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Complex Numbers, EXT2 N2 2025 HSC 15c

  1. Show that
  2.     \(\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.\)   (3 marks)

  3. Use De Moivre's theorem to show that the sixth roots of \(-1\) are given by 
  4.    \(\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)\)  for  \(k=0,1,2,3,4,5\).   (2 marks)

  5. Hence, or otherwise, show the solutions to  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)  are 
  6. \(z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)\), and \(i \cot \left(\dfrac{11 \pi}{12}\right)\).   (2 marks)

Show Answers Only

i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

Show Worked Solution

i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)

♦♦♦ Mean mark (iii) 23%.

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

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

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

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

♦♦ Mean mark (i) 36%.
  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 

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

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

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

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

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

♦♦♦ Mean mark (ii) 18%.

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

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

 

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

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

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

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

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

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

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

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

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

Complex Numbers, EXT2 N2 2023 HSC 12d

Find the cube roots of  \(2-2 i\). Give your answer in exponential form.  (3 marks)

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\(\sqrt[3]{2-2i}= \sqrt2 e^{- \frac{i\pi}{12}}, \sqrt2 e^{\frac{i7\pi}{12}}, \sqrt2 e^{- \frac{i3\pi}{4}} \)

Show Worked Solution

\(\text{Express}\ \ 2-2i\ \ \text{in exponential form:}\)
 

\(\big{|}2-2i\big{|}=2\sqrt2 \)

\(\arg(2-2i) = – \dfrac{\pi}{4} \)

\(2-2i=2\sqrt2 e^{-\frac{i\pi}{4}} \)
 

\(\text{Find}\ \ z=re^{i\theta},\ \ \text{where}\ \ z^3=r^3e^{i 3\theta}=2\sqrt2 e^{-\frac{i\pi}{4}} \)

\(r^3=2\sqrt2\ \ \Rightarrow\ \ r=\sqrt2 \)

\( i3\theta\) \(= -\dfrac{i\pi}{4} \)  
\(\theta_1\) \(=-\dfrac{\pi}{12} \)  

 
\(\text{Since roots are found symmetrically around Argand diagram:}\)

\(\theta_2=-\dfrac{\pi}{12}+\dfrac{2\pi}{3}=\dfrac{7\pi}{12} \)

\(\theta_3=-\dfrac{\pi}{12}-\dfrac{2\pi}{3}=-\dfrac{3\pi}{4} \)

\(\therefore \sqrt[3]{2-2i}= \sqrt2 e^{- \frac{i\pi}{12}}, \sqrt2 e^{\frac{i7\pi}{12}}, \sqrt2 e^{- \frac{i3\pi}{4}} \)

Complex Numbers, EXT2 N2 2022 HSC 16d

Find all the complex numbers `z_1, z_2, z_3` that satisfy the following three conditions simultaneously.  (3 marks)

`{[|z_(1)|=|z_(2)|=|z_(3)|],[z_(1)+z_(2)+z_(3)=1],[z_(1)z_(2)z_(3)=1]:}`

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`z_1,z_2,z_3=1,i,-i \ \ text{(in any order)}`

Show Worked Solution

`z_1z_2z_3=1\ \ =>\ \ abs(z_1) abs(z_2) abs(z_3)=1`

`text{Given}\ \ abs(z_1) = abs(z_2) = abs(z_3)`

`=>abs (z_1)^3=1 \ \ => \ \ abs(z_1)=1`

`:.abs(z_1) = abs(z_2) = abs(z_3)=1`
 


♦♦♦ Mean mark 12%.

`text{Consider}\ \ z_1=e^(itheta):`

`1/z_1=e^(-itheta)=bar(e^(itheta))=bar(z)_1`

`text{Similarly,}`

`1/z_2=bar(z)_2, \ 1/z_3=bar(z)_3`

`=>z_1z_2z_3=1`

  
`text{Consider}\ \ z_1z_2:`

`z_1z_2=1/z_3=barz_3`

`text{Similarly,}`

`z_2z_3=1/z_1=barz_1, \ z_1z_3=1/z_2=barz_2`

`z_1z_2+z_1z_3+z_2z_3` `=barz_3+barz_2+barz_1`  
  `=bar(z_1+z_2+z_3)`  
  `=1`  

 
`z_1, z_2,z_3\ \ text{are zeros of polynomial:}`

`z^3-z^2+z-1` `=0`  
`(z-1)(z^2+1)` `=0`  

 
`z_1,z_2,z_3=1,i,-i \ \ text{(in any order)}`

Complex Numbers, EXT2 N2 2021 HSC 14c

  1. Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that
  2.    `cos5theta = 16cos^5theta - 20cos^3 theta + 5cos theta`.  (3 marks)

