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Complex Numbers, EXT2 N2 2021 HSC 14c*

Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that

      `cos5theta = 16cos^5theta-20cos^3 theta + 5cos theta`.   (3 marks)

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`text(See Worked Solution)`

Show Worked Solution

`(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`

Complex Numbers, EXT2 N2 2025 HSC 9 MC

The points \(U, V, W\) and \(Z\) represent the complex numbers \(u, v, w\) and \(z\) respectively. It is given that  \(v+z=u+w\)  and  \(u+k i z=w+k i v\)  where  \(k \in \mathbb{R} , k>1\).

Which quadrilateral best describes \(UVWZ\) ?

  1. Parallelogram
  2. Rectangle
  3. Rhombus
  4. Square
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Quadrilateral}\ UVWZ\ \ \Rightarrow\ \ \text{Diagonals are \(UW\) and \(VZ\)} \).

\(\text{Given}\ \ v+z=u+w\ \ \Rightarrow\ \ \dfrac{v+z}{2}=\dfrac{u+w}{2}\)

\(\text{Mid-points of diagonals are equal (diagonals bisect).}\)

\(u+kiz\) \(=w+kiv\)  
\(u-w\) \(=ki(v-z)\)  

 
\(\therefore UW\ \text{and}\ VZ\ \text{are perpendicular.}\)

\(\Rightarrow C\)

♦♦ Mean mark 36%.

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)

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

\(\text{Method 1:}\)

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

 
\(\text{Method 2 (exponential form – ex-syllabus in 2027):}\)

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

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

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

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

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

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

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

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

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

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

Show Worked Solution

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

\(\text{Method 1:}\)

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

 
\(\text{Method 2 (exponential form – ex-syllabus in 2027):}\)

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

 
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%.

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

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

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

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

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

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

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

Complex Numbers, EXT2 N2 2025 HSC 14c

Let \(w\) be a complex number such that  \(1+w+w^2+\cdots+w^6=0\).

  1. Show that \(w\) is a 7th root of unity.   (1 mark)

The complex number  \(\alpha=w+w^2+w^4\)  is a root of the equation  \(x^2+b x+c=0\), where \(b\) and \(c\) are real and \(\alpha\) is not real.

  1. Find the other root of  \(x^2+b x+c=0\)  in terms of positive powers of \(w\).   (2 marks)

  2. Find the numerical value of \(c\).   (1 mark)

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i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \text{(given)}\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7=1\) \(=1\)

ii.   \(w^6+w^5+w^3\)

iii.  \(2\)

Show Worked Solution

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \ \text{(given,}\ w\neq 1)\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7\) \(=1\)
♦♦ Mean mark (i) 35%.

ii.    \(\text {Find the other root of:} \ \ x^2+b x+c=0\)

\(\text{Since \(b, c\) are real (given),}\)

\(\text{Using conjugate root theory, other root}\ =\bar{\alpha}\)

\(\bar{\alpha}\) \(=\overline{w+w^2+w^4}\)
  \(=\overline{w}+\overline{w^2}+\overline{w^4}\)
  \(=\dfrac{1}{w}+\dfrac{1}{w^2}+\dfrac{1}{w^4} \quad\left( \bar{w}=\dfrac{1}{w} \ \text{since} \ \ \abs{w}=1\right)\)
  \(=\dfrac{w^7}{w}+\dfrac{w^7}{w^2}+\dfrac{w^7}{w^4}\)
  \(=w^6+w^5+w^3\)

 

iii.    \(\text{Product of roots}=\dfrac{c}{a}=c\)

\(c\) \(=\left(w+w^2+w^4\right)\left(w^6+w^5+w^3\right)\)
  \(=w^7+w^6+w^4+w^8+w^7+w^5+w^{10}+w^9+w^7\)
  \(=1+w^6+w^4+\left(w^7 \cdot w\right)+1+w^5+\left(w^7 \cdot w^3\right)+\left(w^7 \cdot w^2\right)+1\)
  \(=2+\underbrace{1+w+w^2+w^3+w^4+w^5+w^6}_{=0}\)
  \(=2\)
♦♦ Mean mark (iii) 32%.

Complex Numbers, EXT2 N2 2024 HSC 9 MC

Consider the solutions of the equation  \(z^4=-9\).

What is the product of all of the solutions that have a positive principal argument?

  1. \(3\)
  2. \(-3\)
  3. \(3 i\)
  4. \(-3 i\)
Show Answers Only

\(B\)

Show Worked Solution

\(z^4=-9\)

\(\text{Convert}\ z^4 \ \text{to Mod/Arg form:}\)

\(\left|z^4\right|=9, \ \ \arg \left(z^4\right)=\pi \ \text{(\(-9\) is on negative real axis})\)

Mean mark 57%.

\(\text{By De Moivre:}\)

   \(\abs{z}=\sqrt[4]{9}=\sqrt{3}\)

   \(\arg (z)=\dfrac{\pi}{4}\)

\(\text{Roots are} \ \ \dfrac{\pi}{2} \ \ \text{rotations of}\ \  z=\sqrt{3} \, \text{cis}\left(\dfrac{\pi}{4}\right)\)

\(z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{\pi}{4}\right), z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{3 \pi}{4}\right)\)

\(\sqrt{3}\, \text{cis}\left(\dfrac{\pi}{4}\right) \cdot \sqrt{3} \, \text{cis}\left(\dfrac{3 \pi}{4}\right)=3 \, \text{cis}(\pi)=-3\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N1 2024 HSC 11e

  1. Write the number  \(\sqrt{3}+i\)  in modulus-argument form.   (2 marks)

  2. Hence, or otherwise, write  \((\sqrt{3}+i)^7\)  in exact Cartesian form.   (2 marks)

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i.     \(2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)

ii.    \(-64 \sqrt{3}-64 i\)

Show Worked Solution

i.     \(z=\sqrt{3}+i\)

\(|z|=\sqrt{3+1}=2\)

\(\arg (z)=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}\)

\(z=2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)
 

ii.     \((\sqrt{3}+i)^7\) \(=2^7\left(\cos \left(\dfrac{7 \pi}{6}\right)+i \sin \left(\dfrac{7 \pi}{6}\right)\right)\)
    \(=128\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} i\right)\)
    \(=-64 \sqrt{3}-64 i\)

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{Since}\ abs(z_1)=1\ \ =>\ \ z_1= text{cis} theta`

`1/z_1 = 1/(text{cis} theta) = text{cis}(- theta) = \cos theta-i\sin theta = 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 2022 HSC 16a

A square in the Argand plane has vertices

        `5+5i,quad5-5i,quad-5-5i`  and  `-5+5i`.

