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Complex Numbers, EXT2 N2 2023 14a*

Let \(z\) be the complex number  \(z=\text{cis}\dfrac{\pi}{6} \)  and \(w\) be the complex number  \(w=\text{cis}\dfrac{3\pi}{4} \).

  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.    \(\text{See Worked Solutions}\)

ii.   \(\text{See Worked Solutions}\)

iii.  \(\text{See Worked Solutions}\)

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

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

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

 
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%.

Complex Numbers, EXT2 N2 2025 HSC 6 MC

The complex numbers \(z\) and \(w\) lie on the unit circle. The modulus of  \(z+w\)  is \(\dfrac{3}{2}\).

What is the modulus of  \(z-w\) ?

  1. \(\dfrac{1}{8}\)
  2. \(\dfrac{\sqrt{7}}{2}\)
  3. \(\dfrac{3}{2}\)
  4. \(\dfrac{7}{4}\)
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\(B\)

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\(|z|=|w|=1, \quad \abs{z+w}=\dfrac{3}{2} \ \text{(given)}\)

\(\text{Find}\ \ \abs{z-w}:\)

\(\abs{z+\omega}^2=(z+\omega)(\bar{z}+\bar{\omega})=z \bar{z}+z \bar{\omega}+\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z-\omega}^2=(z-\omega)(\bar{z}-\bar{\omega})=\bar{z} z-z \bar{\omega}-\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z+\omega}^2+|z-\omega|^2\) \(=2 z \bar{z}+2 \omega \bar{\omega}=2\abs{z}^2+2\abs{\omega}^2=4\)
\(\dfrac{9}{4}+\abs{z-\omega}^2\) \(=4\)
\(\abs{z-w}^2\) \(=\dfrac{7}{4}\)
\(\abs{z-\omega}\) \(=\dfrac{\sqrt{7}}{2}\)

 
\(\Rightarrow B\)

Mean mark 56%.

Complex Numbers, EXT2 N2 2024 HSC 14c

For the complex numbers \(z\) and \(w\), it is known that  \(\arg \left(\dfrac{z}{w}\right)=-\dfrac{\pi}{2}\).  

Find \(\left|\dfrac{z-w}{z+w}\right|\).   (2 marks)

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\(\abs{\dfrac{z-w}{z+w}}=1\)

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\(\arg \left(\dfrac{z}{w}\right)=\arg \, z-\arg \, w=-\dfrac{\pi}{2}\)

\(\text{Graphically, \(\arg \, z\) is a \(90^{\circ}\) anticlockwise rotation from \(\arg \, w\).}\)

♦ Mean mark 49%.

\(\abs{z-w}=\abs{z+w} \ \text{(diagonals of rectangle)}\)

\(\therefore \abs{\dfrac{z-w}{z+w}}=1\)