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Complex Numbers, SPEC1 2024 VCAA 1

Consider the function with rule  \(f(z)=3 z^3+2 i z^2+3 z+2 i\),  where  \(z \in C\).

  1. Verify that  \(3 z+2 i\)  is a factor of \(f(z)\).   (1 mark)

  2. Hence or otherwise, solve the equation  \(f(z)=0\).   
  3. Give your answers in Cartesian form.  (2 marks)

  4. Plot the solutions of  \(f(z)=0\)  on the Argand diagram below.  (1 mark)

Show Answers Only

a.   \(\text{Method 1}\)

\(f(z)=3 z^3+2 i z^2+3 z+2 i = (3z+2i)(z^2+1)\)
 

\(\text{Method 2}\)

\(f\left( -\dfrac{2i}{3} \right)=\dfrac{8}{9}-\dfrac{8}{9}-2i+2i=0\)

\(\text{By factor theorem:}\)

\(z+ \dfrac{2}{3} i\ \ \text{is a factor of}\ f(z) \)

\(\text{Hence,}\ 3\left( z+\dfrac{2i}{3} \right) = 3z+2i\ \ \text{is a factor.}\)
 

b.   \(f(z)= (3z+2i)(z^2+1)\)

\(z=-\dfrac{2i}{3} , z= \pm i\)
 

c.   
         

Show Worked Solution

a.   \(\text{Method 1}\)

\(f(z)=3 z^3+2 i z^2+3 z+2 i = (3z+2i)(z^2+1)\)
 

\(\text{Method 2}\)

\(f\left( -\dfrac{2i}{3} \right)=\dfrac{8}{9}-\dfrac{8}{9}-2i+2i=0\)

\(\text{By factor theorem:}\)

\(z+ \dfrac{2}{3} i\ \ \text{is a factor of}\ f(z) \)

\(\text{Hence,}\ 3\left( z+\dfrac{2i}{3} \right) = 3z+2i\ \ \text{is a factor.}\)
 

b.   \(f(z)= (3z+2i)(z^2+1)\)

\(z=-\dfrac{2i}{3} , z= \pm i\)
 

c.   
         

♦ Mean mark (c) 49%.

Complex Numbers, SPEC2 2023 VCAA 2

Let \(w=\text{cis}\left(\dfrac{2 \pi}{7}\right)\).

  1. Verify that \(w\) is a root of  \(z^7-1=0\).   (1 marks)
  1. List the other roots of  \(z^7-1=0\)  in polar form.   (1 mark)
  1. On the Argand diagram below, plot and label the points that represent all the roots of  \(z^7-1=0\).   (2 marks)

  1.  i. On the Argand diagram below, sketch the ray that originates at the real root of  \(z^7-1=0\)  and passes through the point represented by \( \text{cis}\left(\dfrac{2 \pi}{7} \right)\).   (1 mark)

     

  1. ii. Find the equation of this ray in the form \(\text{Arg}\left(z-z_0\right)=\theta\), where \(z_0 \in C\), and \(\theta\) is measured in radians in terms of \(\pi\).   (1 mark)

  2. Verify that the equation  \(z^7-1=0\)  can be expressed in the form 
  3.     \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\).   (1 mark)
  4.   i. Express  \(\text{cis}\left(\dfrac{2 \pi}{7}\right)+\operatorname{cis}\left(\dfrac{12 \pi}{7}\right)\) in the form \(A \cos (B \pi)\), where \(A, B \in R^{+}\).   (1 mark)

  5. ii. Given that  \(w=\operatorname{cis}\left(\dfrac{2 \pi}{7}\right)\)  satisfies  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\),

use De Moivre's theorem to show that

      1. \(\cos \left(\dfrac{2 \pi}{7}\right)+\cos \left(\dfrac{4 \pi}{7}\right)+\cos \left(\dfrac{6 \pi}{7}\right)=-\dfrac{1}{2}\).   (2 marks)

Show Answers Only

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Show Worked Solution

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Complex Numbers, SPEC2 2019 VCAA 2

  1. Show that the solutions of  `2z^2 + 4z + 5 = 0`, where  `z ∈ C`, are  `z = −1 ± sqrt6/2 i`.   (1 mark)

  2. Plot the solutions of  `2z^2 + 4z + 5 = 0`  on the Argand diagram below.   (1 mark)

     

     

Let  `|z + m| = n`, where  `m, n ∈ R`, represent the circle of minimum radius that passes through the solutions of  `2z^2 + 4z + 5 = 0`.