The complex numbers `z_A=5+i, z_B` and `z_C` lie on the square and form the vertices of an equilateral triangle, as shown in the diagram.
 
 
                 
 
Find the exact value of the complex number `z_B`.   (4 marks)

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`(5-16/sqrt3)+5i`

Show Worked Solution

`z_A=5+i, \ \ z_B=b+5i, \ \ z_C=c-5i`

`z_B-z_A=(b-5)+4i`

`z_C-z_A=(c-5)-6i`
 


♦♦♦ Mean mark 21%.

`text{Internal angles of equilateral triangle} = pi/3:`

`=>\ (z_B-z_A)\ text{is an anti-clockwise rotation of}\ (z_C-z_A)\ text{by}\ pi/3`
 

`e^(i pi/3)(z_B-z_A)=z_C-z_A`

`(1/2+i sqrt3/2)((b-5)+4i)=(c-5)-6i`

`(b-5)/2+2i+((b-5)sqrt3)/2 i-2sqrt3` `=(c-5)-6i`  
`(b-5-4sqrt3)/2 + i((4+(b-5)sqrt3)/2)` `=(c-5)-6i`  

 
`text{Equating imaginary parts:}`

`(4+(b-5)sqrt3)/2` `=-6`  
`(b-5)sqrt3` `=-16`  
`b-5` `=-16/sqrt3`  
`b` `=5-16/sqrt3`  

 
`:.z_B=(5-16/sqrt3)+5i`

Complex Numbers, EXT2 N2 2022 HSC 13c

Consider the equation  `z^5+1=0`, where `z` is a complex number.

  1. Solve the equation  `z^5+1=0`  by finding the 5th roots of `-1`.   (2 marks)

  2. Show that if `z` is a solution of  `z^5+1=0`  and  `z !=-1`, then  `u=z+(1)/(z)`  is a solution of  `u^2-u-1=0`.   (2 marks)

  3. Hence find the exact value of `cos\ (3pi)/(5)`.   (3 marks)

Show Answers Only
  1. `z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  2. `text{Proof (See Worked Solutions)}`
  3. `(1-sqrt5)/4`
Show Worked Solution

i.   `z^5+1=0\ \ =>\ \ z^5=-1`

`text{cis}\ pi=-1\ \ =>\ \ z=text{cis} ((pi+2k pi)/5),\ k=0,1,2,3,4`

`:.z=text{cis} ((pi)/5), text{cis} ((3pi)/5),-1,text{cis} (-(3pi)/5),text{cis} (-(pi)/5)`
  

ii.   `z^5+1=(z+1)(z^4-z^3+z^2-z+1)`

`text{Given}\ \ z!=-1,`

`z^4-z^3+z^2-z+1=0`
 

`text{Divide by}\ z^2\ \ (z!=0)`

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

 
`text{Let}\ \ u=z+1/z:`

`:.u^2-u-1=0`
 


Mean mark (ii) 53%.

iii.  `u^2-u-1=0`

`text{By quadratic formula:}`

`u=(1+-sqrt(1-4xx1xx(-1)))/(2)(1+-sqrt5)/2`


♦ Mean mark (iii) 43%.

`z+1/z=(1+-sqrt5)/2`

`text{Let}\ \ z=text{cis} ((3pi)/5)\ \ =>\ \ 1/z=text{cis} (-(3pi)/5)`

`text{cis}((3pi)/5)+text{cis} (-(3pi)/5)` `=(1-sqrt5)/2,\ \ (cos((3pi)/5) <0)`  
`2cos((3pi)/5)` `=(1-sqrt5)/2`  
`cos((3pi)/5)` `=(1-sqrt5)/4`  

Complex Numbers, EXT2 N2 2021 HSC 10 MC

Consider the two non-zero complex numbers `z` and `w` as vectors.

Which of the following expressions is the projection of `z` onto `w` ? 

  1.  `{text{Re} (zw)}/{|w|} w`
  2.  `|z/w| w`
  3.  `text{Re} (z/w) w`
  4. `{text{Re}(z)}/{|w|} w`
Show Answers Only

`C`

Show Worked Solution

`text{Let} \ \ z = a + i b \ \ =>\ underset~z = ((a),(b))`

♦♦♦ Mean mark 30%.

`text{Let} \ \ w = c + i d \ \ =>\ underset~w = ((c),(d))`

`underset~z * underset~w = ac + bd`

`|underset~w|^2 = c^2 + d^2`

`text{proj}_(underset~w) underset~z = (underset~z*underset~w)/|underset~w|^2 *underset~w= (ac+bd)/(c^2+d^2)*underset~w`

`z/w` `=(a+ib)/(c+id) xx (c-id)/(c-id)`  
  `=(ac+bd + i(bc-ad))/(c^2+d^2)`  

 
`text{Re}(z/w) =(ac+bd)/(c^2+d^2)`

`:.\ text{proj}_(underset~w) underset~z = text{Re} (z/w) w`
 

`=> C`

Complex Numbers, EXT2 N2 2021 HSC 13a

Indicate the locations of all of the fourth roots of the complex number  `a + ib`.   (2 marks)
 

Show Answers Only

Show Worked Solution

`4 \ text{roots:} \ z_1 , z_2 , z_3 , z_4`

♦ Mean mark 47%.

`text{arg}(z_1) = 1/4 text{arg}(a + ib)`

`|z| > 1 \ text{but less than} \ |a + ib|`

`text{Rotations between roots} = pi/2`

Complex Numbers, EXT2 N1 EQ-Bank 14

Let  `z = sqrt3-3 i`

  1. Express `z` in modulus-argument form.   (2 marks)

  2. Find the smallest integer `n`, such that  `z^n + (overset_z)^n = 0`.   (3 marks)

Show Answers Only

a.    `2 sqrt3 text{cis} (frac{-pi}{3})`

b.    `3`

Show Worked Solution
a.     `z` `= sqrt3-3 i`
  `|z|` `= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3`

 

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

`text{arg} (z)=-frac{pi}{3}`

`therefore  z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`
 

b.    `z^n + (overset_z)^n = 0`

`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3})-i sin (frac{-pi}{3}) ]^n = 0`

`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3})-i sin (frac{-n pi}{3}) = 0`

`2 \ cos (frac{-n pi}{3})` `= 0`
`cos (frac{n pi}{3})` `= 0`
`frac{n pi}{3}` `= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …`
`frac{n}{3}` `= frac{(2k + 1)}{2}`
`n` `= frac{3 (2k + 1)}{2}`

 
`text{Numerator will always be odd  ⇒  no solution exists}`

Complex Numbers, EXT2 N1 2005 HSC 2b

Let  `beta = 1-i sqrt3`.

