SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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

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 2020 VCAA 6 MC

For the complex polynomial  `P(z) = z^3 + az^2 + bz + c`  with real coefficients  `a, b` and `c, P(−2) = 0`  and  `P(3i) = 0`.

The values of  `a, b` and `c`  are respectively

  1. `− 2, 9,− 18`
  2. `3, 4, 12`
  3. `2, 9, 18`
  4. `−3, −4, 12`
  5. `2, −9, −18`
Show Answers Only

`C`

Show Worked Solution

`text(S)text(ince coefficients are real,)`

`P(3i) = 0 => P(−3i) = 0\ \ (text(conjugate root theorem))`

`P(z)` `= (z + 2)(z – 3i)(z + 3i)`
  `= z^3 + 2z^2 + 9z + 18`

 
`:. a = 2, b = 9, c = 18`

`=> C`

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-NHT 2019 VCAA 2

A cubic polynomial has the form  `p(z) = z^3 + bz^2 + cz + d, \ z ∈ C`, where  `b, c, d ∈ R`.

Given that a solution of  `p(z) = 0`  is  `z_1 = 3 - 2i`  and that  `p(–2) = 0`, find the values of  `b, c` and `d`.   (4 marks)

Show Answers Only

`b =-4 \ , \ c = 1 \ , \ d = 26`

Show Worked Solution

`text(Roots:)\  \ z_1 = 3 – 2i \ , \ z_2 = 3 + 2i \ , \ z_3 = -2`
 

`p(z)` `= (z – 3 + 2i )(z – 3 – 2i )(z + 2)`
  `= ((z-3)^2 – (2i)^2)(z+2)`
  `= (z^2 – 6z + 9 + 4)(z + 2)`
  `= (z^2 – 6z + 13)(z + 2)`
  `= z^3 + 2z^2 – 6z^2 – 12z + 13z + 26`
  `= z^3 – 4z^2 + z + 26`

 

`:. \ b =-4 \ , \ c = 1 \ , \ d = 26`

Complex Numbers, SPEC2 2016 VCAA 4 MC

One of the roots of   `z^3 + bz^2 + cz = 0`  is  `3 - 2i`, where `b` and `c` are real numbers.

The values of `b` and `c` respectively are

A.  `6, 13`

B.  `3, -2`

C.  `-3, 2`

D.  `2, 3`

E.  `-6, 13` 

Show Answers Only

`E`

Show Worked Solution

`z(z^2 + bz + c) = 0,\ b, c in RR`

`text(Given)\ \ z= 3 – 2i \ \ text(is a root,)`

`=> z = 3 + 2i\ \ text{is a root (conjugate).}`
 

`z(z^2 + bz + c)` `= 0`
`z((z – 3) + 2i)((z – 3) – 2i)` `= 0`
`z((z – 3)^2 – 4i^2)` `= 0`
`z(z^2 – 6z + 9 + 4)` `= 0`
`z^3 – 6z^2 + 13z` `= 0`

 
`b = -6,\ \ c = 13` 

`=>  E`

Complex Numbers, SPEC1-NHT 2017 VCAA 4

Find the values of `a` and `b` given that  `z - 1 - i`  is a factor of  `z^3 + (a + b)z^2 + (b^2 - a)z - 4 = 0`, where `a` and `b` are real constants.  (4 marks)

Show Answers Only

`(a, b) = (-5, 1) or (-2, -2)`

Show Worked Solution

`text(All coefficients are real):`

`=> z – 1 + i \ \  text{is a factor (conjugate pair)}`

`(z – 1 – i) (z – 1 + i)` `= ((z – 1) – i)((z – 1) + i)`
  `= (z – 1)^2 – i^2`
  `= z^2 – 2z + 1 + 1`
  `= z^2 – 2z + 2`

 
`(z – alpha)(z^2 – 2z + 2)`

`=z^3 -2z^2 + 2z -alpha z^2 +2 alpha z – 2 alpha, \ \ \ alpha in RR`

`=z^3 + (-alpha – 2)z^2 + (2 alpha + 2)z – 2 alpha`

`= z^3 + (a + b)z^2 + (b^2 – a) z – 4`
 

`text(Equating coefficients:)`

`2 alpha = -4 \ \ =>\ \ alpha = 2`

`a + b = -4 \ \ …\ (1)`

`b^2 – a = 6 \ \ …\ (2)`

 
`text{Substitute}\ \ a=-b-4\ \ text{from (1) into (2):}`

`b^2 – (-4 – b)` `= 6`
`b^2 + b + 4` `= 6`
`b^2 + b – 2` `=0`
`(b + 2)(b – 1)` `=0`

 
`b = 1\ \ => \ \ a=-5, \ text(or)`

`b=-2\ \ => \ \ a = -4 + 2=-2`

`:. (a, b) = (-5, 1) or (-2, -2)`

Complex Numbers, SPEC2 2014 VCAA 7 MC

The sum of the roots of  `z^3 - 5z^2 + 11z - 7 = 0`,  where  `z ∈ C`,  is

  1. `1 + 2sqrt3i`
  2. `5i`
  3. `4 - 2sqrt3i`
  4. `2sqrt3i`
  5. `5`
Show Answers Only

`E`

Show Worked Solution

`text(Coefficients are real)`

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

`z_1=alpha + beta i,\ \ z_2=alpha – beta i, \ \ z_3=gamma`

`z_1+z_2+z_3` `= alpha + beta i + alpha – beta i + gamma`
  `= 2 alpha + gamma\ \ \ (∈R)`

 

`=> E`

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, SPEC1 VCAA 2017 3

Let  `z^3 + az^2 + 6z + a = 0, \ z ∈ C`, where `a` is a real constant.

Given that  `z = 1 - i`  is a solution to the equation, find all other solutions.  (3 marks)

Show Answers Only
`z_2` `= 1 + i`
`z_3` `= 2`
Show Worked Solution

`z_1 = 1 – i`

`=>\ z_2 = 1 + i\ \ text{(conjugate pair)}`

`=>\ z_3 ∈ R`
 

`z^3 + az^a + 6z + a`

  `= (z – z_1)(z – z_2)(z – z_3)`
  `= (z – (1 – i))(z – (1 + i))(z – z_3)`
  `= ((z – 1) + i)((z – 1) – i)(z – z_3)`
  `= ((z – 1)^2 – i^2)(z – z_3)`
  `= (z^2 – 2z + 1 + 1)(z – z_3)`
  `= (z^2 – 2z + 2)(z – z_3)`
  `= z^3 – (z + z_3)z^2 + (2z_3 + 2)z – 2z_3`

 
`text(Equating the co-efficients of)\ \ z:`

`6` `= 2z_3 + 2`
`4` `= 2z_3`
`z_3` `= 2`

 
`:.\ text(Other solutions are:)`

`z_2` `= 1 + i`
`z_3` `= 2`

Complex Numbers, SPEC2-NHT 2018 VCAA 6 MC

Given that  `(z - 3i)`  is a factor of  `P(z) = z^3 + 2z^2 + 9z + 18`, which one of the following statements is false?

  1. `P(3i) = 0`
  2. `P(-3i) = 0`
  3. `P(z)`  has three linear factors over `C`
  4. `P(z)`  has no real roots
  5. `P(z)`  has two complex conjugate roots
Show Answers Only

`D`

Show Worked Solution

`z – 3i\ \ text(is a factor) \ => \ z + 3i\ text(is a factor)`

`text(deg) (P) = 3`

`text(Complex roots occur in conjugate pairs.)`

`text(S)text(ince 3 roots → 3rd root must be real.)`

`=>   D`