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2019 Paper 3 Q6
D: 1500.0 B: 1500.0
complex numbers Argand diagram circle locus modulus conjugate Möbius transformation geometric proof

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

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2018 Paper 3 Q6
D: 1700.0 B: 1516.0
complex numbers Argand diagram collinearity unit circle conjugate geometric proof intersection of lines real and imaginary parts

  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 \[ p(q+q^*) = 2 \,. \]

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*}

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2013 Paper 3 Q6
D: 1700.0 B: 1500.0
complex numbers triangle inequality modulus conjugate Argand diagram inequality algebraic manipulation

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.

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2007 Paper 3 Q6
D: 1700.0 B: 1472.0
complex numbers Argand diagram circle perpendicular chords conjugate geometric proof modulus-argument form

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.

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1994 Paper 1 Q6
D: 1500.0 B: 1516.0
complex numbers real and imaginary parts modulus conjugate function composition Möbius transformation algebraic manipulation

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*}

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1993 Paper 3 Q6
D: 1700.0 B: 1484.0
complex numbers Argand diagram circle locus conjugate Mobius transformation geometric proof

The point in the Argand diagram representing the complex number \(z\) lies on the circle with centre \(K\) and radius \(r\), where \(K\) represents the complex number \(k\). Show that $$ zz^* -kz^* -k^*z +kk^* -r^2 =0. $$ The points \(P\), \(Q_1\) and \(Q_2\) represent the complex numbers \(z\), \(w_1\) and \(w_2\) respectively. The point \(P\) lies on the circle with \(OA\) as diameter, where \(O\) and \(A\) represent \(0\) and \(2i\) respectively. Given that \(w_1=z/(z-1)\), find the equation of the locus \(L\) of \(Q_1\) in terms of \(w_1\) and describe the geometrical form of \(L\). Given that \(w_2=z^*\), show that the locus of \(Q_2\) is also \(L\). Determine the positions of \(P\) for which \(Q_1\) coincides with \(Q_2\).

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