  1. Express  `beta`  in modulus-argument form.    (2 marks)

  2. Express  `beta^5`  in modulus-argument form.    (2 marks)

  3. Hence express  `beta^5`  in the form  `x+iy`.    (1 mark)

Show Answers Only

a.    `2 \ text{cis} (-frac{pi}{3})`

b.    `32 \ text{cis} (frac{pi}{3})`

c.    `16 + i 16 sqrt3`

Show Worked Solution

a.    `beta = 1-i sqrt3`
 

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

`tan theta` `= frac{sqrt3}{1} = sqrt3`
`theta` `= frac{pi}{3}`
`text{arg} (beta)` `= -frac{pi}{3}`

`therefore \ beta = 2 \ text{cis} (-frac{pi}{3})`

 

b.     `beta^5` `= 2^5 \ text{cis} (-frac{pi}{3} xx5)`
    `= 32 \ text{cis} (-frac{5pi}{3} + 2 pi)`
    `= 32 \ text{cis} (frac{pi}{3})`

 

c.     `beta^5` `= 32 ( cos (frac{pi}{3}) + i sin (frac{pi}{3}) )`
    `= 32 ( frac{1}{2} + i  frac{sqrt3}{2})`
    `= 16 + i 16 sqrt3`

Complex Numbers, EXT2 N2 2019 HSC 16b

Let  `P(z) = z^4-2kz^3 + 2k^2z^2 + mz + 1`, where `k` and `m` are real numbers.

The roots of `P(z)` are `alpha, bar alpha, beta, bar beta`.

It is given that  `|\ alpha\ | = 1`  and  `|\ beta\ | = 1`.

  1. Show that  `(text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`.   (3 marks)

  2. The diagram shows the position of `alpha`.
     


 

On the diagram, accurately show all possible positions of `beta`.   (2 marks)

Show Answers Only

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

ii.   `text(See Worked Solutions)`

Show Worked Solution

i.    `P(z) = z^4-2kz^3 + 2k^2z^2 + mz + 1,\ \ k, m in RR`

`text(Roots):\ \ alpha, bar alpha, beta, bar beta and |\ alpha\ | = 1, |\ beta\ | = 1`

`text(Show)\ \ (text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`

♦♦ Mean mark part (i) 26%.

`alpha + bar alpha + beta + bar beta` `= 2k`
`2 text{Re} (alpha) + 2 text{Re} (beta)` `= 2k`
`text{Re} (alpha) + text{Re} (beta)` `= k`

 

`alpha bar alpha + alpha beta + alpha bar beta + bar alpha beta + bar alpha bar beta + beta bar beta` `= 2k^2`
`|\ alpha\ |^2 + alpha(beta + bar beta) + bar alpha(beta + bar beta) + |\ beta\ |^2` `= 2k^2`
`1 + (alpha + bar alpha)(beta + bar beta) + 1` `= 2k^2`
`2 + 2 text{Re} (alpha) ⋅ 2 text{Re} (beta)` `= 2 (text{Re} (alpha) + text{Re} (beta))^2`
`2 + 4 text{Re} (alpha) text{Re} (beta)` `= 2 text{Re} (alpha)^2 + 4 text{Re} (alpha) text{Re} (beta) + 2 text{Re} (beta)^2`
`2` `= 2(text{Re} (alpha)^2 + text{Re} (beta)^2)`
`:. 1` `= text{Re} (alpha)^2 + text{Re} (beta)^2`

 

ii.    `|\ alpha\ | = |\ beta\ |\ \ \ text{(given)}`
  `text{Re}(alpha)^2 + text{Re}(beta)^2 = 1\ \ \ text{(see part (i))}`
  `text{Re}(alpha)^2 + text{Im}(alpha)^2 = 1\ \ \ (|\ alpha\ | = 1)`
  `=> text{Re}(beta)^2 = text{Im} (alpha)^2`
  `\ \ \ \ \ \ text{Re}(beta) = +-text{Im}(alpha)`

 

♦♦♦ Mean mark part (ii) 10%.

Complex Numbers, EXT2 N1 2019 HSC 11e

Let  `z = -1 + i sqrt 3`.

  1. Write  `z`  in modulus-argument form.   (2 marks)

  2. Find  `z^3`, giving your answer in the form  `x + iy`, where `x` and `y` are real numbers.   (2 marks)

Show Answers Only

i.    `z = 2 text(cis) (2 pi)/3`

ii.   `8 + 0i`

Show Worked Solution

i.    `|\ z\ |= -1 + i sqrt 3= sqrt((-1)^2 + (sqrt 3)^2)= 2`

  `tan theta` `= -sqrt 3`
  `text(arg)(z)` `= (2 pi)/3`
  `:. z` `= 2 text(cis) (2 pi)/3`

 

ii.   `z^3 = 2^3 [cos(3 xx (2 pi)/3) + i sin (3 xx (2 pi)/3)]\ \ \ text{(by De Moivre)}`

`= 8(cos 2 pi + i sin 2 pi)`

`= 8(1 + 0i)`

`= 8 + 0i`

Complex Numbers, EXT2 N2 2018 HSC 15b

  1. Use De Moivre's theorem and the expansion of `(costheta + isintheta)^8` to show that
  2. `sin8theta = ((8),(1)) cos^7thetasintheta-((8),(3)) cos^5thetasin^3theta`
  3.                  `+ ((8),(5)) cos^3thetasin^5theta-((8),(7)) costhetasin^7theta`   (2 marks)

  4. Hence, show that
  5. `(sin8theta)/(sin2theta) = 4(1-10sin^2theta + 24sin^4theta-16sin^6theta)`.   (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(By De Moivre)`

`costheta + isintheta^8 = cos8theta + isin8theta\ \ …\ (text{*})`
 

`text(Using Binomial Expansion)`

`(costheta + isintheta)^8`

`= cos^8theta + ((8),(1))cos^7theta * isintheta + ((8),(2)) cos^6theta *i^2sin^2theta`

`+ ((8),(3)) cos^5theta *i^3sin^3theta + ((8),(4)) cos^4theta *i^4sin^4theta + ((8),(5)) cos^3theta *i^5sin^5theta`

`+ ((8),(6)) cos^2theta *i^6sin^6theta + ((8),(7)) costheta *i^7sin^7theta + i^8sin^8theta`

