49 problems found
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 \(y\).
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 \(u = \frac{2x}{x^2+(y+1)^2}, v = \frac{x^2+y^2-1}{x^2+(1+y)^2}\)
Solution:
The points \(S\), \(T\), \(U\) and \(V\) have coordinates \((s,ms)\), \((t,mt)\), \((u,nu)\) and \((v,nv)\), respectively. The lines \(SV\) and \(UT\) meet the line \(y=0\) at the points with coordinates \((p,0)\) and \((q,0)\), respectively. Show that \[ p = \frac{(m-n)sv}{ms-nv}\,, \] and write down a similar expression for \(q\). Given that \(S\) and \(T\) lie on the circle \(x^2 + (y-c)^2 = r^2\), find a quadratic equation satisfied by \(s\) and by \(t\), and hence determine \(st\) and \(s+t\) in terms of \(m\), \(c\) and \(r\). Given that \(S\), \(T\), \(U\) and \(V\) lie on the above circle, show that \(p+q=0\).
The point \(P(a\cos\theta\,,\, b\sin\theta)\), where \(a>b>0\), lies on the ellipse \[\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1\,.\] The point \(S(-ea\,,\,0)\), where \(b^2=a^2(1-e^2)\,\), is a focus of the ellipse. The point \(N\) is the foot of the perpendicular from the origin, \(O\), to the tangent to the ellipse at \(P\). The lines \(SP\) and \(ON\) intersect at \(T\). Show that the \(y\)-coordinate of \(T\) is \[\dfrac{b\sin\theta}{1+e\cos\theta}\,.\] Show that \(T\) lies on the circle with centre \(S\) and radius \(a\).
Solution: Find the gradient of the tangent of the ellipse at \(P\): \begin{align*} && \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= - \frac{2xb^2}{2ya^2} \\ &&&=- \frac{a \cos \theta b^2}{b \sin \theta a^2} \\ &&&=-\frac{b}{a} \cot \theta \end{align*} Therefore the gradient of \(ON\) is \(\frac{a}{b} \tan \theta\). \begin{align*} && y &= \frac{a}{b} \tan \theta x \\ && \frac{y-0}{x-(-ea)} &= \frac{b\sin \theta-0}{a\cos \theta -(-ea)} \\ && y &= \frac{b \sin \theta}{a(e+\cos \theta)}(x+ea) \\ \Rightarrow && y &= \frac{b \sin \theta}{a(\cos \theta+e)}\frac{b}{a} \cot \theta y+ \frac{eb \sin \theta}{\cos \theta + e} \\ &&&= \frac{b^2 \cos \theta}{a^2(\cos \theta +e)}y + \frac{eb \sin \theta}{\cos \theta + e} \\ \Rightarrow && (\cos \theta+e)y &= (1-e^2)\cos \theta y +eb \sin \theta\\ && e(1+e\cos \theta)y &= eb \sin \theta \\ \Rightarrow && y &= \frac{b \sin \theta}{1+e\cos \theta} \\ && x &= \frac{b \sin \theta}{1+e\cos \theta} \frac{b}{a} \cot \theta \\ &&&= \frac{b^2 \cos \theta}{a(1+e\cos \theta)} \end{align*} Therefore \(\displaystyle T\left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}, \frac{b \sin \theta}{1+e\cos \theta} \right)\). Finally, we can look at the distance of \(T\) from \(S\) \begin{align*} && d^2 &= \left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}-(-ea) \right)^2 + \left (\frac{b \sin \theta}{1+e\cos \theta} -0\right)^2 \\ &&&= \frac{\left (b^2 \cos \theta+ea^2(1+e\cos\theta)\right)^2 + \left ( ab \sin \theta\right)^2}{a^2(1+e\cos \theta)^2} \\ &&&= \frac{b^4\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2b^2(1+e\cos\theta)+a^2b^2\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= \frac{a^4(1-e^2)^2\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2a^2(1-e^2)(1+e\cos\theta)+a^4(1-e^2)\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= a^2 \left ( \frac{(1-e^2)^2\cos^2\theta+e^2(1+e\cos\theta)^2+2e(1-e^2)(1+e\cos\theta)+(1-e^2)(1-\cos^2\theta)}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)((1-e^2)\cos^2\theta+2e(1+e\cos\theta)+(1-\cos^2\theta))}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)(1+e\cos\theta)^2}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \end{align*} Therefore a circle radius \(a\) centre \(S\).
