Complex numbers 2

Eulers formulae, de moivre, roots of unity

Showing 26-32 of 32 problems
1990 Paper 3 Q1
D: 1700.0 B: 1516.0

Show, using de Moivre's theorem, or otherwise, that \[ \tan9\theta=\frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)},\qquad\mbox{ where }t=\tan\theta. \] By considering the equation \(\tan9\theta=0,\) or otherwise, obtain a cubic equation with integer coefficients whose roots are \[ \tan^{2}\left(\frac{\pi}{9}\right),\qquad\tan^{2}\left(\frac{2\pi}{9}\right)\qquad\mbox{ and }\qquad\tan^{2}\left(\frac{4\pi}{9}\right). \] Deduce the value of \[ \tan\left(\frac{\pi}{9}\right)\tan\left(\frac{2\pi}{9}\right)\tan\left(\frac{4\pi}{9}\right). \] Show that \[ \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right)=33273. \]

Show Solution
Writing \(c = \cos \theta, s = \sin \theta\) then de Moivre states that: \begin{align*} && \cos 9 \theta + i \sin 9 \theta &= (c +i s)^9 \\ &&&= c^9 + 9ic^8s - 36c^7s^2-84ic^6s^3+126c^5s^4 + 126ic^4s^5 -84c^3s^6 -36ic^2s^7+9cs^8+is^9 \\ &&&= (c^9-36c^7s^2+126c^5s^3-84c^3s^6+8cs^8)+i(9c^8s-75c^6s^3+126c^4s^5-36c^2s^7+s^9) \\ \Rightarrow && \tan 9\theta &= \frac{(9c^8s-75c^6s^3+126c^4s^5-36s^2c^7+s^9)}{(c^9-36c^7s^2+126c^5s^4-84c^3s^6+8cs^8)} \\ &&&= \frac{9t-75t^3+126s^5-36t^7+t^9}{1-36t^2+126t^4-84t^6+8t^8} \\ &&&= \frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)} \end{align*} If we consider \(\tan 9\theta = 0\) it will have the roots \(\theta = \frac{n \pi}{9}, n \in \mathbb{Z}\), in particular, the numerator of our fraction for \(\tan 9 \theta\) will be zero for \(t = 0, \tan \frac{\pi}{9}, \tan \frac{2\pi}{9}, \tan \frac{3\pi}{9}, \tan \frac{4 \pi}{9}, \tan \frac{5\pi}{9}, \tan \frac{6 \pi}{9}, \tan \frac{7 \pi}{9}, \tan \frac{8\pi}{9}\). It's worth noting all other values of \(\theta\) will repeat these values. Also note that \(0,\tan \frac{\pi}{3}, \tan \frac{2\pi}{3}\) are the roots of \(t\) and \(t^2-3\) respectively. Therefore the other values are the roots of our sextic. However, also note that \(\tan \frac{8\pi}{9} = - \tan \frac{\pi}{9}\) and similar, therefore we can notice that all the roots in pairs can be mapped to \(\tan \frac{\pi}{9}, \tan \frac{2 \pi}{9}\) and \(\tan \frac{4 \pi}{9}\) and all those values are squared, so the roots of: \(x^3 - 33x^2+27x-3\) will be \(\tan^2 \frac{\pi}{9}, \tan^2 \frac{2 \pi}{9}\) and \(\tan^2 \frac{4 \pi}{9}\). The product of the roots will be \(3\), so \begin{align*} && \tan^2 \frac{\pi}{9} \tan^2 \frac{2 \pi}{9} \tan^2 \frac{4 \pi}{9} &= 3 \\ \Rightarrow && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \pm \sqrt{3} \\ \underbrace{\Rightarrow}_{\text{all positive}} && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \sqrt{3} \\ \end{align*} Notice that \(x^3 + y^3 +z^3 - 3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))\) Therefore \begin{align*} \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right) &= 33(33^2-3\cdot27) + 3 \cdot 3 \\ &= 33\,273 \end{align*}
1990 Paper 3 Q4
D: 1700.0 B: 1516.0

Given that \(\sin\beta\neq0,\) sum the series \[ \cos\alpha+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta) \] and \[ \cos\alpha+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta). \] Given that \(\sin\theta\neq0,\) prove that \[ 1+\cos\theta\sec\theta+\cos2\theta\sec^{2}\theta+\cdots+\cos r\theta\sec^{r}\theta+\cdots+\cos n\theta\sec^{n}\theta=\frac{\sin(n+1)\theta\sec^{n}\theta}{\sin\theta}. \]