 
`text(Equating imaginary parts of the expansion equation (*)):`

`isin8theta = ((8),(1)) cos^7theta* isintheta + ((8),(3)) icos^5theta* i^3sin^3theta`

`+ ((8),(5)) cos^3theta* i ^5sintheta + ((8),(7)) costheta *i^7sin^7theta`

`:. sin8theta = ((8),(1)) cos^7theta sintheta-((8),(3)) cos^5theta sin^3theta`

`+ ((8),(5)) cos^3theta sin^5theta-((8),(7)) costheta sin^7theta`
 

ii.    `sin8theta` `= 8cos^7theta sintheta-56cos^5 sin^3theta + 56cos^3theta sin^5theta-8costheta sin^7theta`
    `= 2sinthetacostheta (4cos^6theta-28cos^4theta sin^2theta + 28cos^2theta sin^4theta-4sin^6theta)`

 
`:. (sin8theta)/(sin2theta)`

  `= 4cos^6theta-28cos^4theta sin^2theta + 28cos^2theta sin^4theta-4sin^6theta`

  `= 4(1-sin^2theta)^3-28(1-sin^2theta)^2 sin^2theta + 28(1-sin^2theta) sin^4theta-4sin^6theta`

  `= 4(1-3sin^2theta + 3sin^4theta + sin^6theta)-28sin^2theta (1-2sin^2theta + sin^4theta)`

`+ 28sin^4theta (1-sin^2theta)-4sin^6theta`

  `= 4-40sin^2theta + 96sin^4theta-56sin^6theta`

  `= 4(1-10sin^2theta + 24sin^4theta-16sin^6theta)`

Complex Numbers, EXT2 N2 2018 HSC 11d

The points `A`, `B` and `C` on the Argand diagram represent the complex numbers `u`, `v` and `w` respectively.

The points `O`, `A`, `B` and `C` form a square as shown on the diagram.
 

It is given that  `u = 5 + 2i`.

  1.  Find  `w`.   (1 mark)

  2.  Find  `v`.   (1 mark)

  3.  Find  `text(arg)(w/v)`.   (1 mark)

Show Answers Only

i.    `−2 + 5i`

ii.   `3 + 7i`

iii.  `pi/4`

Show Worked Solution

i.    `w= iu= i(5 + 2i)= -2 + 5i`
 

ii.    `v` `= u + w`
    `= 5 + 2i + (-2 + 5i)`
    `= 3 + 7i`

 

iii.    `text(arg)(w/v)` `= text(arg)(w)-text(arg)(v)`
    `= pi/4\ \ (text(diagonal of square bisects corner))`

Complex Numbers, EXT2 N2 2018 HSC 6 MC

Which complex number is a 6th root of `i`?

  1. `−1/sqrt2 + 1/sqrt2i`
  2. `−1/sqrt2-1/sqrt2i`
  3. `−sqrt2 + sqrt2i`
  4. `−sqrt2-sqrt2i`
Show Answers Only

`A`

Show Worked Solution

`text(Consider option A:)`
 

`|−1/sqrt2 + 1/sqrt2i|= sqrt((−1/sqrt2)^2 + (1/sqrt2)^2) = 1`

`text(arg)(z)` `= (3pi)/4`
`z` `= 1(cos\ (3pi)/4 + i sin\ (3pi)/4)`
`z^6` `= cos\ (18pi)/4 + i sin\ (18pi)/4\ \ \ text{(De Moivre)}`
  `= i`

 
`=> A`

Complex Numbers, EXT2 N2 2017 HSC 13e

The points `A, B, C` and `D` on the Argand diagram represent the complex numbers `a, b, c` and `d`respectively. The points form a square as shown on the diagram.
 

By using vectors, or otherwise, show that  `c = (1 + i) d-ia`.   (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

♦ Mean mark 49%.
`c-d` `= i (d-a)\ \ \ text{(rotation of}\ pi/2 text{)}`
`:. c ` `= d + id-ia`
  `= (1 + i) d-ia\ text(… as required.)`

Complex Numbers, EXT2 N2 2017 HSC 1 MC

The complex number `z` is chosen so that  `1, z, …, z^7`  form the vertices of the regular polygon shown.

Which polynomial equation has all of these complex numbers as roots?

  1. `x^7 - 1 = 0`
  2. `x^7 + 1 = 0`
  3. `x^8 - 1 = 0`
  4. `x^8 + 1 = 0`
Show Answers Only

`C`

Show Worked Solution

`P(x)\ \ text(has 8 separate roots.)`

`:.\ text(Must be of degree at least 8.)`

`text(S) text(ince 1 is also a root,)`

`=>  C`

Complex Numbers, EXT2 N2 2016 HSC 16b

  1. The complex numbers `0, \ u` and `v` form the vertices of an equilateral triangle in the Argand diagram.
  2. Show that  `u^2 + v^2 = uv.`   (2 marks)
  3. Give an example of non-zero complex numbers `u` and `v`, so that `0, \ u` and `v` form the vertices of an equilateral triangle in the Argand diagram.   (1 mark)

Show Answers Only

a.    `text(Show Worked Solutions)`

b.    `u = 1/2 + sqrt3/2 i`

Show Worked Solution

a.    `0,u,v\ text(are vertices of an equilateral triangle.)`

♦♦ Mean mark (a) 23%.

 

ext2-2016-hsc-16b-answer

`=> u` `= v text(cis)(±pi/3)`
`u^3` `= v^3text(cis)(±pi)`
`u^3` `= − v^3`
`u^3 + v^3` `= 0`

 
`(u + v)(u^2-uv + v^2) = 0`

`text(S)text(ince)\ u != v:`

`u^2-uv + v^2` `= 0`
`:. u^2 + v^2` `= uv`

 

b.    `text(Let)\ \ v = 1,`

♦♦ Mean mark (b) 33%.
`:. u` `= text(cis)(pi/3)`
  `= cos\ pi/3 + isin\ pi/3`
  `= 1/2 + sqrt3/2 i`

Complex Numbers, EXT2 N2 2016 HSC 16a

  1. The complex numbers  `z = cos theta + i sin theta`  and  `w = cos alpha + i sin alpha`, where  `-pi < theta < pi`  and  `-pi < alpha <= pi`, satisfy
  2.    `1 + z + w = 0.`
  3. By considering the real and imaginary parts of  `1 + z + w`, or otherwise, show that `1,  z` and `w` form the vertices of an equilateral triangle in the Argand diagram.   (3 marks)

