Problems

Solution:

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1. \(\,\) \begin{align*} \overset{\curvearrowright}{A}:&& W \cos \theta \cdot a + kW \cos \theta \cdot 2a - T \cos \theta \sin \theta \cdot 2a - T \sin \theta \cos \theta \cdot 2a &= 0 \\ && (2k+1) \cos \theta W &= T \cos \theta \cdot 4 \sin \theta \\ \Rightarrow && T &= \frac{2k+1}{4 \sin \theta} W \\ \Rightarrow && \text{breaks if }\quad \quad 2k+1 &> 4 \sin \theta \end{align*}
2. \(\,\) \begin{align*} \text{N2}(\uparrow): && R - W - kW - T \sin \theta &= 0 \\ \Rightarrow && R &= (k+1)W - T \sin \theta \\ &&&= (k+1)W - \frac{2k+1}{4} W \\ &&&= \frac{2k+3}{4}W \\ \text{N2}(\leftarrow): && F_A - T \cos \theta &= 0 \\ \Rightarrow && F_A &= \frac{2k+1}{4 }\cot \theta \\ \Rightarrow && \text{slips if }\quad \quad\quad \quad\quad \quad F_A &> \mu R \\ \Rightarrow && \text{slips if }\quad \quad \frac{2k+1}{4 }\cot \theta &> \mu \frac{2k+3}{4}W \\ \Rightarrow && 2k+1 &> (2k+3) \mu \tan \theta \end{align*}
3. The condition for breaking is \(k > 2\sin \theta -\frac12\). The condition for slipping is equivalent to: \begin{align*} && 2k+1 &> (2k+3) \mu \tan \theta \\ \Leftrightarrow && 2k(1- \mu \tan \theta) &> 3 \mu \tan \theta-1 \\ \Leftrightarrow && k &> \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} \end{align*} Therefore we will slip first if: \begin{align*} && \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} &< 2 \sin \theta - \frac12 \\ \Leftrightarrow && 3 \mu \tan \theta-1 &< 4 \sin \theta (1- \mu \tan \theta) - (1- \mu \tan \theta) \\ &&&=4 \sin \theta - 1 + \mu \tan \theta (1-4 \sin \theta) \\ \Leftrightarrow && 3 \mu \tan \theta &< 4 \sin \theta + \mu \tan \theta (1- 4 \sin \theta) \\ \Leftrightarrow && \mu \tan \theta(3-1+4\sin \theta) &< 4 \sin \theta \\ \Leftrightarrow && \mu &< \frac{2 \cos \theta}{1+2 \sin \theta} \end{align*}

Solution:

1. If \(n_3 = 9\) and we are looking for \(0 < n_1 < n_2 < n_3\) we can consider values for each \(n_2\). \begin{array}{clc|c} n_2 & \text{range} & \text{count} \\ \hline 6 & 4-5 & 2 \\ 7 & 3-6 & 4 \\ 8 & 2-7 & 6 \\ \hline & & 12 \end{array} When \(n_3 = 10\) \begin{array}{clc|c} n_2 & \text{range} & \text{count} \\ \hline 6 & 5 & 1 \\ 7 & 4-6 & 3 \\ 8 & 3-7 & 5 \\ 9 & 2-8 & 7 \\ \hline & & 16 \end{array} When \(n_3 = 2n+1\) we can have \(2 + 4 + \cdots + 2n-2 = n(n-1)\) When \(n_3 = 2n\) we can have \(1 + 3 + \cdots + 2n-3 = (n-1)^2\)
2. For the 3 rods to form a triangle, it suffices for the sum of the lengths of the shorter rods to be larger than \(N\). When \(N = 2n+1\) there are \(n(n-1)\) ways this can happen, out of \(\binom{2n}{2}\) ways to choos the numbers, ie \begin{align*} && P &= \frac{n(n-1)}{\frac{2n(2n-1)}{2}} \\ &&&= \frac{n-1}{2n-1} \end{align*} When \(N = 2n\) there are \((n-1)^2\) ways this can happen, out of \(\binom{2n-1}{2}\) ways, ie \begin{align*} && P &= \frac{(n-1)^2}{\frac{(2n-1)(2n-2)}{2}} \\ &&&= \frac{n-1}{2n-1} \end{align*}
3. The number of ways this can happen is: \begin{align*} C &= \sum_{k=3}^{2M+1} \# \{ \text{triangles where }k\text{ is largest} \} \\ &= \sum_{k=1}^{M} \# \{ \text{triangles where }2k+1\text{ is largest} \} +\sum_{k=1}^{M} \# \{ \text{triangles where }2k\text{ is largest} \}\\ &= \sum_{k=1}^{M} n(n-1)+\sum_{k=1}^{M} (n-1)^2\\ &= \sum_{k=1}^{M} (2n^2-3n+1)\\ &= \frac26M(M+1)(2M+1) - \frac32M(M+1) + M \\ &= \frac16 M(4M+1)(M-1) \end{align*} Therefore the probability is \begin{align*} && P &= \frac{M(4M+1)(M-1)}{6 \binom{2M+1}{3}} \\ &&&= \frac{M(4M+1)(M-1)}{(2M+1)2M(2M-1)} \\ &&&= \frac{(4M+1)(M-1)}{2(2M+1)(2M-1)} \end{align*}