Show Solution
\begin{align*} \sum_{r = 0}^n \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha} \sum_{r = 0}^n \ (e^{i 2 \beta})^r\right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{2(n+1)\beta i}-1}{e^{2\beta i}-1} \right) \\ &= \textrm{Re} \left (e^{i \alpha} \frac{e^{(n+1)\beta i} (e^{(n+1)\beta i}-e^{-(n+1)\beta i})}{e^{\beta i}(e^{\beta i}-e^{-\beta i})} \right) \\ &= \textrm{Re} \left (\frac{e^{i \alpha} e^{(n+1)\beta i}}{e^{\beta i}} \frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \textrm{Re} \left ( e^{i(\alpha + n \beta)}\frac{\sin (n+1) \beta}{\sin \beta} \right) \\ &= \frac{\cos (\alpha + n \beta)\sin (n+1) \beta}{\sin \beta} \end{align*} \begin{align*} \sum_{r = 0}^n \binom{n}{r} \cos (\alpha + 2r \beta) &= \sum_{r = 0}^n \textrm{Re} \left ( \binom{n}{r}\exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (\sum_{r = 0}^n \binom{n}{r} \exp(i(\alpha + 2r \beta)) \right) \\ &= \textrm{Re} \left (e^{i \alpha}(e^{2\beta i}+1)^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}(e^{\beta i}+e^{-\beta i})^n \right) \\ &= \textrm{Re} \left (e^{i \alpha}e^{n\beta i}2^n \cos^n \beta \right) \\ &= 2^n \cos(\alpha + n \beta) \cos ^n \beta \end{align*} \begin{align*} \sum_{r = 0}^n \cos r \theta \sec^r \theta &= \sum_{r = 0}^n \textrm{Re} ( e^{i r \theta})\sec^r \theta \\ &= \textrm{Re} \left ( \sum_{r=0}^n e^{i r \theta} \sec^r \theta\right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n+1} \theta -1}{e^{i \theta}\sec \theta -1} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{e^{i \theta} -\cos \theta} \right) \\ &= \textrm{Re} \left ( \frac{e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta}{i \sin \theta} \right) \\ &= \frac{1}{\sin \theta} \textrm{Im} \left ( e^{i (n+1) \theta}\sec^{n} \theta -\cos \theta \right) \\ &= \frac{\sin(n+1) \theta \sec^{n} \theta}{\sin \theta} \end{align*}
1989 Paper 1 Q8
D: 1500.0 B: 1516.0

By using de Moivre's theorem, or otherwise, show that

  1. \(\cos4\theta=8\cos^{4}\theta-8\cos^{2}\theta+1;\)
  2. \(\cos6\theta=32\cos^{6}\theta-48\cos^{4}\theta+18\cos^{2}\theta-1.\)
Hence, or otherwise, find all the real roots of the equation \[ 16x^{6}-28x^{4}+13x^{2}-1=0. \] [No credit will be given for numerical approximations.]

