) can be positive or negative, but \(|r| = |OP|\). \begin{align*} && |PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| &= |(p-1)(p-\omega) \cdots (p-\omega^{n-1})| \\ &&&= |p^n - 1| \\ &&&= |r^ne^{\pi i} - 1| \\ &&&= |-|OP|^n - 1| \tag{since \(n\) even} \\ &&&= |OP|^n+1 \end{align*} If \(n\) is odd, depending on the sign of \(r\) we get \(|OP|^n+1\) or \(||OP|^n-1|\).
  2. \(\,\) \begin{align*} && (z-\omega) \cdots(z-\omega^{n-1}) &= \frac{z^n-1}{z-1} \\ &&&= 1 + z +\cdots + z^{n-1} \\ && |X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| &= |(1 - \omega)\cdots(1-\omega^{n-1})| \\ &&&= 1+1+1^2+\cdots + 1^{n-1} \\ &&&= n \end{align*}

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 \(\vert w \vert <1\)?


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.

2009 Paper 3 Q6
D: 1700.0 B: 1473.1

Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.

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 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 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 \(\vert\,z\,\vert\le 1/4\).


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.

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 \(-1\).


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 \(\mathrm{f}(\mathrm{f}(\mathrm{f}(x))).\)


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*}