  3. By using part (i), or otherwise, show that  `text(Re)(e^((ipi)/10)) = sqrt((5 + sqrt5)/8)`.  (3 marks)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `(cos theta + i sin theta)^5 = cos5theta + i sin 5theta\ \ text{(by De Moivre)}`

`text(Using binomial expansion:)`

`(cos theta + i sin theta)^5`

`= cos^5theta + 5cos^4theta · isin theta + 10cos^3theta · i^2sin^2theta + 10 cos^2theta · i^3sin^3theta`

`+ 5costheta · i^4sin^4theta + i^5sin^5theta`

`= cos^5theta – 10cos^3thetasin^2theta + 5costhetasin^4theta + i\ \ text{(imaginary part)}`
 

`text(Equating real parts:)`

`cos5theta` `= cos^5theta – 10cos^3thetasin^2theta + 5costhetasin^4theta`
  `= cos^5theta – 10cos^3theta(1 – cos^2theta) + 5costheta(1 – cos^2theta)sin^2theta`
  `= cos^5theta – 10cos^3theta + 10cos^5theta + (5costheta – 5cos^3theta)(1 – cos^2theta)`
  `= 11cos^5theta – 10cos^3theta + 5costheta – 5cos^3theta – 5cos^3theta + 5cos^5theta`
  `= 16cos^5theta – 20cos^3theta + 5costheta`

 

ii.   `text(Re)(e^((ipi)/10)) = cos\ pi/10`

♦♦ Mean mark 31%.

`text(Using part a:)`

`16 cos^5(pi/10) – 20cos^3(pi/10) + 5cos(pi/10) = cos((5pi)/10) = cos(pi/2) = 0`

`text(Solutions to)\ cos(5theta) = 0`

`5theta` `= pi/2, (3pi)/2, (5pi)/2, …`
`theta` `= pi/10, (3pi)/10, (5pi)/10, …`

  
`text(Let)\ cos(pi/10) = x`

`16x^5 – 20x^3 + 5x` `= 0`
`16x^4 – 20x^2 + 5` `= 0`
`x^2` `= (20 ± sqrt(20^2 – 4 · 16 · 5))/(2 xx 16)`
  `= (20 ± sqrt80)/32`
  `= (5 ±sqrt5)/8`
`x` `= ±sqrt((5 ± sqrt5)/8)`

 
`cos\ pi/10 > 0 \ => \ x = sqrt((5 ± sqrt5)/8)`

`1 > cos\ pi/10 > cos\ (3pi)/10 > 0`

`:.\ text(Re)\ (e^((ipi)/10)) = cos\ pi/10 = sqrt((5 + sqrt5)/8)`

Complex Numbers, EXT2 N2 SM-Bank 1

Find all solutions for `z`, in exponential form, given  `z^4 = -2 sqrt3 - 2 i`.   (3 marks)

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`sqrt2 e^(-frac{i 17 pi}{24}) \ , \  sqrt2 e^(-frac{i 5 pi}{24}) \ , \  sqrt2 e^(frac{i 7 pi}{24}) \ , \  sqrt2 e^(frac{i 19 pi}{24})`

Show Worked Solution

`text{Convert} \ \ z^4\ \ text{to Mod/Arg:}`

`| z^4 | = sqrt{(2 sqrt3)^2 + 2^2} = 4`
 

 

`tan \ theta` `= frac{2 sqrt3}{2} = sqrt3`
`theta` `= frac{pi}{3}`

`text{arg} (z^4) = – (frac{pi}{2} + frac{pi}{3}) = – frac{5 pi}{6}`
 

`text{By De Moivre:}`

`| z | = root4 (4) = sqrt2`

`text{arg}(z) = -frac{5 pi}{24} + frac{ k pi}{2} \ , \ k = 0 , ± 1 , ± 2`

`k = 0:\ text{arg} (z)=-frac{5 pi}{24}`

`k = 1:\ text{arg} (z)= -frac{5 pi}{24} + frac{pi}{2} = frac{7 pi}{24}`

`k = – 1:\ text{arg} (z)= -frac{5 pi}{24} – frac{pi}{2} = frac{-17 pi}{24}`

`k = 2:\ text{arg} (z)= -frac{5 pi}{24} + pi = frac{19 pi}{24}`

 
`therefore \ text{Solutions are:} \ sqrt2 e^(- frac{i 17 pi}{24}) \ , \  sqrt2 e^(frac{ -i 5 pi}{24}) \ , \  sqrt2 e^(frac{i 7 pi}{24}) \ , \  sqrt2 e^(frac{i 19 pi}{24})`

Complex Numbers, EXT2 N2 SM-Bank 8

If  `(x + iy)^3 = e^( - frac{i pi}{2}),\ \ x, y ∈ R`, find a solution in the form  `x + i y, \ x, y ≠ 0`.   (2 marks)

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`frac{sqrt3}{2} – i frac{1}{2}`

Show Worked Solution
`text{Let}`   `(x + i y)` `= cos theta + i sin theta`
  `(x + i y)^3` `= cos (3 theta) + isin (3 theta)`

 
`e^(- frac{i pi}{2}) = cos (frac{-pi}{2}) + i sin (frac{-pi}{2})`

COMMENT: Read the question fully and save time – only one solution is required here.