    1. Find  `m`  and  `n`.   (2 marks)

    2. Find the cartesian equation of the circle  `|z + m| = n`.   (1 mark)

    3. Sketch the circle on the Argand diagram in part a.ii. Intercepts with the coordinate axes do not need to be calculated or labelled.   (1 mark)

  2. Find all values of  `d`, where  `d ∈ R`, for which the solutions of  `2z^2 + 4z + d = 0`  satisfy the relation  `|z + m| <= n`.   (2 marks)

  3. All complex solutions of  `az^2 + bz + c = 0`  have non-zero real and imaginary parts.

     

    Let  `|z + p| = q`  represent the circle of minimum radius in the complex plane that passes through these solutions, where  `a, b, c, p, q ∈ R`.

     

    Find  `p`  and  `q`  in terms of  `a, b`  and  `c`.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.   
  3. `m = 1, n = sqrt6/2`
  4. `(x + 1)^2 + y^2 = 3/2`
  5. `−1 <= d <= 5\ \ (text(by CAS))`
  6. `p = b/(2a), \ q = |sqrt(b^2-4ac)/(2a)|`
Show Worked Solution
a.i.    `z` `= (−b ± sqrt(b^2-4ac))/(2a)`
    `= (−4 ± sqrt(16-4 · 2 · 5))/(4)`
    `= (−4 ± 2sqrt6 i)/(4)`
    `= −1 ± sqrt6/2 i\ \ …\ text(as required)`

 

a.ii.   

 

b.i.   `text(Radius of circle = )sqrt6/2`

 `text(Centre) = (0, −1)`

`:. m = 1, \ n = sqrt6/2`

 

b.ii.    `|z + 1|` `= sqrt6/2`
  `|x + iy + 1|` `= sqrt6/2`
  `(x + 1)^2 + y^2` `= 3/2`

 

b.iii.   

 

c.   `text(Solve:)\ 2z^2 + 4z + d = 0`

`z = −1 ± sqrt(4-2d)/2 = −1 ± sqrt((2-d)/2)`

`z + 1 = ± sqrt((2-d)/2)`
 

`text(Solve for)\ d\ text(such that:)`

`|sqrt((2-d)/2)| <= sqrt6/2`

`−1 <= d <= 5\ \ (text(by CAS))`

 

d.   `z = (−b ± sqrt(b^2-4ac))/(2a) = (−b)/(2a) ± sqrt(b^2-4ac)/(2a)`

`z + b/(2a)` `= ± sqrt(b^2-4ac)/(2a)`
`|z + b/(2a)|` `= |sqrt(b^2-4ac)/(2a)|`

 
`:. p = b/(2a), \ q = |sqrt(b^2-4ac)/(2a)|`

Complex Numbers, SPEC2 2011 VCAA 8 MC

On the argand diagram below, the twelve points  `P_1, P_2, P_3 …, P_12`  are evenly spaced around the circle of radius 3.
 

SPEC2 2011 VCAA 8 MC
 

The points which represent complex numbers such that  `z^3 = -27i`  are

A.   `P_10` only

B.   `P_4` only

C.   `P_2, P_6, P_10`

D.   `P_3, P_7, P_11`

E.   `P_4, P_8, P_12`

Show Answers Only

`E`

Show Worked Solution

`z^3 = 27text(cis)(−pi/2)`

♦ Mean mark 49%.

`:. z_1 = 3text(cis)(−pi/6)\ \ text(is one root.)`

`text(The other two are evenly spaced around the circle)`

`text(which separates them by)\ \ (2pi)/3.`

`=> E`

Complex Numbers, SPEC1 2012 VCAA 3

Consider the equation  `z^3-z^2-2z-12 = 0, \ z in C.`

  1. Given that  `z = 2 text(cis) ((2pi)/3)`  is a root of the equation, find the other two roots in the form  `a + ib`, where  `a, b in R.`   (3 marks)

  2. Plot all of the roots clearly on the Argand diagram below.   (1 mark)

     

     


Show Answers Only
  1. `z_2 = -1-i sqrt 3`
    `z_3 = 3 + 0i`
     

  2.  

