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Complex Numbers, EXT2 N2 2023 HSC 16a

Let \(w\) be the complex number  \(z=e^{\small{\dfrac{2i \pi}{3}}} \).

  1. Show that  \(1+w+w^2=0\).   (2 marks)

The vertices of a triangle can be labelled \(A, B\) and \(C\) in anticlockwise or clockwise direction, as shown.
 

Three complex numbers \(a, b\) and \(c\) are represented in the complex plane by points \(A, B\) and \(C\) respectively.

  1. Show that if triangle \(A B C\) is anticlockwise and equilateral, then  \(a+b w+c w^2=0\).   (2 marks)

  2. It can be shown that if triangle \(A B C\) is clockwise and equilateral, then \(a+b w^2+c w=0\). (Do NOT prove this.)
  3. Show that if \(A B C\) is an equilateral triangle, then

\(a^2+b^2+c^2=a b+b c+c a .\)  (2 marks)

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i.    \(w=e^{\small{\dfrac{2i \pi}{3}}} \)

\(w^2=e^{\small{-\dfrac{2i \pi}{3}}} = \overline w \)

\(\text{Re}(w)=\cos \Big{(}\dfrac{2\pi}{3} \Big{)}=-\dfrac{1}{2} \)

\(1+w+w^2\) \(=1+w+\overline w \)  
  \(=1+ 2 \times \text{Re}(w) \)  
  \(=1+2 \times -\dfrac{1}{2} \)  
  \(=0\)  

 
ii.    \(\text{Rotate}\ \ b-c\ \ \text{anticlockwise by}\ \ \dfrac{2\pi}{3} \)

\(\Rightarrow (b-c)w = c-a\)

\(\therefore bw-cw-c+a = a+bw-c(1+w) = 0 \)

\(\text{Substitute}\ \ 1+w=-w^2\ \ \text{(using part (i))} \)

\(a+bw-c(-w^2) \) \(=0\)  
\(a+bw+cw^2\) \(=0\ \ …\ \text{as required} \)  
♦♦♦ Mean mark (ii) 18%.

iii.   \(a+b w^2+c w=0\ \ \text{(if}\ \Delta ABC\ \text{is clockwise – given)} \)

\(a+bw+cw^2=0\ \ \text{(if}\ \Delta ABC\ \text{is anticlockwise)} \)

\(\Rightarrow (a+b w^2+c w)(a+bw+cw^2) = 0\ …\ (1)\)
 

\(\text{Expand (1) using part (i):}\ \ w+w^2=-1\ \ \text{and}\ \ w^3 = z^3 = 1 \)

\(a^2+abw^2+acw+bwa+b^2w^3+bcw^2+cw^2a+cbw^4+c^2w^3\) \(=0\)  
\(a^2+b^2+c^2+ab(w+w^2)+bc(w+w^2)+ca(w+w^2)\) \(=0\)  
\(a^2+b^2+c^2+ab(-1)+bc(-1)+ca(-1) \) \(=0\)  

 
\(\therefore a^2+b^2+c^2 = ab+bc+ca\ …\ \text{as required} \)

♦♦♦ Mean mark (iii) 11%.

Complex Numbers, EXT2 N2 2023 14a

Let \(z\) be the complex number  \(z=e^{\small{\dfrac{i \pi}{6}}} \)  and \(w\) be the complex number  \(w=e^{\small{\dfrac{3 i \pi}{4}}} \).

  1. By first writing \(z\) and \(w\) in Cartesian form, or otherwise, show that
  2.    \(|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}\).  (3 marks)

  3. The complex numbers \(z, w\) and \(z+w\) are represented in the complex plane by the vectors \(\overrightarrow{O A},\overrightarrow{O B}\) and \(\overrightarrow{O C}\) respectively, where \(O\) is the origin.
  4. Show that  \(\angle A O C=\dfrac{7 \pi}{24}\).  (2 marks)

  5. Deduce that  \(\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}\).  (1 mark)

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i.    \(z=e^{\small\dfrac{i \pi}{6}} = \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i \)

\(w=e^{\small\dfrac{3i \pi}{4}} = \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} \)

\(|z+w|^2\) \(=\Bigg{|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{|}\)  
  \(=\Bigg{|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{|}\)  
  \(=\Bigg{|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{|}\)  
  \(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)  
  \(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)  
  \(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)  
  \(=\dfrac{4-\sqrt6+\sqrt2}{2} \)  

 
ii.   

\(\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} \)

\( |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} \)

\(\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC \)

\(\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB \)

\(\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} \)
  

iii.   \(\text{In}\ \triangle AOC: \)

\( \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} \)

\(\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. \)
 

\(\text{Using the cos rule in}\ \triangle AOC: \)

\(\cos\,\dfrac{7\pi}{24}\) \(=\dfrac{|z|^2+|z+w|^2-|w|^2}{2|z||z+w|}\)  
  \(=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1  \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} \)  
  \(=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} \)  
  \(=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} \)  
  \(=\dfrac{8-2\sqrt6+2\sqrt2}{4} \)  
♦♦ Mean mark (iii) 26%.