Problems

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Solution:

\(z\) 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 \(-\frac12\)

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Solution:

1. The line \(PO\) has gradient \(\frac{p^2}{p} = p\) and teh line \(QO\) has gradient \(q\), therefore we must have that \(pq = -1\). Therefore, \(R\) is the point \begin{align*} && R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\ &&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\ &&&= \left ( t, 2t^2+1\right) \end{align*} So we are looking at another parabola.
   

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2. The tangents are \(y = 2px+c\), ie \(p^2 = 2p^2+c\), ie \(y = 2px -p^2\) so we have \begin{align*} && y - 2px &= -p^2 \\ && y - 2qx &= -q^2 \\ \Rightarrow && (2p-2q)x &= p^2-q^2 \\ \Rightarrow && x &= \frac12 (p+q)\\ && y &= p(p+q)-p^2 \\ && y &= pq = -1 \end{align*} Therefore \(x = \frac12(p - \frac1p), y= -1\), so we have the line \(y = -1\) (the directrix)
   

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Solution:

1. Multiplication by \(1+i\) rotates by \(45^{\circ}\) anticlockwise
   

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2. \(|z| \leq \sqrt{2}\), \(\textrm{Re}(z^2) \geq 0\) means \(\textrm{Re}{z} \geq \textrm{Im}{z}\)
   

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3. These are all points within \(1\) unit from a circle radius \(2\) units.
   

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Solution: \(\mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a})+\mu(\mathbf{c}-\mathbf{a})\) Therefore it is the plane through \(\mathbf{a}\) with direction vectors \(\mathbf{b}-\mathbf{a}\) and \(\mathbf{c}-\mathbf{a}\), ie it is the plane through \(\mathbf{a},\mathbf{b},\mathbf{c}\). The normal to this plane will be \((\mathbf{b}-\mathbf{a} ) \times (\mathbf{c}-\mathbf{a}) = \mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a}\), so we must have: \begin{align*} && \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) &= \mathbf{a} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) \end{align*} Therefore, \begin{align*} && \mathbf{a} \cdot (\mathbf{b}\times \mathbf{c}) &= \mathbf{r} \cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= \left ( (1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c} \right)\cdot \left (\mathbf{b}\times \mathbf{c}-\mathbf{a} \times \mathbf{c}-\mathbf{b}\times \mathbf{a} \right) \\ &&&= (1-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\ \Rightarrow && 0 &= (-\lambda- \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})-\lambda \mathbf{b}\cdot(\mathbf{a} \times \mathbf{c})-\mu \mathbf{c}\cdot(\mathbf{b} \times \mathbf{a}) \\ &&&= -(\lambda+ \mu) \mathbf{a}\cdot (\mathbf{b} \times \mathbf{c})+\lambda \mathbf{b}\cdot(\mathbf{c} \times \mathbf{a})+\mu \mathbf{c}\cdot(\mathbf{a} \times \mathbf{b}) \\ \end{align*} The result follows from setting \(\mu = 0, \lambda = 1\) and \(\mu = 1, \lambda = 0\). If they all lie in the same plane then the plane described is through the origin, and those values are all the same, but equal to \(0\).

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Solution:

\begin{align*} \arg w &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{4}-z_{1})(z_{2}-z_{3})} \\ &= \arg \frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{2}-z_{3})(z_{4}-z_{1})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})}\frac{(z_{3}-z_{4})}{(z_{1}-z_{4})} \\ &= \arg \frac{(z_{1}-z_{2})}{(z_{3}-z_{2})} + \arg \frac{(z_{3}-z_{4})}{(z_{1}-z_{4})}\\ &= \beta + \pi - \beta = \pi \end{align*} Therefore \(w\) is real