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.
Solution:
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.
The line \(y=d\,\), where \(d>0\,\), intersects the circle \(x^2+y^2=R^2\) at \(G\) and \(H\). Show that the area of the minor segment \(GH\) is equal to \begin{equation} R^2\arccos \left({d \over R}\right) -d\sqrt{R^2 - d^2}\;. \tag {\(*\)} \end{equation} In the following cases, the given line intersects the given circle. Determine how, in each case, the expression \((*)\) should be modified to give the area of the minor segment.
In the \(x\)--\(y\) plane, the point \(A\) has coordinates \((a\,,0)\) and the point \(B\) has coordinates \((0\,,b)\,\), where \(a\) and \(b\) are positive. The point \(P\,\), which is distinct from \(A\) and \(B\), has coordinates~\((s,t)\,\). \(X\) and \(Y\) are the feet of the perpendiculars from \(P\) to the \(x\)--axis and \(y\)--axis respectively, and \(N\) is the foot of the perpendicular from \(P\) to the line \(AB\,\). Show that the coordinates \((x\,,y)\) of \(N\) are given by \[ x= \frac {ab^2 -a(bt-as)}{a^2+b^2} \;, \ \ \ y = \frac{a^2b +b(bt-as)}{a^2+b^2} \;. \] Show that, if $\ds \ \left( \frac{t-b} s\right)\left( \frac t {s-a}\right) = -1\;\(, then \)N$ lies on the line \(XY\,\). Give a geometrical interpretation of this result.
Show that the equation of any circle passing through the points of intersection of the ellipse \((x+2)^2 +2y^2 =18\) and the ellipse \(9(x-1)^2 +16y^2 = 25\) can be written in the form \[ x^2-2ax +y^2 =5-4a\;. \]
Solution: \begin{align*} && (x+2)^2 +2y^2 &=18 \\ && 9(x-1)^2 +16y^2 &= 25 \\ \Rightarrow && 2y^2 &= 18 - (x+2)^2 \\ && 16y^2 &= 25 - 9(x-1)^2 \\ \Rightarrow && 25-9(x-1)^2 &= 8 \cdot 18 - 8(x+2)^2 \\ \Rightarrow && 25 -9+18x-9x^2 &= 144 -32- 32x +8x^2 \\ \Rightarrow && 0 &= 96 - 50x+x^2 \\ &&&= (x-48)(x-2) \\ \Rightarrow && x &= 2,48 \\ \Rightarrow && 2y^2 &= 2, 18-50^2 \\ \Rightarrow && (x,y) &= (2,\pm1) \end{align*} Therefore any circle must have it's centre on there perpendicular bisector of \((2, \pm 1)\), ie on the \(x\)-axis. Therefore it will have equation \((x-a)^2+y^2 = r^2\) and also contain the point \((2,1)\), therefore: \begin{align*} r^2 &= (2-a)^2 + 1^2 \\ &= 4 -2a+a^2 + 1 \\ &= 5-2a+a^2 \end{align*} and the equation is: \begin{align*} && (x-a)^2 + y^2 &= 5-4a+a^2 \\ \Rightarrow && x^2-2ax+a^2 +y^2 &= 5-4a+a^2 \\ \Rightarrow && x^2-2ax+y^2 &= 5-4a \end{align*} as required.
Three ships \(A\), \(B\) and \(C\) move with velocities \({\bf v}_1\), \({\bf v}_2\) and \(\bf u\) respectively. The velocities of \(A\) and \(B\) relative to \(C\) are equal in magnitude and perpendicular. Write down conditions that \(\bf u\), \({\bf v}_1\) and \({\bf v}_2\) must satisfy and show that \[ \left| {\bf u} -{\textstyle\frac12} \l {\bf v}_1 + {\bf v}_2 \r \right|^2 = \left|{\textstyle\frac12} \l {\bf v}_1 - {\bf v}_2 \r \right|^2 \] and \[ \l {\bf u} -{\textstyle\frac12} \l {\bf v}_1 + {\bf v}_2 \r \r \cdot \l {\bf v}_1 - {\bf v}_2 \r = 0 \;. \] Explain why these equations determine, for given \({\bf v}_1\) and \({\bf v}_2\), two possible velocities for \(C\,\), provided \({\bf v}_1 \ne {\bf v}_2 \,\). If \({\bf v}_1\) and \({\bf v}_2\) are equal in magnitude and perpendicular, show that if \({\bf u} \ne {\bf 0}\) then \({\bf u} = {\bf v}_1 + {\bf v}_2\,\).