Show Worked Solution

a.   `text(Given)\ \  z = 2 text(cis)((2 pi)/3)\ \ text(is a root, and)`

`z^3-z^2-2z-12 = 0\ \ text(has real coefficients,)`

`=> z= 2 text(cis)(-(2 pi)/3)\ \ text{is a root (conjugate).}`

`=> text(Roots:)\ \ -1+sqrt3i,\ \ -1-sqrt3i`
 

`(z-alpha) (z-beta) (z-gamma) = z^3-z^2-2z-12`
 

`text(Using)\ \ alpha beta gamma = 12:`

`(-1 + i sqrt 3) (-1-i sqrt 3) gamma` `=12`  
`((-1)^2-(sqrt3i)^2)gamma` `=12`  
`4gamma` `=12`  
`gamma` `=3`  

 
`:.\ text(Other two roots:)`

♦♦ Mean mark part (b) 33%.

`z_2` `= -1-i sqrt 3`
`z_3` `= 3 + 0i`

 

b.   

Complex Numbers, SPEC2 2017 VCAA 4

  1. Express  `−2-2sqrt3 i`  in polar form.  (1 mark)

  2. Show that the roots of  `z^2 + 4z + 16 = 0`  are  `z = −2-sqrt3 i`  and  `z = −2 + 2sqrt3 i`.  (1 mark)

  3. Express the roots of  `z^2 + 4z + 16 = 0`  in terms of  `2-2sqrt3 i`.  (1 mark)

  4. Show that the cartesian form of the relation  `|z| = |z-(2-2sqrt3 i)|`  is  `x-sqrt3 y-4 = 0`  (2 marks)

  5. Sketch the line represented by  `x-sqrt3y -4 = 0`  and plot the roots of  `z^2 + 4z + 16 = 0`  on the Argand diagram below.  (2 marks)

     
       

  6. The equation of the line passing through the two roots of  `z^2 + 4z + 16 = 0`  can be expressed as  `|z-a| = |z-b|`, where  `a, b ∈ C`.

     

    Find `b` in terms of `a`.  (1 mark)

  7. Find the area of the major segment bounded by the line passing through the roots of  `z^2 + 4z + 16 = 0`  and the major arc of the circle given by  `|z| = 4`.  (2 marks)

Show Answers Only
  1. `4text(cis)((−2pi)/3)`
  2. `text(See Worked Solutions)`
  3. `−(2-2sqrt3 i) and-bar((2-2sqrt3 i))`
  4. `text(See Worked Solutions)`
  5.  
  6. `−4-bara`
  7. `4sqrt3 + (32pi)/3`
Show Worked Solution
a.    `r` `= sqrt((−2)^2 + (−2sqrt3)^2)=4`

 

`theta` `= −pi + tan^(−1)((2sqrt3)/2)`
  `= −pi + pi/3`
  `=(-2pi)/3`

 
`:. −2-2sqrt3 i = 4text(cis)((−2pi)/3)`
 

b.   `z^2 + 4z + z^2-4 + 16` `=0`
`(z + 2)^2 + 12` `= 0`
`(z + 2)^2` `= -12`
`(z + 2)^2` `= 12i^2`
`z + 2` `= ±sqrt12 i`
`z + 2` `= ±2sqrt3 i`
`:. z` `= -2 ± 2sqrt3 i`

 

♦♦ Mean mark part (c) 31%.

c.    `z_1` `= -2 + 2sqrt3 i = -(2-2sqrt3 i)`
  `z_2` `=-2-2sqrt3 i =-bar((2-2sqrt3 i))`

 

d. `|z|` `= |z-(2-2sqrt3 i)|`
     `x^2 + y^2` `= (x-2)^2 + (y + 2sqrt3)^2`
    `x^2 + y^2` `= x^2-4x + 4+ y^2 + 4sqrt3 y + 12`
  `0` `= −4x + 4sqrt3 y + 16`
  `0` `= −x + sqrt3 y + 4`

 
`:. x-sqrt3 y-4=0`

 

e.   

 

f.   `x = − 2\ \ text(is equidistant from)\ \ z_1 = a\ \ text(and)\ \ z_2 = b`

♦♦♦ Mean mark 1%!