Show Solution
Given that \(e^{i \theta} = \cos \theta + i \sin \theta\) we must have that
  1. \begin{align*} \cos 4 \theta &= \textrm{Re} \l e^{i 4 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^4 \r \\ &= \cos^4 \theta - \binom{4}{2}\cos^2 \theta \sin^2 \theta +\sin^4 \theta \\ &= \cos^4 \theta - 6\cos^2 \theta (1-\cos^2 \theta) +(1-\cos^2 \theta)^2 \\ &= 8\cos^4 \theta - 8\cos^2 \theta + 1 \end{align*}
  2. Similarly, \begin{align*} \cos 6 \theta &= \textrm{Re} \l e^{i 6 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^6 \r \\ &= \cos^6 \theta -\binom{6}{2}\cos^4 \theta \sin^2 \theta +\binom{6}{4} \cos^2\theta \sin^4 \theta - \sin^6 \theta \\ &= \cos^6 \theta - 15 \cos^4 \theta (1-\cos^2 \theta) + 15\cos^2 \theta (1-\cos^2\theta)^2 - (1-\cos^2 \theta)^3\\ &= 31\cos^6 \theta-45\cos^4\theta+15\cos^2\theta-1+3\cos^2 \theta-3\cos^4 \theta+\cos^6 \theta \\ &= 32 \cos^6 \theta-48\cos^4 \theta+18\cos^2 \theta-1 \end{align*}
\begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1\\ &= \frac12 (32x^6-56x^4+26x^2-1) \\ &= \frac12(32x^6-48x^4+18x^2-1-(8x^4-8x^2+1)) \end{align*} Therefore if \(x = \cos \theta\) then we are looking at solving \(\cos 6 \theta = \cos 4 \theta\). \(\cos 6 \theta - \cos 4 \theta = -2 \sin 5\theta \sin \theta = 0\). So we should be looking at \(\sin 5 \theta = 0\) and \(\sin \theta = 0\). \(\sin \theta = 0 \Rightarrow x = \cos \theta = \pm 1\) both of which are roots. The other roots will be \(\cos \frac{\pi}{5}, \cos \frac{2\pi}{5}\) etc but it's unclear this is an acceptable form. Alternatively, given our two roots, we can factorize \begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1 \\ &= (x^2-1)(16x^4-12x^2+1) \end{align*} We can solve \(16y^2-12y+1=0\) to see that \(x^2 = \frac{3 \pm \sqrt{5}}{8}\) so our roots are: \(x = -1, 1, \pm \sqrt{\frac{3 + \sqrt{5}}{8}}, \pm \sqrt{\frac{3 -\sqrt{5}}{8}}\) (We might notice that \(3+\sqrt{5} =\l \frac{1+\sqrt{5}}{\sqrt{2}} \r^2\) so our final answer could be: \(x = -1, 1, \pm \frac{1+\sqrt{5}}{4}, \pm \frac{\sqrt{5}-1}{4}\))
1988 Paper 2 Q5
D: 1600.0 B: 1484.0

By considering the imaginary part of the equation \(z^{7}=1,\) or otherwise, find all the roots of the equation \[ t^{6}-21t^{4}+35t^{2}-7=0. \] You should justify each step carefully. Hence, or otherwise, prove that \[ \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7}=\sqrt{7}. \] Find the corresponding result for \[ \tan\frac{2\pi}{n}\tan\frac{4\pi}{n}\cdots\tan\frac{(n-1)\pi}{n} \] in the two cases \(n=9\) and \(n=11.\)

Show Solution
Suppose \(z^7 = 1\), then we can write \(z = \cos \theta + i \sin \theta\) and we must have that: \begin{align*} 0 &= \textrm{Im}((\cos \theta + i \sin \theta)^7) \\ &= \binom{7}{6}\cos^6 \theta \sin \theta - \binom{7}{4} \cos^4 \theta \sin^3 \theta + \binom{7}{2} \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= 7 \cos^6 \sin \theta - 35 \cos^4 \theta \sin ^3 \theta + 21 \cos^2 \theta \sin^5 \theta - \sin^7 \theta \\ &= -\cos^7 \theta \l \tan^7 \theta - 21 \tan^5 \theta + 35 \tan^3 \theta - 7 \tan \theta\r \\ &= \cos^7 \theta \cdot t (t^7-21t^4+35t^2-7) \end{align*} Where \(t = \tan \theta\). So if \(z\) is a root of \(z^7 = 1\) and \(\cos \theta \neq 0, \tan \theta \neq 0\) then \(t\) is a root of the equation. Thererefore the roots are: \(\tan \frac{2\pi k}{7}\) where \(k = 1, 2, \ldots 6\). Noting that \(\tan \frac{\pi}7 = -\tan \frac{6\pi}{7}, \tan \frac{3\pi}{7} = -\tan \frac{4 \pi}{7}, \tan \frac{5\pi}{7} = -\tan \frac{2 \pi}{7}\) we can conclude that: \begin{align*} && 7 &= \prod_{k=1}^k \tan \frac{k \pi}{6} \\ &&&= \l \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \r^2 \\ \Rightarrow&& \pm \sqrt{7} &= \tan\frac{2\pi}{7}\tan\frac{4\pi}{7}\tan\frac{6\pi}{7} \end{align*} However, we know that \(\tan \frac{2\pi}{7}\) is positive, \(\tan \frac{4\pi}{7},\tan \frac{6\pi}{7}\) are negative, therefore the result must be positive, ie \(+\sqrt{7}\) Using a similar method, we notice that: \begin{align*} 0 &= \textrm{Im} \l (\cos \theta + i \sin \theta)^n \r \\ &= \cos^n \theta \cdot t (t^{n-1} + \cdots - \binom{n}{n-1}) \end{align*} Therefore \(\prod_{k=0}^{n-1} \tan \frac{k \pi}{n} = n\) and since \(\tan \frac{(2k+1) \pi}{n} = \tan \frac{(n-2k-1)\pi}{n}\) is a map of all the odd numbers to the even numbers (and vice versa) when \(n\) is odd. We also know that the terms less where \(\tan \theta\) has \(\theta < \frac{\pi}{2}\) are positive, and the others even, we can determine the signs: \begin{align*} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} \tan \frac{6 \pi}{9} \tan \frac{8 \pi}{9} & = 3 \\ \tan \frac{2 \pi}{11} \tan \frac{4 \pi}{11} \tan \frac{6 \pi}{11} \tan \frac{8 \pi}{11} \tan \frac{10 \pi}{11} &= -\sqrt{11} \end{align*}
1987 Paper 1 Q7
D: 1500.0 B: 1500.0