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Solution: The perpendicular bisector of the chords runs through the origin, therefore \(LM\) is perpendicular to \(PQ\) if and only if \(\frac{l+m}{2}\) is perpendicular to \(\frac{p+q}{2}\), ie \begin{align*} && (l+m) &= it (p+q) \\ \Leftrightarrow && \frac{l+m}{p+q} & \in i \mathbb{R} \\ \Leftrightarrow && 0 &= \frac{l+m}{p+q} + \frac{l^*+m^*}{p^*+q^*} \\ &&&= \frac{l+m}{p+q} + \frac{\frac{r^2}{l}+\frac{r^2}{m}}{\frac{r^2}{p}+\frac{r^2}{q}} \\ &&&=\frac{l+m}{p+q} + \frac{l+m}{p+q} \frac{pq}{lm} \\ &&&= \frac{l+m}{p+q} \left ( \frac{lm+pq}{lm} \right) \end{align*} Therefore as long as \(l+m, p+q \neq 0\) \(lm+pq = 0\) is equivalent to the chords being perpendicular. In the case where (say) \(l,m\) is a diameter, then the condition for the chords to be perpendicular is that \(p,q\) is also a diameter and at right angles, but clearly this is also equivalent to our condition. Suppose \(A_1, A_2, A_3\) are distinct points on \(S\), and \(A_1'\) is given and suppose \(a_i, a_i'\) are the corresponding complex numbers, then the conditions are: \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_1'' \perp A_3A_1: && 0 &= a_3'a_1'' + a_3a_1 \\ \\ \Rightarrow && a_2' &= -\frac{a_1a_2}{a_1'} \\ && a_3' &= -\frac{a_2a_3}{a_2'} \\ &&&= \frac{a_1'a_2a_3}{a_1a_2} \\ &&&= \frac{a_1'a_3}{a_1} \\ && a_1'' &= - \frac{a_3a_1}{a_3'} \\ &&&= \frac{a_3a_1a_1}{a_1'a_3} \\ &&&= \frac{a_1^2}{a_1'} \\ \Rightarrow && a_1'a_1'' &= a_1^2 \end{align*} Therefore \(a_1' = a_1''\) if \(a_1' = \pm a_1\) Suppose we have \(4\) points, then \begin{align*} A_1'A_2' \perp A_1A_2: && 0 &= a_1'a_2' + a_1a_2 \\ A_2'A_3' \perp A_2A_3: && 0 &= a_2'a_3' + a_2a_3 \\ A_3'A_4' \perp A_3A_4: && 0 &= a_3'a_4' + a_3a_4 \\ A_4'A_1'' \perp A_4A_1: && 0 &= a_4'a_1'' + a_4a_1 \\ \\ \Rightarrow && a_4' &= -\frac{a_3a_4}{a_3'} \\ &&&= -\frac{a_1a_3a_4}{a_1'a_3} \\ &&&= -\frac{a_1a_4}{a_1'} \\ \Rightarrow && a_1'' &= -\frac{a_4a_1}{a_4'} \\ &&&= \frac{a_4a_1a_1'}{a_1a_4} \\ &&&= a_1' \end{align*} So they coincide. For \(n\) points if there are an even number of points they coincide, an odd number and there are two points when they coincide.

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Solution:

The initial velocity is \(\begin{pmatrix} v \cos \theta \\ v \sin \theta \end{pmatrix}\). The trajectory will be: \(\begin{pmatrix} x_0 + (v \cos \theta) t \\ (v \sin \theta)t -\frac12 g t^2 \end{pmatrix}\) we must have that for some time \(t\), this is equal to \(\begin{pmatrix} 0 \\ h \end{pmatrix}\) So \(t = -\frac{x_0}{v \cos \theta}\) and so \begin{align*} &&h &= (v \sin \theta)t -\frac12 g t^2 \\ &&&= -x_0\tan \theta - \frac12 g \frac{x_0^2}{v^2 \cos^2 \theta} \\ &&&= -x_0\tan \theta - \frac{g}{2v^2 \cos^2 \theta}x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} \sec^2 \theta x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} (1+\tan^2 \theta )x_0^2 \\ &&&= -\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2+\frac{v^2}{2g}-\frac{g}{2v^2}x_0^2 \\ \Rightarrow && \frac{g}{2v^2}x_0^2 &= \frac{v^2}{2g}-h-\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2 \\ \Rightarrow && x_0^2 &= \frac{v^2(v^2-2gh)}{g^2}-K^2 \end{align*} Therefore \(\displaystyle |x_0| \leq \frac{v}{g}\sqrt{v^2-2gh}\)

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Solution:

1. Suppose \(w = u+ vi\) then \(w^2 - 2w = u^2-v^2-2u+(2uv-2v)i\) so to be purely real we must have \(2uv-2v = 2v(u-1) = 0\) ie either \(v = 0\) or \(u = 1\). Therefore the locus is the real axis and the line \(1 + ti\):
   

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2. If \(w^2 = x+yi\) then we must have \(x = u^2-v^2\) and \(y = 2uv\), so either \(v = 0, y = 0, x = u^2-2u \geq -1\) or \(u = 1, x = 1-v^2, y = 2v\) which is a parabola:
   

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If \(t = u+iv\) then \(t^2-2t = u^2-v^2-2u + (2uv-2v)i\). For this to be purely imaginary, we need \(u^2-v^2 - 2u = 0 \Rightarrow (u-1)^2-v^2 = 1\), ie points on a hyperbola. Then \(p = u^2-v^2\) and \(q = 2uv\). We can parameterise our hyperbola as \(u = 1 \pm \cosh s, v = \sinh s\) and so \(p = 1 + 2 \cosh s\) and \(q = \sinh 2s\) or \(q = \pm (p-1) \sqrt{(\frac{p-1}{2})^2-1}\) where \(p \geq 3\)

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Solution: The plane is the same as the plane \((\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0\), ie the plane through \(\mathbf{a}\) whose normal is parallel to \(\mathbf{a}\) The distance from \(\mathbf{r}\) to the plane therefore is \(\lambda\) where \(\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}\) must be on the plane, ie \((\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}\) But if \(\mathbf{a} \cdot \mathbf{r} = a^2 - ar\) then \(\lambda = r\), ie the distance to the plane is the same as the distance to the origin. \(\mathbf{q} = k \mathbf{p}\) and so \(\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2\) if \(k > 0\) we will find \(k = 1\) the position vector we already know about, therefore suppose \(k < 0\) so: \begin{align*} && \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\ \Rightarrow && k(a^2-ap)-kap &= a^2 \\ \Rightarrow && k(a^2-2ap) &= a^2 \\ \Rightarrow && k &= \frac{a^2}{a^2-2ap} \end{align*} Therefore \(\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}\) The line through \(O\) orthogonal to \(\mathbf{p}\) and \(\mathbf{a}\) will be parallel to \(\mathbf{a} \times \mathbf{p}\). Therefore we should consider points of the from \(s \mathbf{a} \times \mathbf{p}\) on the surface \(S\). \begin{align*} && s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2 \end{align*} The angle between \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}\) Therefore \(sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}\) and so the points are as required. Noting that \(|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa\) The area of triangle \(PQT\) is : \begin{align*} \frac12 | (\mathbf{p} - \mathbf{t}) \times (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\ &= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\ &= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\ &= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\ &= \frac{ap^2}{a^2-ap} \end{align*}

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Solution: There are many possible parametric forms, for example \(z = i + 2e^{it}, z = 2\ cos \theta + (1 + 2\sin \theta)i\) etc. It is a circle radius \(2\) about the point \(i\).

1. \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{2i + 2e^{it}}{2e^{it}} \\ &= 2 + ie^{-it} \end{align*} This is obvious a circle radius \(1\) about the point \(2\).
   

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2. If \(z\) is real, then \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{(z+i)^2}{z^2+1} \\ &= \frac{z^2-1 + 2zi}{z^2+1} \end{align*} We can quickly notice this describes a circle radius \(1\) about \(0\). Alternatively, \(|z+i| = |z-i| \Rightarrow |\frac{z+i}{z-i}| = 1\) so we must be talking about points on the unit circle. Since this is a Mobius transform we know it maps lines and circles to lines and circles, therefore it must map to the unit circle;
3. If \(z\) is purely imaginary, say \(it\) then: \begin{align*} w &= \frac{z+i}{z-i} \\ &= \frac{(it+i)(i-it)}{(-1+t)^2} \\ &= \frac{t^2-1}{(t-1)^2} \end{align*} Which is purely real, and can take all real values.
   

