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Complex Numbers, EXT2 N2 2024 HSC 16b

The number  \(w=e^{\small{\dfrac{2 \pi i}{3}}}\)  is a complex cube root of unity. The number \(\gamma\) is a cube root of \(w\).

  1. Show that  \(\gamma+\bar{\gamma}\)  is a real root of  \(z^3-3 z+1=0\).   (3 marks)

  2. By using part (i) to find the exact value of  \(\cos \dfrac{2 \pi}{9} \cos \dfrac{4 \pi}{9} \cos \dfrac{8 \pi}{9}\), deduce the value(s) of  \(\cos \dfrac{2^n \pi}{9} \cos \dfrac{2^{n+1} \pi}{9} \cos \dfrac{2^{n+2} \pi}{9}\)  for all integers  \(n \geq 1\). Justify your answer.   (3 marks)

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i.     \(w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}\)

\(\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1\)

\(\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 
\(\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}\)

\(\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})\)

\(\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \  z^3-3 z+1=0.\)
 

ii.   \(e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}\)

\(\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}\)

\(\text{Roots of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})_1\) \(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_2\) \(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_3\) \(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

 

\(\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:\)

  \(2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)\) \(=-1\)
  \(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)\) \(=-\dfrac{1}{8}\)

 
\(\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):\)

\(\text{If} \ \ n=1,\)

\(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}\)

\(\text {If} \ \ n=2,\)

\(\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}\)

 
\(\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}\)

\(\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}\)

Show Worked Solution

i.     \(w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}\)

\(\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1\)

\(\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 

\(\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}\)

\(\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})\)

\(\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \  z^3-3 z+1=0.\)
 

ii.   \(e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}\)

\(\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}\)

\(\text{Roots of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})_1\) \(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_2\) \(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_3\) \(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

 

\(\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:\)

  \(2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)\) \(=-1\)
  \(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)\) \(=-\dfrac{1}{8}\)

 
\(\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):\)

\(\text{If} \ \ n=1,\)

\(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}\)

\(\text {If} \ \ n=2,\)

\(\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}\)

 
\(\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}\)

\(\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}\)

Complex Numbers, EXT2 N1 2023 HSC 7 MC

Which of the following statements about complex numbers is true?

  1. For all real numbers \(x, y, \theta\)  with  \(x \neq 0\),

\(\tan \theta=\dfrac{y}{x} \ \Rightarrow \ x+i y=r e^{i \theta}\), for some real number \(r\).

  1. For all non-zero complex numbers \(z_1\) and \(z_2\),

\(\operatorname{Arg}\left(z_1\right)=\theta_1\)  and  \(\operatorname{Arg}\left(z_2\right)=\theta_2 \ \Rightarrow \ \operatorname{Arg}\left(z_1 z_2\right)=\theta_1+\theta_2,\)

where \(\operatorname{Arg}\) denotes the principal argument.

  1. For all real numbers \(r_1, r_2, \theta_1, \theta_2\)  with  \(r_1, r_2>0\),

\(r_1 e^{i \theta_1}=r_2 e^{i \theta_2} \ \Rightarrow \ r_1=r_2\)  and  \(\theta_1=\theta_2 \text {. }\)

  1. For all real numbers \(x, y, r, \theta\)  with  \(r>0\)  and  \(x \neq 0\),

\(x+i y=r e^{i \theta} \ \Rightarrow \ \theta=\arctan  \Big(\dfrac{y}{x} \Big)\)

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\(A\)

Show Worked Solution

\(\text{Eliminating options by contradiction}\)

\(\text{Option}\ B:\)

\(\text{If}\ \ \theta_1= \pi\ \ \text{and}\ \ \theta_2=\dfrac{\pi}{2}, \ \theta_1 + \theta_2 = \dfrac{3\pi}{2} > \pi \)

\( -\pi < \operatorname{Arg}\left(z_1 z_2\right) < \pi\ \ \ \  \text{(Eliminate}\ B) \)

♦♦♦ Mean mark 18%.

\(\text{Option}\ C:\)

\(\text{If}\ \ \theta_1= \pi\ \ \text{and}\ \ \theta_2=3\pi, \ \operatorname{Arg}(e^{i\pi}) =  \operatorname{Arg}(e^{3i\pi}) \)

\( \text{However,}\ \ \theta_1 \neq \theta_2\ \ \ \  \text{(Eliminate}\ C) \)
 

\(\text{Option}\ D:\)

\(\text{If}\ \ x=y=-1, \ \theta=-\dfrac{3\pi}{4} \ \ (r>0) \)

\( \text{However,}\ \ \arctan\Big(\dfrac{-1}{-1}\Big)=\dfrac{\pi}{4} \ \ \  \text{(Eliminate}\ D) \)

\(\Rightarrow A\)

Complex Numbers, EXT2 N1 2020 HSC 9 MC

What is the maximum value of  `|e^(i theta) - 2| + |e^(i theta) + 2|`  for  `0 ≤ theta ≤ 2 pi`?

  1. `sqrt5`
  2. `4`
  3. `2 sqrt5`
  4. `10`
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`C`

Show Worked Solution

`e^(i theta) = a + ib \ \ text{where}\ \ a^2 + b^2 =1`

Mean mark 56%.
`X` `= |e^(i theta) – 2| + |e^(i theta) + 2|`
  `= | (a – 2) + ib | + | (a + 2) + i b |`
  `= sqrt((a-2)^2 + b^2) + sqrt((a+2)^2 + b^2)`
  `= sqrt(a^2 – 4a + 4 + b^2) + sqrt(a^2 + 4a + 4 + b^2`
  `= sqrt(5 – 4a) + sqrt(5 + 4a)`

 

`frac{dX}{da}` `= -frac{1}{2} xx 4 xx frac{1}{sqrt(5 – 4a)} + frac{1}{2} xx 4 xx frac{1}{sqrt(5 + 4a)}`
  `= frac{2}{sqrt(5 + 4a)} – frac{2}{sqrt(5 – 4a)}`

 
`text(When)\ \ frac{dX}{da} = 0:`

`5 + 4a` `= 5 – 4a`
`a` `= 0`

 
`:. X_text(max) = 2 sqrt5`
 
`=> \ C`