Solution:

1. \(\,\) \begin{align*} && \mu &= \E[X] \\ &&&= \int_0^1 x f(x) \d x \\ &&&= \int_0^1 nx^n \d x \\ &&&= \frac{n}{n+1} \\ \\ && \var[X] &= \sigma^2 \\ &&&= \E[X^2] - \mu^2 \\ &&&= \int_0^1 x^2 f(x) \d x - \mu^2 \\ &&&= \int_0^1 nx^{n+1} \d x - \mu^2 \\ &&&= \frac{n}{n+2} - \frac{n^2}{(n+1)^2} \\ &&&= \frac{n(n+1)^2 - n^2(n+2)}{(n+1)^2(n+2)} \\ &&&= \frac{n}{(n+1)^2(n+2)} \end{align*}
2. \(\,\) \begin{align*} && \frac14 &= \int_0^{Q_1} 2x \d x \\ &&&= Q_1^2 \\ \Rightarrow && Q_1 &= \frac12 \\ && \frac34 &= \int_0^{Q_3} 2x \d x \\ &&&= Q_3^2 \\ \Rightarrow && Q_3 &= \frac{\sqrt{3}}2 \\ \\ \Rightarrow && IQR &= Q_3 - Q_1 = \frac{\sqrt{3}-1}{2} \\ && 2 \sigma &= 2\sqrt{\frac{2}{3^2 \cdot 4}} \\ &&&= \frac{\sqrt{2}}{3} \\ \\ && 2\sigma - IRQ &= \frac{\sqrt{2}}{3} - \frac{\sqrt{3}-1}{2} \\ &&&= \frac{2\sqrt{2}-3\sqrt{3}+3}{6} \\ && (3+2\sqrt{2})^2 &= 17+12\sqrt{2} > 29 \\ && (3\sqrt{3})^2 &= 27 \end{align*} Therefore \(2\sigma > IQR\)
3. \[ (1+x)^n = 1 + nx + \frac{n(n-1)}2 x^2 + \cdots + \binom{n}{k} x^k+ \cdots \] \begin{align*} && Q_1^{-n} &= 4 \\ && Q_2^{-n} &= 2\\ && \mu &=\frac{n}{n+1} \\ \Rightarrow && \mu^{-n} &= \left (1 + \frac1n \right)^n\\ &&&\geq 1 + n \frac1n + \cdots > 2 \\ \Rightarrow && \mu &< Q_2 \\ \\ && \mu^{-n} &= \left (1 + \frac1n \right)^n\\ &&&= 1 + n \frac1n + \frac{n(n-1)}{2!} \frac{1}{n^2} + \cdots + \frac{n(n-1) \cdots (n-k+1)}{k!} \frac{1}{n^k} + \cdots \\ &&&= 1 + 1 + \left (1 - \frac1n \right ) \frac1{2!} + \cdots + \left (1 - \frac1n \right)\cdot\left (1 - \frac2n \right) \cdots \left (1 - \frac{k-1}n \right) \frac{1}{k!} + \cdots \\ &&&< 1 + 1 + \frac1{2!} + \cdots + \frac1{k!} \\ &&&< 4 \\ \Rightarrow && \mu &> Q_1 \end{align*}

Solution:

1. Let \(k = 1\), then \begin{align*} \dot{x} &= - x - y \\ \dot{y} &= x - y \\ \dot{x}-\dot{y} &= -2x \\ \ddot{x} &= -\dot{x}-\dot{y} \\ &= -\dot{x} - (\dot{x}+2x) \\ &= -2\dot{x}- 2x \\ \dot{x}+\dot{y} &= -2y \\ \ddot{y} &= \dot{x}-\dot{y} \\ &= -2y-2\dot{y} \end{align*} So we have an auxiliary equation for \(x\) and \(y\) which is \(\lambda^2 + 2 \lambda+2 = 0 \Rightarrow \lambda = -1 \pm i\). Therefore \(x = Ae^{-t} \cos t + B e^{-t} \sin t, y = Ce^{-t}\cos t + De^{-t} \sin t\). We also must have that, \(A = 1, C = 0\), so \(x = e^{-t} \cos t + Be^{-t} \sin t\) and \(y = De^{-t} \sin x\). \begin{align*} \dot{y} &= -De^{-t} \sin t +De^{-t} \cos t \\ &= e^{-t} \cos x + Be^{-t} \sin t- De^{-t} \sin t \\ \end{align*} therefore \(B = 0, D = 1\) and \(x = e^{-t} \cos t, y = e^{-t} \sin t\)
   

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   \begin{align*} y &= e^{-t} \sin t \\ \dot{y} &= -e^{-t} \sin t + e^{-t} \cos t \\ \dot{x} &= e^{-t} \cos t -e^{-t} \sin t \end{align*}
   

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2. \begin{align*} \dot{x} = -x \\ \dot{y} = x-y \end{align*} So \(x = e^{-t}\). \(\dot{y} + y = e^{-t}\) so \(y = (t+B)e^{-t}\) and so \(y =te^{-t}\).
   

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Solution:

1. \(\,\) \begin{align*} && f(0+x) &= f(0)f(x) \\ \Rightarrow && f(0) &= 0, 1\\ &&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\ \Rightarrow && f(0) &= 1 \end{align*} \begin{align*} && f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\ &&&= f(x) \cdot f'(0) \\ &&&= kf(x) \end{align*} Since \(f'(x) = kf(x)\) we must have \(\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}\) but \(f(0) = 1\) so \(f(x) = e^{kx}\)
2. Consider \begin{align*} && g(0+0) &= \frac{g(0)+g(0)}{1+(g(0))^2} \\ \Rightarrow && g(0)(1+(g(0))^2)&= 2g(0) \\ \Rightarrow && 0 &= g(0)\left (1- (g(0))^2 \right) \\ \Rightarrow && g(0) &= -1, 0, 1 \\ \Rightarrow && g(0) &= 0 \tag{\(|g(0)| < 1\)} \end{align*} \begin{align*} && g'(x) &=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)}-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{g(x)+g(h)-g(x)(1+g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= \lim_{h\to 0} \frac{g(h)-g(x)(g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= (1-(g(x))^2) \cdot \lim_{h \to 0} \frac{1}{1+g(x)g(h)} \cdot \lim_{h \to 0} \frac{g(h)}{h} \\ &&&= (1-(g(x))^2) \cdot \frac{1}{1+g(x)\cdot 0} \cdot \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ &&&= (1-(g(x))^2) \cdot g'(0)\\ &&&= k (1-(g(x))^2) \\ \end{align*} Let \(y = g(x)\) so \begin{align*} && y' &= k(1-y^2) \\ \Rightarrow && kx &= \int \frac{1}{1-y^2} \d y \\ \Rightarrow &&&= \int \frac12\left ( \frac{1}{1-y} + \frac{1}{1+y} \right) \d y \\ &&&= \frac12\ln \left ( \frac{1+y}{1-y} \right) + C \\ x = 0, y = 0: && 0 &= \ln 1 + C \\ \Rightarrow && C &= 0 \\ \Rightarrow && \frac{1+y}{1-y} &= e^{2kx} \\ \Rightarrow && 1+y &= e^{2kx} - e^{2kx}y \\ \Rightarrow && y &= \frac{e^{2kx}-1}{e^{2kx}+1} \\ &&&= \tanh kx \end{align*}

Solution:

1. Suppose \((x,y)\) is on the line of invariant points, then \begin{align*} &&\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ &&&= \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} \\ \Rightarrow && \begin{cases} (a-1)x + by = 0 \\ (cx + (d-1)y = 0 \end{cases} \tag{*} \end{align*} Therefore either \(x = 0, y = 0\) or \((a-1)(d-1)-bc = 0\) \(\Rightarrow ((a-1)(d-1)-bc)xy = 0\). We also know this is true for all values \(x,y\) on the line of invariant points. If there is one where both \(x \neq 0, y \neq 0\) we are done, otherwise the line of invariant points must be one of the axes. ie but then one of \(\begin{pmatrix} a \\ c \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) or \(\begin{pmatrix} b \\ d \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) is true and we'd also be done. If the line doesn't go through the origin then there are points on every line, not equal to the origin which are fixed. But then every point on those lines is fixed (since \(\mathbf{A}\) is a linear operator) and so every point is fixed. ie \(\mathbf{A} = \mathbf{I}\).
2. Suppose \((a-1)(d-1) -bc = 0\) and \(b \neq 0\) then I claim that \(y = \frac{1-a}{b}x\) is a line of invariant points. It's clear that the first equation will be satisfied in \((*)\) so it suffices to check the second, but the first condition is equivalent to the equations being linearly dependent, ie both equations are satisfied. If \(b = 0\) then \((a-1)(d-1) = 0\), so our matrix must look like \(\begin{pmatrix} 1 & 0 \\ c & d\end{pmatrix}\) (if \(d \neq 1\))or \(\begin{pmatrix} * & 0 \\ * & 1\end{pmatrix}\). In the first case, the line \(y = \frac{c}{1-d}x\) and in the second \(x = 0\) is an invariant line.
3. Suppose the invariant line is \(y = mx+k\) then we must have that \begin{align*} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} &= \begin{pmatrix} (a + mb)x + bk \\ (c+dm)x + dk \end{pmatrix} \end{align*} and \((c+dm)x + dk = m((a + mb)x + bk) +k \Rightarrow k(d-mb-1) = x(-c+(a-d)m+m^2b)\) Since this equation must be true for all values of \(x\), and \(k \neq 0\) we can say that \(mb = d-1\) and \(-c+(a-d)m+m^2b = 0\), ie \(-c + (a-d)m + m(d-1) = 0 \Rightarrow (a-1)m-c = 0\) if \(m \neq 0\) then \((a-1)\frac{(d-1)}{b} - c = 0\) ie our desired relation is true. If \(m = 0\) then we must have that \(y = k\) is an invariant line, ie \(d-1=0\) and \(c=0\) which also satisfies our relation.

Solution:

1. Suppose \(n = 1\), then all polynomials are reflexive (since \(x - a_1\) has the root \(a_1\). Suppose \(n = 2\), then we want \begin{align*} && x^2-a_1x+a_2 &= (x-a_1)(x-a_2) \\ &&&= x^2-(a_1+a_2)x+a_1a_2 \\ \Rightarrow && a_2 &= 0 \\ \end{align*} So all polynomials of the form \(x^2-a_1x\) work and no others. Suppose \(n = 3\) then we want \begin{align*} && x^3-a_1x^2+a_2x-a_3 &= (x-a_1)(x-a_2)(x-a_3) \\ &&&= x^3-(a_1+a_2+a_3)x+(a_1a_2+a_1a_3+a_2a_3)x-a_1a_2a_3 \\ \Rightarrow && a_2+a_3 &= 0 \\ && a_2a_3 &= a_2 \\ \Rightarrow && -a_2^2 &= a_2 \\ \Rightarrow && a_2 &= 0, -1 \\ && -a_1a_2^2 &= -a_2 \\ \Rightarrow && a_2 &= 0, a_2 = 1/a_1 \end{align*} So we need either \(x^3-a_1x\) or \((x+1)^2(x-1) = x^3+x^2-x-1\)
2. Suppose \(n > 3\) then \begin{align*} && \sum a_i^2 &= \left (\sum a_i \right)^2 - 2 \sum_{i < j} a_i a_j \\ && &= a_1^2 - 2a_2 \\ \Rightarrow && 2a_2 &= a_1^2 - \sum a_i^2 \\ &&&= -a_2^2 - a_3^2 - \cdots - a_n^2 \end{align*} So \((a_2+1)^2 = 1-a_3^2 -\cdots -a_n^2\) so if \(a_n > 0\) (or any other \(a_i, i > 2\) for that matter) then we must have \(a_n = \pm 1, a_{3}, \ldots a_{n-1} = 0\), but if \(a_n = \pm 1\) \(x = 0\) is not a root. Therefore we must have \(a_0\) and \(a_i = 0\) for all \(i > 3\)
3. The only reflexive polynomials therefore must be \(x^n - kx^{n-1}\) and \(x^{n+3}+x^{n+2}-x^{n+1}-x^n\)