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Solution:

The shown isoceles triangle has base \(2r\), and the two side lengths are \(R\). The angle at the center of the circle is \(\frac{2\pi}{n}\). The height of the triangle (by Pythagoras) is \(\sqrt{R^2-r^2}\) and so the area enclosed in the triangle is \(\frac12 2r \sqrt{R^2-r^2}\). The area in the three sectors are: \(\frac{\pi}{n}(R-r)^2\), two sets of \(\frac12(\frac12 \l \pi - \frac{2\pi}{n}\r)r^2 = \l\frac{1}{2} - \frac{1}{n} \r \frac12 \pi r^2\). Therefore the remaining area \(Y/n\) is \(r \sqrt{R^2-r^2} -\frac{\pi}{n}(R-r)^2 - \l\frac{1}{2} - \frac{1}{n} \r \pi r^2\). Multiplying this by \(n\) we get the desired result. For \(X\) we can look at the larger sector, which we obtain using the extension of this triangle. This has area \(\frac12 \frac{2 \pi}{n} (R+r)^2\). The areas to be removed are the area of the triangle: \(r \sqrt{R^2-r^2}\) and the areas of the two sectors, which have radii \(r\) and angles \(\pi - \l \frac12 -\frac1{n} \r\pi = \l \frac{1}{2} + \frac{1}{n} \r \pi\). Therefore the area for \(X/n\) is \(\frac{\pi}{n} (R+r)^2 - r \sqrt{R^2-r^2} - \l \frac{1}{2} + \frac{1}{n} \r \pi r^2\) and so \(X\) has area: \[ X = \pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r \] \(Z = n\pi r^2\) \begin{align*} \frac{(X+Y)}{Z} &= \frac{\pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r}{n \pi r^2 }\\ &= \qquad \frac{nr\sqrt{R^{2}-r^{2}}-\pi(R-r)^{2}-n\pi r^{2}\left(\frac{1}{2}-\frac{1}{n}\right)}{n \pi r^2} \\ &= \frac{4\pi R r - n \pi r^2}{n\pi r^2} \\ &= \frac{4R-nr}{n r} \\ &= \frac{4R}{nr} - 1 \end{align*} Since \(\frac{r}{R} = \sin \frac{\pi}{n}\) we have: \begin{align*} \frac{(X+Y)}{Z} &= \frac{4R}{nr} - 1 \\ &= \frac{4}{n \sin \frac{\pi}n} - 1 \\ & \to \frac{4}{\pi} - 1 \end{align*}

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Solution:

Suppose \(P\) is a fraction \(0 \leq k\leq 1\) along \(AB\), then \(\overrightarrow{OP} = \overrightarrow{OA} +\overrightarrow{AP} = \overrightarrow{OA} +k\overrightarrow{AB} = \mathbf{a} + k(\mathbf{b} - \mathbf{a}) = \lambda \mathbf{b} + (1-k) \mathbf{a}\), ie an arbitrary point on \(AB\) has the position vector where \((1-k) = \rho \geq 0\) and \(k= \sigma \geq 0\) and \((1-\lambda) + \lambda = 1\). Any point on the segment \(PC\) will be of the form \(l\mathbf{c} + (1-l) (k \mathbf{b} + (1-k) \mathbf{a})\) which has the form \(\lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}\) where \(\lambda + \mu + \nu = (1-l)(1-k) + (1-l)k + l = 1\) and all coefficients are positive.

1. This is equivalent to the area inside the triangle where every point (\(\mathbf{a}, \mathbf{b}, \mathbf{c}\)) has been send to it's negative (\(-\mathbf{a}, -\mathbf{b}, -\mathbf{c}\)), ie
   

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2. Writing points as: \begin{align*} \lambda\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} &= (1-\mu - \nu)\mathbf{a}+\mu\mathbf{b}+\nu\mathbf{c} \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) \\ &= \mathbf{a} + (-\mu)(\mathbf{a}-\mathbf{b}) + (-\nu)(\mathbf{a} - \mathbf{c}) + (1+\mu+\nu)\mathbf{0}\\ \end{align*} So this is a translation of \(\mathbf{a}\) of the triangle with vertices at \(\mathbf{0}, \mathbf{a-b}, \mathbf{a-c}\), or a triangle with vertices \(\mathbf{a}, 2\mathbf{a-b}, 2\mathbf{a-c}\).
   

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