The line \(l\) has vector equation \({\bf r} = \lambda {\bf s}\), where \[ {\bf s} = (\cos\theta+\sqrt3\,) \; {\bf i} +(\surd2\;\sin\theta)\;{\bf j} +(\cos\theta-\sqrt3\,)\;{\bf k} \] and \(\lambda\) is a scalar parameter. Find an expression for the angle between \(l\) and the line \mbox{\({\bf r} = \mu(a\, {\bf i} + b\,{\bf j} +c\, {\bf k})\)}. Show that there is a line \(m\) through the origin such that, whatever the value of \(\theta\), the acute angle between \(l\) and \(m\) is \(\pi/6\). A plane has equation \(x-z=4\sqrt3\). The line \(l\) meets this plane at \(P\). Show that, as \(\theta\) varies, \(P\) describes a circle, with its centre on \(m\). Find the radius of this circle.
A point moves in the \(x\)-\(y\) plane so that the sum of the squares of its distances from the three fixed points \((x_{1},y_{1})\), \((x_{2},y_{2})\), and \((x_{3},y_{3})\) is always \(a^{2}\). Find the equation of the locus of the point and interpret it geometrically. Explain why \(a^2\) cannot be less than the sum of the squares of the distances of the three points from their centroid. [The centroid has coordinates \((\bar x, \bar y)\) where \(3\bar x = x_1+x_2+x_3,\) $3\bar y =y_1+y_2+y_3. $]
Solution: \begin{align*} && a^2 &= d_1^2 + d_2^2 + d_3^2 \\ &&&= (x-x_1)^2+(y-y_1)^2 + (x-x_2)^2+(y-y_2)^2 + (x-x_3)^2+(y-y_3)^2 \\ &&&= \sum (x-\bar{x}+\bar{x}-x_i)^2 + \sum (y-\bar{y}+\bar{y}-y_i)^2 \\ &&&= \sum \left ( (x-\bar{x})^2+(\bar{x}-x_i)^2 + 2(x-\bar{x})(\bar{x}-x_i) \right)+ \sum \left ( (y-\bar{y})^2+(\bar{y}-y_i)^2 + 2(y-\bar{y})(\bar{y}-y_i) \right)\\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2 + 6x\bar{x} -6\bar{x}^2-2x\sum x_i+2\bar{x}\sum x_i + \\ &&&\quad\quad\quad 3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 + 6y\bar{y} -6\bar{y}^2-2y\sum y_i+2\bar{y}\sum y_i \\ &&&= 3(x-\bar{x})^2 + \sum (\bar{x}-x_i)^2+3(y-\bar{y})^2 + \sum (\bar{y}-y_i)^2 \\ \\ \Rightarrow && (x-\bar{x})^2+(y-\bar{y})^2 &= \frac13\left ( a^2- \sum \left((\bar{x}-x_i)^2+(\bar{y}-y_i)^2 \right) \right) \end{align*} Therefore the locus is a circle, centre \((\bar{x}, \bar{y})\). radius \(\sqrt{\frac13(a^2 - \text{sum of squares distances of centroid to vertices}})\). \(a^2\) cannot be less than this distance, because clearly the right hand side is always bigger than it!
Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is a square. The result changes if, instead of maximising the sum of lengths of sides of the rectangle, we seek to maximise the sum of \(n\)th powers of the lengths of those sides for \(n\geqslant 2\). What happens if \(n=2\)? What happens if \(n=3\)? Justify your answers.
Solution: We can always rotate the circle so that sides are parallel to the \(x\) and \(y\) axes. Therefore if one corner is \((a,b)\) the other coordinates are \((-a,b), (a,-b), (-a,-b)\) and the perimeter will be \(4(a+b)\). Therefore we wish to maximise \(4(a+b)\) subject to \(a^2+b^2 = \text{some constant}\). Notice that \(\frac{a+b}{2} \leq \sqrt{\frac{a^2+b^2}{2}}\) with equality when \(a = b\), therefore the maximum is a square. If \(n = 2\) then we are looking at \(2((2a)^2+(2b)^2) = 8(a^2+b^2)\) which is constant for all rectangles. If \(n=3\) we are maximising \(16(a^3+b^3) = 16(a^3+(c^2-a^2)^{3/2})\) which is maximised when \(a = 0, c\)