Sum each of the series \[ \sin\left(\frac{2\pi}{23}\right)+\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)+\cdots+\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right) \] and \[ \sin\left(\frac{2\pi}{23}\right)-\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)-\cdots-\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right), \] giving each answer in terms of the tangent of a single angle. {[}No credit will be given for a numerical answer obtained purely by use of a calculator.{]}

Show Solution
\(\sin x = \frac{e^{ix} - e^{-ix}}{2i}\). Also let \(z = e^{ \frac{2\pi i}{23}}\) \begin{align*} \sum_{k=0}^{10} \sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} \exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} z^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}-1}{z^2-1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}-z^{-11})}{z(z-z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2i \sin \frac{22 \pi}{23} }{2i \sin \frac{2 \pi}{23}} \r \r \\ &= \frac{\sin \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\sin^2 \frac{22 \pi}{23}}{\sin \frac{2 \pi}{23}} \\ &= \frac{\sin^2 \frac{\pi}{23}}{2\sin \frac{\pi}{23}\cos \frac{\pi}{23}} \\ &= \frac12 \tan \frac{\pi}{23} \end{align*} Similarly, \begin{align*} \sum_{k=0}^{10} (-1)^k\sin \l \frac{(4k +2)\pi}{23} \r &= \sum_{k=0}^{10} \textrm{Im} \l (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l \sum_{k=0}^{10} (-1)^k\exp\l \frac{(4k +2)\pi i}{23} \r \r \\ &= \textrm{Im} \l e^{ \frac{2\pi i}{23}} \sum_{k=0}^{10} (-1)^kz^{2k} \r \\ &= \textrm{Im} \l z \l \frac{z^{22}+1}{z^2+1} \r \r \\ &= \textrm{Im} \l z \l \frac{z^{11}(z^{11}+z^{-11})}{z(z+z^{-1})} \r \r \\ &= \textrm{Im} \l \frac{z^{11}2 \cos \frac{22 \pi}{23} }{2 \cos\frac{2 \pi}{23}} \r \r \\ &= \frac{\cos\frac{22 \pi}{23}}{\cos \frac{2 \pi}{23}} \textrm{Im} ( z^{11}) \\ &= \frac{\cos \frac{22 \pi}{23}\sin \frac{22 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{\sin \frac{44 \pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= \frac12 \frac{-\sin \frac{2\pi}{23}}{\cos\frac{2 \pi}{23}} \\ &= -\frac12 \tan \frac{2\pi}{23} \end{align*}
1987 Paper 2 Q4
D: 1500.0 B: 1500.0

Explain the geometrical relationship between the points in the Argand diagram represented by the complex numbers \(z\) and \(z\mathrm{e}^{\mathrm{i}\theta}.\) Write down necessary and sufficient conditions that the distinct complex numbers \(\alpha,\beta\) and \(\gamma\) represent the vertices of an equilateral triangle taken in anticlockwise order. Show that \(\alpha,\beta\) and \(\gamma\) represent the vertices of an equilateral triangle (taken in any order) if and only if \[ \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta=0. \] Find necessary and sufficient conditions on the complex coefficients \(a,b\) and \(c\) for the roots of the equation \[ z^{3}+az^{2}+bz+c=0 \] to lie at the vertices of an equilateral triangle in the Argand digram.

