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1998 Paper 1 Q5
D: 1484.0 B: 1529.9
complex numbers Argand diagram regular hexagon rotation rationalising denominators real and imaginary parts

  1. In the Argand diagram, the points \(Q\) and \(A\) represent the complex numbers \(4+6i\) and \(10+2i\). If \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) are the vertices, taken in clockwise order, of a regular hexagon (regular six-sided polygon) with centre \(Q\), find the complex number which represents \(B\).
  2. Let \(a\), \(b\) and \(c\) be real numbers. Find a condition of the form \(Aa+Bb+Cc=0\), where \(A\), \(B\) and \(C\) are integers, which ensures that \[\frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i}\] is real.

Solution:

  1. TikZ diagram
    We are looking for \((10+2i) - (4+6i) = 6 - 4i\) rotated by \(\frac{\pi}{3}\) and then added to \(4+6i\), which is \begin{align*} (6-4i)(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) &= (6-4i)\left(\tfrac12 +\tfrac{\sqrt{3}}2i\right) \\ &= 3+2\sqrt{3} + (3\sqrt{3}-2)i \end{align*}
  2. \begin{align*} &&& \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &\in \mathbb{R} \\ \Longleftrightarrow && \frac{a}{1+i}+\frac{b}{1+2i}+\frac{c}{1+3i} &= \frac{a}{1-i}+\frac{b}{1-2i}+\frac{c}{1-3i} \\ && 0 &= a\left ( \frac{1}{1+i} - \frac{1}{1-i} \right)+ b\left ( \frac{1}{1+2i} - \frac{1}{1-2i} \right)+ c\left ( \frac{1}{1+3i} - \frac{1}{1-3i} \right) \\ &&&= a\left ( \frac{(1-i)-(1+i)}{1^2+1^2} \right) + b\left ( \frac{(1-2i)-(1+2i)}{1^2+2^2} \right) + c\left ( \frac{(1-3i)-(1+3i)}{1^2+3^2} \right) \\ &&&= -\frac{2i}{2}a-\frac{4i}{5}b-\frac{-6i}{10}c \\ \Longleftrightarrow && 0 &= a+\tfrac45b+\tfrac35c \end{align*}

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