Solution:

1. \(\,\)
   

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2. \(\,\) \begin{align*} && y &= \frac{cx}{\sqrt{x^2+p}} \\ && \d y &= \frac{c(x^2+p)-cx^2}{(x^2+p)^{3/2}} \d x \\ &&&= \frac{cp^2}{(x^2+p)^{3/2}} \d x\\ && I &= \int \frac1{(b^2-y^2)\sqrt{c^2-y^2}} \d y \\ &&&= \int \frac{1}{\left ( b^2 - \frac{c^2x^2}{x^2+p} \right) \sqrt{c^2 - \frac{c^2x^2}{x^2+p} }} \d y \\ &&&= \int \frac{(x^2+p)^{3/2}}{((b^2-c^2)x^2+pb^2)\sqrt{c^2p}}\frac{cp}{(x^2+p)^{3/2}} \d x \\ &&&= \int \frac{\sqrt{p}}{((b^2-c^2)x^2+pb^2)} \d x \\ p=1: &&&= \int \frac{1}{(b^2-c^2)x^2+b^2} \d x \end{align*} When \(b = \sqrt{3}, c = \sqrt{2}\) \begin{align*} && I_1 &= \int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \d y\\ &&&= \int_{x =1 }^{x=\infty} \frac{1}{3+x^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{3}} \tan^{-1} \frac{x}{\sqrt{3}} \right]_1^\infty \\ &&&= \frac{\pi}{2\sqrt{3}} - \frac{1}{\sqrt{3}} \frac{\pi}{6} \\ &&&= \frac{\pi}{3\sqrt{3}} \end{align*} \begin{align*} && I_2 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = \frac1y, \d x = -\frac1{y^2} \d y &&&= \int_{x=\sqrt{2}}^{x=1} \frac{x^2}{(3-x^2)\sqrt{2-x^2}}\cdot \left ( -\frac{1}{x^2} \right ) \d x \\ &&&= \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \d x \\ &&&= I_1 = \frac{\pi}{3\sqrt{3}} \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \d y \\ x = 1/y, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{2}} \frac{x}{(3-x^2)\sqrt{2-x^2}} \d x \\ u = x^2, \d u = 2x \d x &&&= \int_{u=1}^{u=2} \frac{\frac12}{(3-u)\sqrt{2-u}} \d u \\ v=2-u, \d v = -\d u &&&= \frac12\int_{v=0}^{v=1} \frac{1}{(1+v)\sqrt{v}} \d v \\ &&&=\left [\tan^{-1}\sqrt{v}\right]_0^1 \\ &&&= \frac{\pi}{4} \end{align*}

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

Solution:

1. Suppose \(a=b\), ie \begin{align*} && y^2(y^2-a^2) &= x^2(x^2-a^2) \\ \Rightarrow && 0 &= x^4-y^4-a^2(x^2-y^2) \\ &&&= (x^2-y^2)(x^2+y^2-a^2) \end{align*} Therefore we have the lines \(y = \pm x\) and a circle radius \(a\).
   

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2. 1. Since \(x^4 - 4x^2 - y^2(y^2-5)= 0\), we must have \(0 \leq \Delta = 16 + 4y^2(y^2-5) \Rightarrow y^4-5y^2+4 = (y^2-4)(y^2-1) \geq 0\), therefore \(0 \leq y \leq 1\) or \(y \geq 2\) (since we are only considering positive values of \(y\)).
   2. When \((x, y) \approx 0\) the equation is more like \(4x^2 \approx 5y^2\) or \(y \approx \frac{2}{\sqrt{5}}x\) If \(|x|, |y|\) are very large, it is more like \(x^4 \approx y^4\), ie \(y \approx x\)
   3. \(\,\) \begin{align*} && (2y(y^2-5)+y^2(2y))y' &= 2x(x^2-4)+2x^3 \\ \Rightarrow && (4y^3-10y)y' &= 4x^3-8x \end{align*} Therefore the gradient is parallel to the \(x\)-axis when \(x = 0, x = \sqrt{2}\). We need \(x = 0, y \neq 0\), ie \(y = \sqrt{5}\), so \((0, \sqrt{5})\) and \((\sqrt{2}, 0)\) It is parallel to the \(y\)-axis when \(y = 0\) or \(y = \sqrt{\frac52}\), ie \((2, 0)\)
   

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3. \(\,\)
   

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