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2000 Paper 2 Q4
D: 1600.0 B: 1500.0
complex numbers De Moivre proof by induction arctan identity trig identity argument of complex number Gaussian integers

Prove that \[ (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) = \cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi) \] and that, for every positive integer \(n\), $$ {(\cos {\theta} + \mathrm{i}\sin {\theta})}^n = \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}. $$ By considering \((5-\mathrm{i})^2(1+\mathrm{i})\), or otherwise, prove that \[ \arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,. \] Prove also that \[ 3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,. \] [Note that \(\arctan\theta\) is another notation for \(\tan^{-1}\theta\).]

Solution: \begin{align*} && LHS &= (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) \\ &&&= \cos \theta \cos \phi - \sin \theta \sin \phi + \mathrm{i}(\sin \theta \cos \phi + \cos \theta \sin \phi) \\ &&&= \cos (\theta + \phi) + \mathrm{i} \sin (\theta + \phi) \\ &&&= RHS \end{align*} Therefore we can see \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n \theta\). \begin{align*} && (5-i)^2(1+i) &= (24-10i)(1+i) \\ &&&= (24+10) + i(24-10) \\ &&&= 34+14i \\ \Rightarrow && 2\arg(5-i) +\arg(1+i) &= \arg(34+14i) \\ \Rightarrow && 2\arctan\left (-\frac{1}{5} \right) + \frac{\pi}{4} &= \arctan \left ( \frac{7}{17} \right) \\ \Rightarrow && 2\arctan\left (\frac{1}{5} \right) +\arctan \left ( \frac{7}{17} \right) &= \frac{\pi}{4} \\ \end{align*} Consider \((1+i)(4-i)^3(20-i)\) \begin{align*} && (1+i)(4-i)^3(20-i) &= (21+19i)(52-47i) \\ &&&= 1985+i \\ \Rightarrow && \frac{\pi}{4} - 3 \arctan \left ( \frac{1}{4} \right) -\arctan \left ( \frac{1}{20} \right) &= \arctan \left ( \frac{1}{1985} \right) \end{align*}

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1987 Paper 3 Q3
D: 1500.0 B: 1500.0
complex numbers modulus inequality argument arctan identity Machin formula complex multiplication proof

  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 \[ \frac{\pi}{4}=3\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{1}{20}\right)+\tan^{-1}\left(\frac{1}{1985}\right). \]

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 \(\displaystyle \frac{\pi}{4}=3\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{20}+\tan^{-1}\frac{1}{1985}\)

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