`=> text(Im)(a) = text(Im)(a)`
 

`text(Let)\ \ a = alpha + betai, \ b = gamma + betaj`

`(alpha + gamma)/2` `= −2`
`alpha + gamma` `= −4`
`gamma` `= -4-alpha`

  

`:. b` `= -4-alpha + betaj`
  `= -4-(alpha + betaj)`
  `= -4-bara`

 

♦♦ Mean mark 31%.

g.    `text(Area)\ DeltaOAB` `= 1/2 xx (4sqrt3 xx 2)`
    `= 4sqrt3`

 

`text(Area of sector)\ AOB` `= pi xx 4^2 xx (2 xx pi/3)/(2pi)`
  `= (16pi)/3`

 
`:.\ text(Area of major segment area)`

`=pi(4)^2-((16pi)/3-4sqrt3)`

`= 4sqrt3 + (32pi)/3`

Complex Numbers, SPEC2-NHT 2017 VCAA 2

One root of a quadratic equation with real coefficients is  `sqrt 3 + i`.

    1. Write down the other root of the quadratic equation.   (1 mark)

    2. Hence determine the quadratic equation, writing it in the form  `z^2 + bz + c = 0`.   (2 marks)

  1. Plot and label the roots of   `z^3-2 sqrt 3 z^2 + 4z = 0`  on the Argand diagram below.   (3 marks)

  

 

  1. Find the equation of the line that is the perpendicular bisector of the line segment joining the origin and the point  `sqrt 3 + i`. Express your answer in the form  `y = mx + c`.   (2 marks)

  1. The three roots plotted in part b. lie on a circle.

     

    Find the equation of this circle, expressing it in the form  `|z-alpha| = beta`,  where  `alpha, beta in R`.   (3 marks)

Show Answers Only
    1. `z_2 = sqrt 3-i`
    2. `z^2-2 sqrt 3 z + 4 = 0`
  2. `text(See Worked Solutions)`
  3. `y = -sqrt 3 x + 2`
  4. `|z-2/sqrt 3 | = 2/sqrt 3`
Show Worked Solution

a.i.   `z_1 = sqrt 3 + i`

 `z_2 = bar z_1 = sqrt 3-i\ \ \ text{(conjugate root)}`

 

a.ii.    `(z-(sqrt 3 + i))(z-(sqrt 3-i))` `= 0`
  `((z-sqrt 3)-i)((z-sqrt 3) + i)` `= 0`
  `(z-sqrt 3)^2-i^2` `= 0`
  `z^2-2 sqrt 3 z + 3 + 1` `= 0`
  `z^2-2 sqrt 3 z + 4` `= 0`

 

b.    `z(z^2-2 sqrt 3 z + 4)` `= 0`
  `z(z-(sqrt 3 + i))(z-(sqrt 3-i))` `= 0`

 
`text(Convert to polar form:)`

  `sqrt 3 + i` `= sqrt((sqrt 3)^2 + 1^2) * text(cis) (tan^(-1)(1/sqrt 3))`
    `= 2 text(cis) (pi/6)`
  `=> sqrt 3-i` `= 2 text(cis) (-pi/6)`

 

   

 

c.   `text(Equidistant from)\ (0, 0) and (sqrt 3, 1)`

`text(Midpoint)\ (x_1, y_1)` `= (sqrt 3/2, 1/2)`

 
`m = (1-0)/(sqrt 3-0) = 1/sqrt 3`

`m_(_|_) = (-1)/m = -sqrt 3`
 

`:.\ text(Equation of ⊥ bisector:)`

`y-1/2` `=-sqrt3(x-sqrt3/2)`  
`y` `=-sqrt3 x +3/2+1/2`  
`:.y` `=-sqrt3 x +2`  

 

d.   `text(Let)\ O = (0, 0),\ P = (sqrt 3, 1),\ Q = (sqrt 3, -1)`

`text(⊥ bisector of two points on arc of a circle passes)`

`text(through the centre of the circle.)`
 

`OP = OQ = PQ = 2`

`=> Delta OPQ\ text(is equilateral)`
 

`text(⊥ bisector of)\ PQ\ text(is)\ y=0.`

`text(Centre of circle occurs when:)`

`0 = -sqrt 3 x + 2\ \ text{(using part c)`

`x=2/sqrt3`

`=>\ text(Radius)\ = 2/sqrt3`
 

`:. |z-2/sqrt 3| = 2/sqrt 3`