 

`text{Equating real parts:}`

`3 theta` `= frac{-pi}{2}`
`theta` `= frac{-pi}{6}`

 

`therefore \ x + i y` `= cos (frac{-pi}{6}) + i sin (frac{-pi}{6})`
  `= frac{sqrt3}{2} – i frac{1}{2}`

Complex Numbers, EXT2 N2 EQ-Bank 1

`z = sqrt2 e^((ipi)/15)`  is a root of the equation  `z^5 = alpha(1 + isqrt3), \ alpha ∈ R`.

  1. Express  `1 + isqrt3`  in exponential form.  (2 marks)

  2. Find the value of `alpha`.  (1 mark)

  3. Find the other 4 roots of the equation in exponential form.  (3 marks)

Show Answers Only
  1. `2e^((ipi)/3)`
  2. `alpha = 2sqrt2`
  3. `e^((i11pi)/15), e^(-(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`
Show Worked Solution

i.   `beta = 1 + isqrt3`

`|beta| = sqrt(1 + (sqrt3)^2) = 2`

`text(arg)(beta) = tan^(−1) (sqrt3/1) = pi/3`

`beta = 2e^((ipi)/3)`

 

ii.    `z` `= sqrt2 e^((ipi)/15)`
  `z^5` `= (sqrt2 e^((ipi)/15))^5`
    `= (sqrt2)^5 e^((ipi)/15 xx 5)`
    `= 4sqrt2 e^((ipi)/3)`

 
`:. alpha = 2sqrt2`

 

iii.   `text(arg)(z^5) = pi/3 + 2kpi, \ \ k = 0, ±1, ±2, …`

`text(arg)(z) = pi/15 + (2kpi)/5`

`k = 1:\ text(arg)(z) = pi/15 + (2pi)/5 = (11pi)/15`

`k = text(−1):\ text(arg)(z) = pi/15 – (2pi)/5 = −pi/3`

`k = 2:\ text(arg)(z) = pi/15 + (4pi)/5 = (13pi)/15`

`k =text(−2):\ text(arg)(z) = pi/15 – (4pi)/5 = −(11pi)/15`
 

`:. 4\ text(other roots are:)`

`e^((i11pi)/15), e^(−(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)` 

Complex Numbers, EXT2 N2 2020 HSC 14a

Let `z_1` be a complex number and let  `z_2 = e^(frac{i pi}{3}) z_1`

The diagram shows points `A` and `B` which represent `z_1` and `z_2`, respectively, in the Argand plane.
 


 

  1. Explain why triangle  `OAB`  is an equilateral triangle.   (2 marks)

  2. Prove that  `z_1 ^2 + z_2^2 = z_1 z_2`.  (3 marks)

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  2. `text{See Worked Solutions}`
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i.

 

`text{Let}`     `z_1` `= r(cos theta + i sin theta)`
  `z_2` `= e^(i frac{pi}{3}) z_1`
    `= r (cos ( theta + frac{pi}{3} ) + i sin (theta + frac{pi}{3}))`

 

`| z_1 | = | z_2 | => OA = OB`

` angle AOB = frac{pi}{3} \ ( z_2 \ text{is a} \ frac{pi}{3} \ text{anti-clockwise rotation of} \ z_1 )`

`=> angle OBA = angle BAO = pi/3\ \ \ text{(angles opposite equal sides)}`

`therefore \ OAB \ text{is equilateral}`

 

ii.     `z_1` `= z_2 e^(i frac{pi}{3})`
  `frac{z_1}{z_2}` `= e^(i frac{pi}{3})`
  `(frac{z_1}{z_2})^3` `= e^((3 xx i frac{pi}{3})) \ \ (text{by De Moivre})`
  `frac{z_1^3}{z_2^3}` `= e^(i pi)`
  `z_1^3` `= -z_2^3`
  `z_1^3 + z_2^3` `= 0`

COMMENT: The identity to prove suggests the addition of 2 cubes is a possible strategy.
`(z_1 + z_2)(z_1^2 – z_1 z_2 + z_2^2)` `= 0`
`z_1^2 – z_1 z_2 + z_2^2` `= 0`
`z_1^2 + z_2^2` `= z_1 z_2`