  4. Hence, or otherwise, show that if the three non-zero complex numbers `2i, z_1` and `z_2` satisfy
  5.    `|\ 2i\ | = |\ z_1\ | = |\ z_2\ |`  AND  `2i + z_1 + z_2 = 0.`
  6. then they form the vertices of an equilateral triangle in the Argand diagram.   (2 marks)

Show Answers Only

i.    `text(See Worked Solutions)`

ii.   `text(See Worked Solutions)`

Show Worked Solution

i.    `z= costheta + isintheta, \ |\ z\ | = 1`

`w= cosalpha + isinalpha, \ |\ w\ | = 1`

`text(S)text(ince)\ 1 + z + w = 0:`

♦♦♦ Mean mark (i) 25%.
STRATEGY: A clear graphical image simplifies both parts of this question significantly.
`text(Im)(1 + z + w)` `= 0`
`sintheta + sinalpha` `= 0`
`sintheta` `= −sinalpha`
`:. theta` `= −alpha\ …\ (1)`

 

`text(Re)(1 + z + w)` `= 0`
`1 + costheta + cosalpha` `= 0`
`2costheta` `= −1qquad(text(S)text(ince)\ cosalpha  = cos(−theta) = costheta)`
`costheta` `= −1/2`
`:. theta` `= (2pi)/3`
`alpha` `= −(2pi)/3`

 
ext2-2016-hsc-16a-answer2 

`=>\ text(All points are on the unit circle separated by)\ (2pi)/3\ text(radians.)`

`:.\ text(They are vertices of an equilateral triangle.)`

 

ii.   `|\ 2i\ | = 2`

`text(Let)\ \ z_1` `= 2(costheta + isintheta), \  |\ z_1\ | = 2`
`z_2`  `= 2(cosalpha + isinalpha), \  |\ z_2\ | = 2`

 

♦♦♦ Mean mark (ii) 5%.
`text(Re)(2i + z_1 + z_2)` `= 0`
`2(costheta + cosalpha)` `= 0`
`:. costheta` `= −cosalpha`
`:. theta` `= pi-alpha`

 

`text(Im)(2i + z_1 + z_2)` `= 0`
`2(1 + sintheta + sinalpha)` `= 0`
`sintheta + sinalpha` `= −1`
`2sintheta` `= −1qquad(text(S)text(ince)\ sinalpha = sin(pi-theta) = sintheta)`
`sintheta` `= −1/2`
`:. theta` `= (7pi)/6`
`:. alpha` `= −pi/6`

 
ext2-2016-hsc-16a-answer3 
 

`=>\ text(All points are on the 2 unit circle separated by)\ (2pi)/3\ text(radians.)`

`:. 2i, z_1\ text(and)\ z_2\ text(are vertices of an equilateral triangle.)`

Complex Numbers, EXT2 N2 2016 HSC 12c

Let  `z = cos theta + i sin theta.`

  1. By considering the real part of `z^4`, show that `cos 4 theta` is
  2. `qquad cos^4 theta-6 cos^2 theta sin^2 theta + sin^4 theta.`   (2 marks)
  3. Hence, or otherwise, find an expression for  `cos 4 theta`  involving only powers of `cos theta.`   (1 mark)

Show Answers Only

i.    `text(See Worked Solutions)`

ii.   `8cos^4theta-8cos^2theta + 1`

Show Worked Solution

i.   `z = costheta + isintheta`

`z^4` `= (costheta + isintheta)^4`
 

`= cos^4theta + 4cos^3theta*(isintheta) + 6cos^2theta*(isintheta)^2 +`

`4costheta*(isintheta)^3 + (isintheta)^4`
 

`= cos^4theta + 4icos^3thetasintheta-6cos^2thetasin^2theta -`

`4icosthetasin^3theta + sin^4theta`

 

`z^4 = cos4theta + isin4theta\ \ text{(by De Moivre)}`
 

`text(Equating real parts:)`

`cos4theta = cos^4theta-6cos^2thetasin^2theta + sin^4theta\ …\ text(as required)`

 

ii.    `cos4theta` `= cos^4theta-6cos^2theta(1-cos^2theta) + (1-cos^2theta)^2`
    `= cos^4theta-6cos^2theta + 6cos^4theta + 1-2cos^2theta + cos^4theta`
    `= 8cos^4theta-8cos^2theta + 1`

Complex Numbers, EXT2 N2 2016 HSC 5 MC

Multiplying a non-zero complex number by  `(1-i)/(1 + i)`  results in a rotation about the origin on an Argand diagram.

What is the rotation?

  1. Clockwise by  `pi/4`
  2. Clockwise by  `pi/2`
  3. Anticlockwise by  `pi/4`
  4. Anticlockwise by  `pi/2`
Show Answers Only

`B`

Show Worked Solution
`(1-i)/(1 + i)` `= ((1-i)^2)/((1 + i)(1-i))`
  `= (−2i)/2`
  `= −i`

 
`:. text(Clockwise rotation by)\ \ pi/2.`

`=> B`

Complex Numbers, EXT2 N1 2016 HSC 4 MC

The Argand diagram shows the complex numbers `z` and `w`, where `z` lies in the first quadrant and `w` lies in the second quadrant.
  

ext2-hsc-2016-4mc

Which complex number could lie in the 3rd quadrant?

  1. `-w`
  2. `2 iz`
  3. `bar z`
  4. `w - z`
Show Answers Only

`=> D`

Show Worked Solution

`text(Using the parallelogram method:)`
 

ext2-hsc-2016-4mc-answer1
 

`text(From the graph above,)`

`w-z\ \ text(could lie in 3rd quadrant.)`

`=> D`

Complex Numbers, EXT2 N2 2007 HSC 8b

  1. Let `n` be a positive integer. Show that if  `z^2 != 1`  then
  2. `1 + z^2 + z^4 + … + z^(2n-2) = ((z^n-z^-n)/(z-z^-1)) z^(n-1)`.   (2 marks)

  3. By substituting  `z = cos theta + i sin theta`  where  `sin theta != 0`, into part (a), show that
  4. `1 + cos 2 theta + … + cos (2n-2) theta + i[sin 2 theta + … + sin (2n-2) theta]`
  5. `= (sin n theta)/(sin theta) [cos (n-1) theta + i sin (n-1) theta].`   (3 marks)

  6. Suppose  `theta = pi/(2n)`.  Using part (ii), show that
  7. `sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n-1) pi)/n = cot\ pi/(2n).`   (2 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.    `1 + z^2 + z^4 + … + z^(2n-2),\ z^2 != 1`

