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Complex Numbers, SPEC1 2025 VCAA 8

Consider the function with rule  \(f(z)=z^4+6 z^2+25\), where \(z \in C\).

  1. Consider  \(z_1=1+2 i\).
  2. Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below.   (1 mark)  

     
  3. Given that  \(1+2 i\)  is a solution of  \(f(z)=0\), find a quadratic factor of \(f(z)\).   (2 marks)  

  4. Hence, find all remaining solutions of  \(f(z)=0\).   (2 marks)

Show Answers Only

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(z^2-2 z+5\)

c.    \(z=-1+2 i, z=-1-2 i\)

Show Worked Solution

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

\(\Rightarrow 1-2 i \ \ \text{is also a solution.}\)

\(\text {Express as a quadratic factor:}\)

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

 

c.    \(f(z)=z^4+6 z^2+25, \quad z \in C\)

\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)

\(f(z)\) \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\)
  \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\)
  \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\)
  \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\)
Mean mark (c) 53%.

\(\text{Equating co-efficients:}\)

\(a-2=0 \ \Rightarrow \ a=2\)

\(5 b=25 \ \Rightarrow \ b=5\)

\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
 

\(\text{Solve:} \ \ z^2+2 z+5=0\)

\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)

\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)

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-NHT 2019 VCAA 6 MC

`P(z)`  is a polynomial of degree  `n`  with real coefficients where  `z ∈ C`. Three of the roots of the equation  `P(z) = 0`  are  `z = 3 - 2i`, `z = 4`  and  `z = −5i`.

The smallest possible value of  `n`  is

  1.  3
  2.  4
  3.  5
  4.  6
  5.  7
Show Answers Only

`C`

Show Worked Solution

`text(Roots:)\ 4, 3 ± 2i, ±5i\ \ \ (text(conjugate root theorem))`

`:.\ text(Minimum roots = 5)`

`=>\ C`

Complex Numbers, SPEC2 2011 VCAA 6 MC

The polynomial `P(z)` has real coefficients. Four of the roots of the equation  `P(z) = 0`  are  `z = 0`,  `z = 1 - 2i`,  `z = 1 + 2i`  and  `z = 3i`.

The minimum number of roots that the equation  `P(z) = 0`  could have is

A.   4

B.   5

C.   6

D.   7

E.   8

Show Answers Only

`B`

Show Worked Solution

`P(z) = 0\ text(has real coefficients,)`

`=>\ text(Conjugate root theorem applies)`

`z = −3i\ \ text(is also a root)`

`:.\ text(Minimum roots = 5)`

`=> B`

Complex Numbers, SPEC1 2013 VCAA 8

Find all solutions of  `z^4 - 2z^2 + 4 = 0,\ \ z in C`  in cartesian form.  (4 marks)

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`sqrt6/2 + i/sqrt2; −sqrt6/2 – i/sqrt2; sqrt6/2 – i/sqrt2; −sqrt6/2 + i/sqrt2`

Show Worked Solution
`(z^2)^2 – 2(z^2) + 4` `= 0`
`(z^2)^2 – 2z^2 + 1^2 – 1 + 4` `= 0`
`(z^2 – 1)^2 + 3` `= 0`
`(z^2 – 1)^2` `= −3`
`z^2 – 1` `= ±isqrt3`
`z^2` `= 1 ± isqrt3`

 

`text(If)\ \ z^2` `= 1 + isqrt3`
  `= sqrt(1 + 3)text(cis)(tan^(−1)(sqrt3))`
  `= 2text(cis)(pi/3)`

 
`z_1 = sqrt2 text(cis)(pi/6)`

♦ Mean mark 50%.

`z_2` `= sqrt2 text(cis)(pi/6 – pi)`
  `= sqrt2 text(cis)((−5pi)/6)`

 

`text(If)\ \ z^2` `= 1 – isqrt3`
  `= sqrt(1 + 3) text(cis)(−pi/3)`
  `= 2 text(cis)(−pi/3)`

 
`z_3 = sqrt2 text(cis)(−pi/6)`

`z_4` `= sqrt2 text(cis)(−pi/6 + pi)`
  `= sqrt2 text(cis)((5pi)/6)`

 
`:. z = sqrt2(sqrt3/2 + i/2), sqrt2(−sqrt3/2 – i/2), sqrt2(sqrt3/2 – i/2), sqrt2(−sqrt3/2 + i/2)`

`= sqrt6/2 +sqrt2/2 i; −sqrt6/2 – sqrt2/2 i; sqrt6/2 – sqrt2/2 i; −sqrt6/2 + sqrt2/2 i`

Complex Numbers, SPEC1 2014 VCAA 3

Let  `f` be a function of a complex variable, defined by the rule  `f(z) = z^4-4z^3 + 7z^2-4z + 6`.

