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Complex Numbers, EXT2 N2 2024 HSC 4 MC

A monic polynomial, \(f(x)\), of degree 3 with real coefficients has \(3\) and  \(2+i\)  as two of its roots.

Which of the following could be \(f(x)\) ?

  1. \(f(x)=x^3-7 x^2-17 x+15\)
  2. \(f(x)=x^3-7 x^2+17 x-15\)
  3. \(f(x)=x^3+7 x^2-17 x+15\)
  4. \(f(x)=x^3+7 x^2+17 x-15\)
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\(B\)

Show Worked Solution

\(\text{Since coefficients are real, roots are:}\ \ 3, 2+i, 2-i\)

\[ \sum \text{roots} = 7 = \dfrac{-b}{1}\ \ \Rightarrow \ b=-7\]

\(\text{Product of roots}\ = 3(2+i)(2-i)=15=\dfrac{-d}{1}\ \ \Rightarrow \ d=-15\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N2 EQ-Bank 20

The polynomial  `p(z) = z^3 + alpha z^2 + beta z + gamma`, where  `z ∈ C`  and  `alpha, beta, gamma ∈ R`, can also be written as  `p(z) = (z-z_1)(z-z_2)(z-z_3)`, where  `z_1 ∈ R`  and  `z_2, z_3 ∈ C`.

  1. State the relationship between `z_2` and `z_3`.   (1 mark)

  2. Determine the values of  `alpha, beta` and `gamma`, given that  `p(2) = -13, |z_2 + z_3| = 0`  and  `|z_2-z_3| = 6`.   (3 marks)

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a.    `z_2 = barz_3`

b.    `alpha = -3, beta = 9, gamma = -27`

Show Worked Solution

a.    `text(By conjugate root theory)`

`z_2 = barz_3`
 

b.    `text(Let)\ \ z_1 = a + bi, \ z_2 = a-bi`

`|z_2 + z_3| = |2a| = 0 \ => \ a = 0`

`|z_2-z_3| = |2b| = 6 \ => \ b = ±3`
 

`text(Using)\ \ p(2) = -13:`

`(2-z_1)(2-3i)(2 + 3i)` `= -13`
`(2-z_1)(4 + 9)` `= -13`
`2-z_1` `= -1`
`z_1` `= 3`

 

`p(z)` `= (z-3)(z-3i)(z + 3i)`
  `= (z-3)(z^2 + 9)`
  `= z^3-3z^2 + 9z-27`

 
`:. alpha = –3, \ beta = 9, \ gamma = –27`

Complex Numbers, EXT2 N2 2021 HSC 6 MC

Which polynomial could have  `2 + i`  as a zero, given that `k` is a real number?

  1. `x^3 − 4 x^2 + k x`
  2. `x^3 − 4 x^2 + k x + 5`
  3. `x^3 − 5 x^2 + k x`
  4. `x^3 − 5 x^2 + k x + 5`
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`A`

Show Worked Solution

`text{Roots:} \ 2 + i \ , \ 2- i \ text{(conjugative roots),} \ alpha\ text{(real)}`

`-b` `= 2 + i + 2 – i + 2 = 4 + alpha`  
`k` `= (2 + i)(2 – i) + alpha(2 + i) + alpha (2 – i)`  
  `= 5 + 4 alpha`  
`-d` `= (2 + i)(2 – 1) α = 5 alpha`  

 

`text{Test coefficients for each option}`

`A: \ 4 + alpha = 4 \ => \ alpha = 0 \ , \ d= 0 \ text{(correct)}` 

`B: \ 4 + alpha = 4 \ => \ alpha = 0 \ , \ d= 5 ≠ 0`

`C: \ 4 + alpha = 5 \ => \ alpha = 1 \ , \ d= 0 ≠ -5`

`D: \ 4 + alpha = 5 \ => \ alpha = 1 \ , \ d= 5 ≠ -5`
 

`=>\ A`

Complex Numbers, EXT2 N2 2020 SPEC1 3

Find the cube roots of  `1/sqrt 2 - 1/sqrt 2 i`. Express your answers in modulus-argument form.  (3 marks)