`text(GP where)\ a = 1,\ \ r = z^2,\ \ n\ text(terms):`

`S_n` `=(1((z^2)^n-1))/(z^2-1)`
  `=(z^(2n)-1)/(z^2-1)`
  `=((z^n-z^-n))/(z-z^-1) xx z^n/z`
  `=((z^n-z^-n)/(z-z^-1))z^(n-1)`

 

b.     `z` `= cos theta + i sin theta`
  `z^n` `= cos n theta + i sin n theta\ \ …\ text(etc)\ \ \ \ text{(De Moivre)}` 
  `z^-n` `= cos( -n theta) + i sin (-n theta)`
    `= cos n theta-i sin n theta`

 

`text(LHS)` `= 1 + (cos 2 theta + i sin 2 theta) + (cos 4 theta + i sin 4 theta) + `
  `… + (cos(2n-2) theta + i sin (2n-2) theta)`
  `= 1 + cos 2 theta + cos 4 theta + … + cos (2n-2) theta + `
  `i (sin 2 theta + sin 4 theta + … + sin (2n-2) theta)`
   

`text{Using part (a):}`

`text(LHS)` `=((cos n theta + i sin n theta-cos n theta + i sin n theta))/(cos theta + i sin theta-cos theta + i sin theta) xx`
  `[cos (n-1) theta + i sin (n-1) theta]`
  `=(2 i sin n theta)/(2 i sin theta) [cos (n-1) theta + i sin (n-1) theta]`
  `=(sin n theta)/(sin theta) [cos (n-1) theta + i sin (n-1) theta]\ \ text(… as required.)`

 

c.    `text{Equating the imaginary parts in part (b):}`

`sin 2 theta + sin 4 theta + … + sin 2 (n-1) theta = (sin (n theta) sin (n-1) theta)/(sin theta)`

`text(When)\ \ theta = pi/(2n):`

`sin\ (2 pi)/(2n) + sin\ (4 pi)/(2n) + … + sin\ (2(n-1) pi)/(2 n) = (sin\ (n pi)/(2n) sin\ ((n-1) pi)/(2n))/(sin\ pi/(2n))`

`:. sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n-1) pi)/n`

`=(sin\ pi/2)/(sin\ pi/(2n)) xx sin\ ((n-1) pi)/(2n)`

`=1/(sin\ pi/(2n)) sin (pi/2-pi/(2n))`

`=(cos\ pi/(2n))/(sin\ pi/(2n))`

`=cot\ pi/(2n)`

Complex Numbers, EXT2 N1 2007 HSC 2b

  1. Write  ` 1 + i`  in the form `r (cos theta + i sin theta).`   (2 marks)

  2. Hence, or otherwise, find `(1 + i)^17` in the form  `a + ib`, where `a` and `b` are integers.   (3 marks)

Show Answers Only

a.    `sqrt 2 ( cos­ pi/4 + i sin­ pi/4)`

b.    `256 + 256i`

Show Worked Solution
a.    
`|\ 1+i\ |` `=sqrt(1^2+1^2)=sqrt2`
`text(arg)(1+i)` `=pi/4`
`:. 1 + i =` `sqrt 2 (cos­ pi/4 + i sin­ pi/4)`

 

b.    `(1 + i)^17` `=(sqrt 2)^17 (cos\ pi/4 + i sin\ pi/4)^17`
  `=2^8 sqrt 2 (cos­ (17 pi)/4 + i sin­ (17 pi)/4)\ \ \ \ text{(De Moivre)}`
  `=2^8 sqrt 2 (cos­ pi/4 + i sin­ pi/4)`
  `=2^8 sqrt2(1/sqrt2 + 1/sqrt2 i)`
  `=2^8 (1 + i)`
  `=256 + 256 i`

Complex Numbers, EXT2 N1 2015 HSC 5 MC

Given that  `z = 1 -i`, which expression is equal to  `z^3 ?`

  1. `sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
  2. `2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
  3. `sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
  4. `2 sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
Show Answers Only

`B`

Show Worked Solution

 HSC 2015 5MC

`z` `=1-i`
`|\ 1-i\ |` `=sqrt2`
`text{arg}(z)` `=-pi/4`
`z` `=sqrt 2 (cos(-pi/4) + i sin(-pi/4))`
`:.z^3` `=2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))\ \ \ \ text{(De Moivre)}`

 
`=>  B`

Complex Numbers, EXT2 N1 2006 HSC 2b

  1. Express  `sqrt 3-i`  in modulus-argument form.   (2 marks)

  2. Express `(sqrt 3-i)^7` in modulus-argument form.   (2 marks)

  3. Hence express `(sqrt 3-i)^7` in the form  `x + iy.`   (1 mark)

Show Answers Only

a.    `2 text(cis) (−pi/6)`

b.    `2^7 text(cis) ((5 pi)/6)`

c.    `64 (−sqrt 3 + i)`

Show Worked Solution
a.    

`|\ sqrt 3-i\ |= sqrt ((sqrt 3)^2+1^2)=2`

`­theta=tan^-1(- 1/sqrt3)=- pi/6`

 `:. sqrt 3-i = 2 text(cis) (- pi/6)`

 

b.   `(sqrt 3-i)^7 =` `2^7 text(cis) (-(7 pi)/6)\ \ \ \ text{(De Moivre)}`
`­=` `128 text(cis) ((5 pi)/6)`

 

c.  `(sqrt 3-i)^7` `=128 (cos\ (5pi)/6 + i sin\ (5pi)/6)`
  `=128 (- sqrt 3/2 + i/2)`
  `=-64 sqrt 3 + 64i`

Complex Numbers, EXT2 N2 2009 HSC 7b

Let  `z = cos theta + i sin theta.`

  1. Show that  `z^n + z^-n = 2 cos(n theta)`, where `n` is a positive integer.   (2 marks)

  2. Let `m` be a positive integer. Show that
  3. `(2 cos theta)^(2m) = 2 [cos (2m theta) + ((2m), (1)) cos (2m-2) theta + ((2m), (2)) cos (2m-4) theta`
  4.           `+ … + ((2m), (m-1)) cos 2 theta] + ((2m), (m)).`   (3 marks)

  5. Hence, or otherwise, prove that
  6. `int_0^(pi/2) cos^(2m) theta\ d theta = pi/(2^(2m + 1)) ((2m), (m))`
  7. where `m` is a positive integer.   (2 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
a.     `z` `= cos theta + i sin theta`
  `z^n` `= cos n theta + i sin n theta\ \ \ \ text{(De Moivre)}`
  `z^-n` `= cos (-n theta) + i sin (-n theta)\ \ \ \ text{(De Moivre)}`
    `= cos n theta-i sin n theta`
  `z^n + z^-n` `= cos n theta + i sin n theta + cos n theta-i sin n theta`
    `= 2 cos n theta,\ \ \ \ n > 0`

