If \(\bar{a}, \bar{b}, \bar{c}\) are the complex conjugates of \(a, b, c\), respectively, and if \(p, q, r\) are real, show that the equation \[ \begin{vmatrix} a-z & p & \bar{b} \\ c & q-z & b \\ \bar{c} & r & \bar{a}-z \end{vmatrix} = 0 \] has either three real roots or one real root and a pair of conjugate complex roots. It is given that \(q\) is a root when \(a=i, b=c=p=r=1\). Find \(q\) and solve the equation completely.
If the polynomial \[ ax^3+x^2-3bx+3b^2 \] has two coincident zeros show that, in general, it is a perfect cube. Hence, or otherwise, show that if \[ x^4+4ax^3+2x^2-4bx+3b^2 \] has three equal zeros then the fourth is identical with them, and find for what values of \(a\) and \(b\) this is the case.
Solution: Suppose \(ax^3+x^2-3bx+3b^2 = a(x-\alpha)^2(x-\beta)\), then if \(F(x) = ax^3+x^2-3bx+3b^2\) we must have \(F(\alpha) = F'(\alpha) = 0\), therefore: \begin{align*} && 0 &= a \alpha^3 + \alpha^2 - 3b\alpha + 3b^2 \\ &&0 &= 3a\alpha^2+2\alpha-3b \\ \Rightarrow && 3b &= \alpha(3a\alpha + 2) \\ \Rightarrow && 0 &= a \alpha^3 + \alpha^2 -\alpha^2(3a\alpha + 2) + \alpha^2(3a\alpha+2)^2/3 \\ &&&= \alpha^2 ((a-3a)\alpha-1 + \frac13 (9a^2\alpha^2+12a \alpha+4)) \\ &&&= \frac13 \alpha^2 (9a\alpha^2+6a\alpha+1) \\ &&&= \frac13 \alpha^2 (3a\alpha+1)^2 \end{align*} Therefore \(\alpha = 0\) or \(\alpha = - \frac{1}{3a}\). If \(\alpha = 0\), then \(b = 0\) and we don't (necessarily) have cubed roots. But in the general form (say when \(b \neq 0\)) we must have \(2\alpha + \beta = -\frac{1}{a} \Rightarrow \beta = -\frac{1}{3a} \Rightarrow \beta = \alpha\), therefore we have repeated roots. Similarly, if \(F(x) = x^4+4ax^3+2x^2-4bx+3b^2\) \(F'(x) = 4x^3+12ax^2+2x-4b = 12a(\frac1{3a} x^3+x^2+\frac{1}{6a} - \frac{b}{3a})\) then we must have \(\alpha\) is a root of \(F''(x) = 12x^2+24ax+4 = 4(3x^2+6ax+1)\)
Find the area common to a circle of radius \(a\) and an ellipse with semi-axes \(b\) and \(c\), where the circle and ellipse have the same centre and \(b>a>c\). If \(b=a(1+\epsilon), c=a(1-\epsilon)\), where \(\epsilon\) is a positive number small compared with unity, show that this area is approximately \[ \pi a^2 - 2\epsilon a^2. \]
Find the stationary values of
The sides of a triangle are \(a, b, c\) and the corresponding angles \(A, B, C\). Prove that
State and prove De Moivre's theorem for a (positive or negative) rational index. Evaluate \[ 32\int_0^\infty e^{-x}\cos^6 x \,dx, \] giving the answer to two decimal places.
Solution: Let \(z = e^{ix}\) \begin{align*} && 2\cos x &= z + \frac{1}{z} \\ \Rightarrow && 64 \cos^6x &= z^6 + 6z^4 + 15z^2 + 20 + 15z^{-2} + 6z^{-4} + z^{-6} \\ &&&= 2\cos 6x + 12\cos 4x + 30 \cos 2x + 20 \end{align*} \begin{align*} &&I &= \int_0^{\infty} e^{-x} \cos k x \d x \\ &&&= \left [ -e^{-x} \cos k x \right]_0^{\infty} -k\int_0^{\infty} e^{-x} \sin k x \d x \\ &&&= 1 - k \left ( [-e^{-x} \sin k x]_0^{\infty} + k\int_0^{\infty} e^{-x} \cos k x \d x\right) \\ \Rightarrow && (1+k^2)I &= 1 \\ \Rightarrow && I &= \frac{1}{1+k^2} \end{align*} Therefore \begin{align*} && I &= 32\int_0^\infty e^{-x}\cos^6 x \,dx \\ &&&= \frac12\int_0^\infty e^{-x}\left ( 2\cos 6x + 12\cos 4x + 30 \cos 2x + 20\right) \,dx \\ &&&= \frac{1}{6^2+1} + \frac{6}{4^2+1} + \frac{15}{2^2+1} + \frac{10}{0^2+1} \\ &&&= \frac{1}{37} + \frac{6}{17} + \frac{15}{5} + 10 \\ &&&= 13 + \frac{6}{17} + \frac{1}{37} \end{align*}
If \(\omega\) is a complex cube root of unity, show that \[ 1+\omega+\omega^2=0. \] It is given that \(a, b\) and \(c\) are real. Show that \((a+\omega b + \omega^2 c)^3\) is real if and only if \(a, b\) and \(c\) are not all different.
Describe, with the help of a rough sketch, the form of the curve \[ x = \cos \frac{y}{x}. \] Prove that all the loops have the same area, namely \(\frac{1}{2}\pi\).
Sum the infinite series
Show that \[ e^{a^2}\int_a^\infty e^{-x^2}\,dx = \frac{1}{2a}\left\{ 1 + \sum_{r=1}^n (-)^r \frac{1 \cdot 3 \cdot 5 \dots (2r-1)}{(2a^2)^r} \right\} + (-)^{n+1}R_n, \] where \[ R_n = \frac{1 \cdot 3 \cdot 5 \dots (2n+1)}{2^{n+1}} e^{a^2} \int_a^\infty \frac{e^{-x^2}}{x^{2n+2}}\,dx. \] Establish that, if \(a\) is positive, \(R_n\) is less than \[ \frac{1 \cdot 3 \cdot 5 \dots (2n-1)}{2^{n+1}} \frac{1}{a^{2n+1}}. \]