Complex Numbers, EXT2 N2 2023 14a*

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

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

--- 9 WORK AREA LINES (style=lined) ---
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.
Show that \(\angle A O C=\dfrac{7 \pi}{24}\). (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Deduce that \(\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}\). (1 mark)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

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

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

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

Show Worked Solution

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%.