 

 

b.    `z + z^-1 = 2 cos theta`

`:.(2 cos theta)^(2m)`

`=(z + z^-1)^(2m)`

`=z^(2m) + ((2m), (1)) z^(2m-1) z^-1 + ((2m), (2)) z^(2m-2) z^-2+`

` … + ((2m), (2m-1)) z^1 z^-(2m-1) + z^-(2m)`

`=z^(2m) + ((2m), (1)) z^(2m-2) + ((2m), (2)) z^(2m-4)+`

` … + ((2m), (2m-1)) z^-(2m-2) + z^(-2m)`

`=z^(2m) + ((2m), (1)) z^(2m-2) + ((2m), (2)) z^(2m-4) + … + ((2m), (m)) z^(2m-2m) …`

`+ ((2m), (2)) z^-(2m-4) + ((2m), (1)) z^-(2m-2) + z^(-2m)`

`=(z^(2m) + z^(-2m)) + ((2m), (1)) (z^(2m-2) + z^-(2m-2)) + ((2m), (2))`

`(z^(2m-4) + z^-(2m-4)) + … + ((2m), (m-1)) (z + z^-1) + ((2m), (m))`

`=2 [cos 2 m theta + ((2m), (1)) cos (2m-2) theta + ((2m), (2)) cos (2m-4) theta`

`+ … + ((2m), (m-1)) cos 2 theta] + ((2m), (m))`

 

c.    `int_0^(pi/2) cos^(2m) d theta`

`=1/(2^(2m))  int_0^(pi/2) (2 cos theta)^(2m)`

`=1/(2^(2m)) int_0^(pi/2)[2(cos 2 m theta + ((2m), (1)) cos (2m-2) theta + ((2m), (2))`

`cos (2m-4) theta + … + ((2m), (m-1)) cos 2 theta) + ((2m), (m))] d theta`

`=1/(2^(2m)) [2((sin 2 m theta)/(2m) + ((2m), (1)) (sin (2m-2) theta)/(2m-2)`

`+ … + ((2m), (m-1)) (sin 2 theta)/2) + ((2m), (m)) theta]_0^(pi/2)`

`=1/(2^(2m)) [2(0 + 0 + … + 0) + ((2m), (m)) pi/2-(0)]`

`=pi/(2^(2m + 1)) ((2m), (m))`

Complex Numbers, EXT2 N2 2009 HSC 2e

  1. Find all the 5th roots of  `–1`  in modulus-argument form.   (2 marks)

  2. Sketch the 5th roots of  `–1`  on an Argand diagram.   (1 mark)

Show Answers Only

a.    `z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis) (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,`

`z_4 = text(cis) (7 pi)/5,\ \ z_5 = text(cis) (9 pi)/5`

b.    
       

Show Worked Solution

a.    `z=cos theta+i sin theta`

`z^5=cos\ 5 theta+i sin\ 5 theta=-1,\ \ text{(De Moivre)}`

  `cos\ 5 theta` `=-1`
  `5 theta` `=pi,\ 3pi,\ 5pi,\ 7pi,\ 9pi`
  `theta` `=pi/5,\ (3pi)/5,\ pi,\ (7pi)/5,\ (9pi)/5`

 

`:.\ text(The roots are)`

`z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis)\ (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,`

`z_4 = text(cis)\ (7 pi)/5,\ \ z_5 = text(cis)\ (9 pi)/5`

 

b.    

Complex Numbers, EXT2 N1 2009 HSC 2c

The points `P` and `Q` on the Argand diagram represent the complex numbers `z` and `w` respectively.
 

 

 

 
 

On the diagram, mark the following points:

  1. the point `R` representing `iz.`   (1 mark)

  2. the point `S` representing `bar w.`   (1 mark)

  3. the point `T` representing  `z + w.`   (1 mark)

Show Answers Only

a, b, and c.

Show Worked Solution

a, b, and c.

Complex Numbers, EXT2 N1 2010 HSC 2b

  1. Express  `-sqrt3-i`  in modulus–argument form.   (2 marks)

  2. Show that  `(-sqrt3-i)^6`  is a real number.   (2 marks)

Show Answers Only

a.    `2text(cis)(-(5pi)/6)`

b.    `-64`

Show Worked Solution

a.    `|-sqrt3-i\ |=sqrt((-sqrt3)^2+sqrt((-1)^2))=2`
 

Complex Numbers, EXT2 2010 HSC 2b 

`text(From the graph)`

`text{arg}(-sqrt3-i)=- (5pi)/6\ \ \ \ text{(for}\  –pi<theta<pi text{)}`

`:.-sqrt3-i= 2text(cis)(-(5pi)/6)`

 

`text{Alternative Solution (to find the argument)}`

`-sqrt3-i= 2(- sqrt3/2-1/2 i)=2text(cis)(-(5pi)/6)`
 

b.     `(-sqrt3-i)^6` `= [2text(cis)(-(5pi)/6)]^6`
    `=2^6[cos((-5pi)/6 xx6) +i sin((-5 pi)/6 xx6)]\ \ \ \ text{(De Moivre)}`
    `= 2^6[cos(-5pi) + i sin(-5pi)]`
    `= 64(-1 + 0i)`
    `= -64`

Complex Numbers, EXT2 N2 2011 HSC 2d

  1. Use the binomial theorem to expand `(cos theta + i sin theta)^3.`   (1 mark)

  2. Use de Moivre’s theorem and your result from part (a) to prove that
  3.      `cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.`   (3 marks)

  4. Hence, or otherwise, find the smallest positive solution of
  5. `4 cos^3 theta-3 cos theta = 1.`   (2 marks)

Show Answers Only
  1. `cos^3 theta + 3 i cos^2 theta sin theta-3 cos theta sin^2 theta-i sin^3 theta`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `theta = (2 pi)/3`
Show Worked Solution

a.    `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

`= cos^3 theta + 3 cos^2 theta (i sin theta)+ 3 cos theta (i sin theta)^2 + (i sin theta)^3`

`= cos^3 theta + 3 i cos^2 theta sin theta- 3 cos theta sin^2 theta-i sin^3 theta`

 

b.     `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`
 

`text(Equate real parts)`

`cos 3 theta` `= cos^3 theta-3 cos theta sin^2 theta`
`cos 3 theta` `= cos^3 theta-3 cos theta (1-cos^2 theta)`
`cos 3 theta` `= 4 cos^3 theta-3 cos theta`
`4 cos^3 theta`  `=cos 3 theta+3cos theta`
`:.cos^3 theta` `= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

 

c.    `text(If)\ \ \ 4 cos^3 theta-3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (b))}`

`3 theta` `= 2 k pi`
`:. theta` `= (2 k pi)/3`

 

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`

Complex Numbers, EXT2 N2 2011 HSC 2b

On the Argand diagram, the complex numbers  `0, 1 + i sqrt 3 , sqrt 3 + i`  and  `z` form a rhombus.
 