  1.  Given that  `z = i`  is a solution of  `f(z) = 0`, write down a quadratic factor of  `f(z)`.   (2 marks)

  2.  Given that the other quadratic factor of  `f(z)`  has the form  `z^2 + bz + c`, find all solutions of  `z^4-4z^3 + 7z^2-4z + 6 = 0`  in a cartesian form.   (3 marks)

Show Answers Only
  1. `z^2 + 1`
  2. ` z = i, quad z = -i, quad z = 2 + i sqrt 2, quad z = 2-i sqrt 2`
Show Worked Solution

a.   `text(Real coefficients) \ => \ z = -i\ \ text(is also a solution)`

`:. (z-i)(z + i)= z^2 + 1\ \ text(is a quadratic factor)`

 

b.    `(z^2 + 1)(z^2 + bz + c)` `= z^4-4z^3 + 7z^2-4z + 6`
  `z^4 + bz^3 + (c + 1)z^2 + bz + c` `= z^4-4z^3 + 7z^2-4z + 6`

 
`text(Equating co-efficients:)`

`bz^3 = -4z^3\ \ =>\ \ b=-4`

`(c + 1)z^2 = 7z^2\ \ =>\ \ c = 6`

`(z^2 + 1)(z^2 + 4z + 6)` `= 0`
`(z^2 + 1)(z^2-4z + 2^2-4 + 6)` `= 0`
`(z^2 + 1)((z-2)^2 + 2)` `= 0`
`(z^2 + 1)((z-2)^2-2i^2)` `= 0`
`(z^2 + 1)(z-2-i sqrt 2)(z-2 + i sqrt 2)` `= 0`

 
`:.z = i, quad z = -i, quad z = 2 + i sqrt 2, quad z = 2-i sqrt 2`

Complex Numbers, SPEC2 2017 VCAA 4 MC

The solutions to  `z^n = 1 + i, \ n ∈ Z^+`  are given by

  1. `2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ R`
  2. `2^(1/n)text(cis)(pi/(4n) + 2pik), k ∈ Z`
  3. `2^(1/(2n))text(cis)(pi/4 + (2pik)/n), k ∈ R`
  4. `2^(1/n)text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
  5. `2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
Show Answers Only

`E`

Show Worked Solution
`z^n` `= sqrt(1^2 + 1^2)text(cis)(tan^(−1)(1/1))`
  `= sqrt2text(cis)(pi/4)`
`z_1` `= (2^(1/2))^(1/n)text(cis)(pi/(4n))`
`z_k` `= (2^(1/2))^(1/n)text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`
  `= 2^(1/(2n))text(cis)(pi/(4n) + (2pik)/n), k ∈ Z`

 
`=>E`

Complex Numbers, SPEC2 2017 VCAA 3 MC

The number of distinct roots of the equation  `(z^4 - 1)(z^2 + 3iz - 2) = 0`, where  `z ∈ C`  is

  1.  2
  2.  3
  3.  4
  4.  5
  5.  6
Show Answers Only

`D`

Show Worked Solution

`text(Solve:)\ \ z^4 = 1`

`z_1 = 1, \ \ z_2 = text(cis)(pi/2) = i, \ \ z_3 = text(cis)(pi)=−1,`

`z_4 = text(cis)((−pi)/4) = -i`

 
`text(Solve:)\ \ z^2 + 3iz – 2=0`

`z^2 + 3iz + ((3i)/2)^2 – (9i^2)/4 – 2` `= 0`
`(z + (3i)/2)^2 + (9 – 8)/4` `= 0`
`(z + (3i)/2)^2 + 1/4` `= 0`
`z + (3i)/2` `= ±i/sqrt4`
`z` `= (−3i ± i)/2`
`z_5` `= (−4i)/2 = −2i`
`z_6` `= (−2i)/2=-i = z_4`

`=>D`