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`z_1 = text(cis)((7 pi)/12)`

`z_2 = text(cis)(-pi/12)`

`z_3 = text(cis)((-3 pi)/4)`

Show Worked Solution

`z^3 = 1/sqrt 2 – 1/sqrt 2 i = text(cis)(-pi/4) = text(cis)(-pi/4 + 2 pi k), k ∈ Z`
 

`text(By De Moivre):`

`z = text(cis)(-pi/12 + (2 pi k)/3)`

`text(If)\ \ k = 1, \ z_1 = text(cis)(-pi/12 + (2 pi)/3) = text(cis)((7 pi)/12)`

`text(If)\ \ k = 0, \ z_2 = text(cis)(-pi/12)`

`text(If)\ \ k = – 1, \ z_3= text(cis)(-pi/12 – (2pi)/3) = text(cis)((-3 pi)/4)`

Complex Numbers, EXT2 N2 2019 SPEC1-N 1

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)

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`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, EXT2 N2 2017 HSC 6 MC

It is given that  `z = 2 + i`  is a root of  `z^3 + az^2 - 7z + 15 = 0`, where `a` is a real number.

What is the value of `a`?

  1. `−1`
  2. `1`
  3. `7`
  4. `−7`
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`A`

Show Worked Solution

`text(S)text(ince)\ \ z_1 = 2 + i\ \ text(is a root.)`

`=> z_2 = 2 – i\ \ text(is a root.)`

 

`text(Roots are)\ \ z_1, z_2, alpha.`

`text(Using product of roots:)`

`z_1z_2alpha` `= −15`
`(2 + i)(2 – i)alpha` `= −15`
`5alpha` `= −15`
`a` `= −3`

 

`text(Using sum of roots:)`

`z_1 + z_2 + alpha` `=-a`
`2 + i + 2 – i + −3` `= −a`
`a` `= −1`

`=> A`

Complex Numbers, EXT2 N2 2006 HSC 2c

Find, in modulus-argument form, all solutions of  `z^3 = -1.`   (2 marks)

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`text(cis)\ pi/3,\ \ \ -1,\ \ \ text(cis)\ -pi/3`

Show Worked Solution
`z^3` `=-1`
`-1` `=\ text(cis)\ (pi+2k pi)`
`:.z` `=\ text(cis)\ (pi+2k pi)/3\ \ \ text{(De Moivre)}`

 

MARKER’S COMMENT: Another successful solution strategy was graphical, using the unit circle and three equal angles generated from pi.

`text(When)\ k=0`

`z=\ text(cis)\ pi/3`

`text(When)\ k=1`

`z=\ text(cis)\ pi = -1`

`text(When)\ k=-1`

`z=\ text(cis)\ -pi/3`
 

`:.\ text(The 3 solutions to)\ \ z^3=-1\ \ text(are)`

`z=\ text(cis)\ pi/3,\ \ \ -1,\ \ \ text(cis)\ -pi/3`

Complex Numbers, EXT2 N2 2011 HSC 2c

Find, in modulus-argument form, all solutions of  `z^3 = 8.`  (2 marks)

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`2text(cis)\ 0,\ 2text(cis)\ (2pi)/3,\ 2text(cis)((-2pi)/3)`

Show Worked Solution

`z^3 = 8`

`z^3 = 8 [cos (2 k pi) + i sin (2 k pi)],\ \ k=0, +-1, +-2`

`z = 2[cos ((2 k pi)/3) + i sin ((2 k pi)/3)]\ \ \ text{(De Moivre)}`
 

`text(When)\ \ k = 0,`

`z` `= 2 (cos 0 + i sin 0)`
  `= 2text(cis)\ 0`

 

`text(When)\ \ k = 1,`

`z` `= 2 [cos ((2 pi)/3) + i sin((2 pi)/3)]`
  `=2text(cis)\ (2pi)/3`

 
`text(When)\ \ k = -1,`

`z` `= 2 [cos ((-2 pi)/3) + i sin ((-2 pi)/3)]`
  `=2 text(cis)((-2 pi)/3)`