  1. Find `z` in the form  `a + ib`, where `a` and `b` are real numbers.   (1 mark)

  2. An interior angle, `theta`, of the rhombus is marked on the diagram.

     

    Find the value of `theta.`   (2 marks)

Show Answers Only

a.    `(1 + sqrt 3) + i(1 + sqrt 3)`

b.    `(5 pi)/6`

Show Worked Solution
a.    `z` `= 1 + i sqrt 3 + sqrt 3 + i`
  `= (1 + sqrt 3) + i (1 + sqrt 3)`

 

b.    `text(arg)\ z = tan^-1 ((1 + sqrt 3)/(1 + sqrt 3)) = pi/4`

`text(arg)\ (sqrt 3 + i) = tan^-1 (1/sqrt 3) = pi/6`

`text(Difference) = pi/4-pi/6 = pi/12`
 

`=>\ text(Opposite angles of a rhombus are equal)`

`=>\ text(The diagonals of a rhombus bisect the angles)`

`:.theta= pi-2 xx pi/12= (5 pi)/6\ \ text{(angle sum of triangle)`

Complex Numbers, EXT2 N2 2012 HSC 12d

On the Argand diagram the points `A_1` and `A_2` correspond to the distinct complex numbers `u_1` and `u_2` respectively. Let `P` be a point corresponding to a third complex number `z`.

Points `B_1` and `B_2` are positioned so that `ΔA_1PB_1` and `ΔA_2B_2P`, labelled in an anti-clockwise direction, are right-angled and isosceles with right angles at `A_1` and `A_2`, respectively. The complex numbers `w_1` and `w_2` correspond to `B_1` and `B_2`, respectively.
 

Complex Numbers, EXT2 2012 HSC 12d1 

  1. Explain why  `w_1 = u_1 + i(z-u_1)`.   (1 mark)

  2. Find the locus of the midpoint of `B_1B_2` as `P`  varies.   (2 marks)

Show Answers Only

a.    `text(See Worked Solutions.)`

b.    `(u_1 + u_2)/2 + (u_2-u_1)/2 i`

Show Worked Solution
a.     `vec (A_1P)` `= z-u_1`
  `vec (A_1B_1)`  `= w_1-u_1` 

`B_1A_1 ⊥ A_1P\ text(and)\ |vec (A_1P)| = |vec (A_1B_1)|`

`vec (A_1B_1)\ text(is an anticlockwise rotation of)\ vec (A_1P)\ text(through)\ 90^@`

`w_1-u_1 = i(z −u_1)\ \ =>\ \ w_1 = u_1+ i(z −u_1)`

 

b.       `vec (A_2B_2)` `= w_2-u_2`
  `vec (A_2P)` `= z-u_2`

`A_2B_2 ⊥ A_2P\ text(and)\ |vec (A_2B_2)| = |vec (A_2P)|`

`vec (A_2P)\ text(is an anticlockwise rotation of)\ vec (A_2B_2)\ text(through)\ 90^@`

`z-u_2` `= i(w_2 −u_2)`
`iw_2` `= z-u_2 + iu_2`
`−w_2` `= iz-iu_2-u_2`
`:. w_2` `= u_2 + i(u_2-z)`

 

`:.\ text(The midpoint of)\ B_1B_2\ text(is)\ (w_1 + w_2)/2`

`= 1/2[u_1 + i(zvu_1) + u_2 + i(u_2-z)]`
`= 1/2[u_1 + u_2 + i(u_2-u_1)]`
`= (u_1 + u_2)/2 + (u_2-u_1)/2 i\ \ \ \ text{(which is a fixed point)}`

Complex Numbers, EXT2 N1 2012 HSC 11d

  1. Write  `z = sqrt3-i`  in modulus-argument form.  (2 marks)

  2. Hence express  `z^9`  in the form  `x + iy`, where `x` and `y` are real.  (1 mark)

Show Answers Only

a.    `2\ text(cis)(-pi/6)`

b.    `i512`

Show Worked Solution
a.     `z` `=sqrt3-i`
  `|\ z\ |` `=sqrt((sqrt3)^2+1^2)=2`

 

`:.z = sqrt3 − i` `= 2(sqrt3/2 − 1/2i)`
  `= 2(cos\ (-pi/6) + i\ sin\ (-pi/6))`  
  `= 2\ text(cis)(-pi/6)`  

 

b.     `z^9` `= 2^9\ (cos\ (-pi/6) + i\ sin\ (-pi/6))^9`
    `= 2^9\ text(cis)(-(9pi)/6)\ \ \ \ text{(by De Moivre)}`
    `=512\ text(cis)(-(3pi)/2)`
    `= 512(0 + i)`
    `=i512`

Complex Numbers, EXT2 N1 2013 HSC 11a

Let  `z = 2-i sqrt 3`  and  `w = 1 + i sqrt 3.`

  1. Find  `z + bar w.`   (1 mark)

  2. Express `w` in modulus–argument form.   (2 marks)

  3. Write `w^24` in its simplest form.   (2 marks)

Show Answers Only

a.    `3-i\ 2 sqrt 3`

b.    `2 text(cis) pi/3`

c.    `2^24`

Show Worked Solution

a.    `z = 2-i sqrt 3\ ,\ \ w = 1 + i sqrt 3`

`bar w = 1-i sqrt 3`

`z + bar w` `= 2-i sqrt 3 + 1-i sqrt 3`
  `= 3-i\ 2 sqrt 3`

 
b.    `|\ w\ |=sqrt(1^2 + (sqrt3)^2)=2`

 `:.w` `= 2 (1/2 + i sqrt 3/2)`
  `=2(cos\ pi/3 + i sin\ pi/3)`
  `= 2 text(cis) pi/3`
MARKER’S COMMENT: The directive “in its simplest form” required students to convert `text(cis)\ 8pi` to 1.

 

c.    `w^24` `= 2^24 text(cis)\ (24 xx pi/3)`
  `= 2^24\ text(cis)(8 pi)`
  `= 2^24`