Show Solution
The point \(ze^{i\theta}\) is obtained by rotating the point \(z\) about \(0\) by an angle \(\theta\) anticlockwise. The complex numbers \(\alpha, \beta\) and \(\gamma\) will form an equilateral triangle iff the angles between each side are \(\frac{\pi}{3}\), ie \begin{align*} \begin{cases}{\gamma - \beta} &= e^{i \frac{\pi}{3}}({\beta - \alpha}) \\ {\alpha- \gamma} &= e^{i \frac{\pi}{3}}({\gamma- \beta}) \\ {\beta- \alpha} &= e^{i \frac{\pi}{3}}({\alpha- \gamma})\end{cases} \end{align*} We don't need all these equations, since the first two are equivalent to the third. Combining the first two equations, we have \begin{align*} && \frac{\gamma - \beta}{\beta-\alpha} &= \frac{\alpha-\gamma}{\gamma - \beta} \\ \Leftrightarrow && (\gamma - \beta)^2 &= (\alpha-\gamma)(\beta-\alpha) \\ \Leftrightarrow && \gamma^2 +\beta^2 - 2\gamma \beta &= \alpha\beta-\alpha^2-\gamma\beta+\gamma\alpha \\ \Leftrightarrow && \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta&=0 \end{align*} as required. If the roots of \(z^{3}+az^{2}+bz+c=0\) are \(\alpha, \beta, \gamma\) then \(\alpha+\beta+\gamma = -a\) and \(\beta\gamma+\gamma\alpha+\alpha\beta = b\). We also have that \(a^2 - 2b = \alpha^2+\beta^2+\gamma^2\). Therefore there roots will lie at the vertices of an equilateral triangle iff \(a^2-3b = 0\)
1987 Paper 3 Q3
D: 1500.0 B: 1500.0

  1. If \(z=x+\mathrm{i}y,\) with \(x,y\) real, show that \[ \left|x\right|\cos\alpha+\left|y\right|\sin\alpha\leqslant\left|z\right|\leqslant\left|x\right|+\left|y\right| \] for all real \(\alpha.\)
  2. By considering \((5-\mathrm{i})^{4}(1+\mathrm{i}),\) show that \[ \frac{\pi}{4}=4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right). \] Prove similarly that \[ \frac{\pi}{4}=3\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{1}{20}\right)+\tan^{-1}\left(\frac{1}{1985}\right). \]

Show Solution
  1. If \(z=x+iy\) then \(|z|^2 = x^2 + y^2 \leq x^2 + y^2 + 2|x||y| \leq (|x|+|y|)^2\). The LHS is Cauchy-Schwarz with the vectors \(\begin{pmatrix} |x| \\ |y| \end{pmatrix}\) and \(\begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix}\), although that's not in the spirit of the question. Consider \(e^{i \alpha}z = (\cos \alpha x - \sin \alpha y) + i(\sin \alpha x + \cos \alpha y)\) then \(\left | \textrm{Re}(e^{i \alpha} z) \right | \leq |z|\) for all values of \(\alpha\) and in particular we can choose \(\alpha\) to match the signs of the \(x\) and \(y\) to prove the result in question.
  2. Consider \((5-\mathrm{i})^{4}(1+\mathrm{i})\), then \begin{align*} \arg \l (5-\mathrm{i})^{4}(1+\mathrm{i}) \r &= \arg (5-i)^4 + \arg (1+i) \\ &= 4 \arg (5-i) + \arg (1+i) \\ &= -4 \tan^{-1} \frac{1}{5} + \tan^{-1} 1 \\ \\ &= \arg ( (24 - 10i)^2 (1+i)) \\ &= \arg (4 (12 - 5i)^2(1+i)) \\ &= \arg ((119 - 120i)(1+i)) \\ &= \arg (239 - i) \\ &= -\tan^{-1} \frac{1}{239} \end{align*} Therefore \(\displaystyle \frac{\pi}{4} =4 \tan^{-1} \frac{1}{5}- \tan^{-1} \frac{1}{239}\) Consider \((4-i)^3(1+i)(20-i)\) then \begin{align*} \arg \l(4-i)^3(1+i)(20-i) \r &= -3 \tan^{-1} \frac14 + \tan^{-1} 1 -\tan^{-1} \frac1{20} \\ \\ &= \arg \l(15-8i)(4-i)(1+i)(20-i) \r \\ &= \arg \l (52 - 47i)(1+i)(20-i) \r \\ &= \arg \l (99 + 5i)(20-i) \r \\ &= \arg (1985+i) \\ &= \tan^{-1} \frac1{1985} \end{align*} Therefore \(\displaystyle \frac{\pi}{4}=3\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{20}+\tan^{-1}\frac{1}{1985}\)