78 problems found
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.
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
In this question, you may use without proof the results \[ 4 \cosh^3 y - 3 \cosh y = \cosh (3y) \ \ \ \ \text{and} \ \ \ \ \mathrm{arcosh} \, y = \ln ( y+\sqrt{y^2-1}). \] \noindent[ {\bf Note: } \(\mathrm{arcosh}y\) is another notation for \(\cosh^{-1}y\,\)] Show that the equation \(x^3 - 3a^2x = 2a^3 \cosh T\) is satisfied by \( 2a \cosh \l \frac13 T \r\) and hence that, if \(c^2\ge b^3>0\), one of the roots of the equation \(x^3-3bx=2c\) is \(\ds u+\frac{b}{u}\), where \(u = (c+\sqrt{c^2-b^3})^{\frac13}\;\). Show that the other two roots of the equation \(x^3-3bx=2c\) are the roots of the quadratic equation \[\ds x^2 + \Big( u+\frac{b}{u}\Big) x + u^2+\frac{b^2}{u^2}-b=0\, ,\] and find these roots in terms of \(u\), \(b\) and \(\omega\), where \(\omega = \frac{1}{2}(-1 + \mathrm{i}\sqrt{3})\). Solve completely the equation \(x^3-6x=6\,\).
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.
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 \[ | \l u_1 - u_2 \r \l u_4 - u_3 \r | + | \l u_1 - u_4 \r \l u_3 - u_2 \r | = | \l u_1 - u_3 \r \l u_4 - u_2 \r |\;. \]
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 $$ - \tfrac{1}{2} (3\sqrt 2+ \sqrt6)(\sqrt3-\mathrm{i})\;. $$
Prove that \[ (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) = \cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi) \] and that, for every positive integer \(n\), $$ {(\cos {\theta} + \mathrm{i}\sin {\theta})}^n = \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}. $$ By considering \((5-\mathrm{i})^2(1+\mathrm{i})\), or otherwise, prove that \[ \arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,. \] Prove also that \[ 3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,. \] [Note that \(\arctan\theta\) is another notation for \(\tan^{-1}\theta\).]
Solution: \begin{align*} && LHS &= (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) \\ &&&= \cos \theta \cos \phi - \sin \theta \sin \phi + \mathrm{i}(\sin \theta \cos \phi + \cos \theta \sin \phi) \\ &&&= \cos (\theta + \phi) + \mathrm{i} \sin (\theta + \phi) \\ &&&= RHS \end{align*} Therefore we can see \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta\). \begin{align*} && (5-i)^2(1+i) &= (24-10i)(1+i) \\ &&&= (24+10) + i(24-10) \\ &&&= 34+14i \\ \Rightarrow && 2\arg(5-i) +\arg(1+i) &= \arg(34+14i) \\ \Rightarrow && 2\arctan\left (-\frac{1}{5} \right) + \frac{\pi}{4} &= \arctan \left ( \frac{7}{17} \right) \\ \Rightarrow && 2\arctan\left (\frac{1}{5} \right) +\arctan \left ( \frac{7}{17} \right) &= \frac{\pi}{4} \\ \end{align*} Consider \((1+i)(4-i)^3(20-i)\) \begin{align*} && (1+i)(4-i)^3(20-i) &= (21+19i)(52-47i) \\ &&&= 1985+i \\ \Rightarrow && \frac{\pi}{4} - 3 \arctan \left ( \frac{1}{4} \right) -\arctan \left ( \frac{1}{20} \right) &= \arctan \left ( \frac{1}{1985} \right) \end{align*}
Given that \(\alpha = \e^{\mathrm{i} \pi/3}\) , prove that \(1 + \alpha^2 = \alpha\). A triangle in the Argand plane has vertices \(A\), \(B\), and \(C\) represented by the complex numbers \(p\), \(q\alpha^2\) and \(- r\alpha\) respectively, where \(p\), \(q\) and \(r\) are positive real numbers. Sketch the triangle~\(ABC\). Three equilateral triangles \(ABL\), \(BCM\) and \(CAN\) (each lettered clockwise) are erected on sides \(AB\), \(BC\) and \(CA\) respectively. Show that the complex number representing \(N\) is \mbox{\(( 1 - \alpha) p- \alpha^2 r\)} and find similar expressions for the complex numbers representing \(L\) and \(M\). Show that lines \(LC\), \(MA\) and \(NB\) all meet at the origin, and that these three line segments have the common length \(p+q+r\).
Solution:
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.
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.]
By considering the solutions of the equation \(z^n-1=0\), or otherwise, show that \[(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=1+z+z^2+\dots+z^{n-1},\] where \(z\) is any complex number and \(\omega={\rm e}^{2\pi i/n}\). Let \(A_1,A_2,A_3,\dots,A_n\) be points equally spaced around a circle of radius \(r\) centred at \(O\) (so that they are the vertices of a regular \(n\)-sided polygon). Show that \[\overrightarrow{OA_1}+\overrightarrow{OA_2}+\overrightarrow{OA_3} +\dots+\overrightarrow{OA_n}=\mathbf0.\] Deduce, or prove otherwise, that \[\sum_{k=1}^n|A_1A_k|^2=2r^2n.\]
Solution:
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 \[\sin\frac{2\pi}{5}=\frac{\surd(10+2\surd 5)}{4} \ \ \hbox{and}\ \ \cos\frac{2\pi}{5}=\frac{-1+\surd 5}{4}.\]
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 \(\sin \frac{2\pi}{5} = \frac{\sqrt{10+2\sqrt{5}}}{4}\)
Show, using de Moivre's theorem, or otherwise, that \[ \tan7\theta=\frac{t(t^{6}-21t^{4}+35t^{2}-7)}{7t^{6}-35t^{4}+21t^{2}-1}\,, \] where \(t=\tan\theta.\)