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Complex Numbers, EXT2 N2 2025 SPEC1 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.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

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

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

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

 

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, 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. \(f(x)=x^3+7 x^2+17 x-15\)
Show Answers Only

\(B\)

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 2023 HSC 12e

The complex number  \(2+i\)  is a zero of the polynomial

\(P(z)=z^4-3 z^3+c z^2+d z-30\)

where \(c\) and \(d\) are real numbers.

  1. Explain why  \(2-i\)  is also a zero of the polynomial \(P(z)\).  (1 marks)

  2. Find the remaining zeros of the polynomial \(P(z)\).  (2 marks)

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i.    \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)

\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)

ii.   \(\text{Remaining zeros:}\ \ -3, 2 \)

Show Worked Solution

i.    \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)

\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)

 

ii.    \(P(z)=z^4-3 z^3+c z^2+d z-30\)

\(\text{Let roots be:}\ \ 2+i, 2-i, \alpha, \beta \)

\( \sum\ \text{roots:}\)

\(2+i+2-i+\alpha + \beta\) \(=-\dfrac{b}{a} \)  
\(4+\alpha+\beta\) \(=3\)  
\(\alpha + \beta\) \(=-1\ \ \ …\ (1) \)  

 
\(\text{Product of roots:} \)

\((2+i)(2-i)\alpha\beta \) \(= \dfrac{e}{a} \)  
\(5\alpha\beta\) \(=-30\)  
\(\alpha \beta \) \(=-6\ \ \ …\ (2) \)  

 
\(\text{Substitute}\ \ \beta=-\alpha-1\ \ \text{into (2):} \)

\(\alpha(-\alpha-1) \) \(=-6 \)  
\(-\alpha^2-\alpha \) \(=-6\)  
\(\alpha^2+\alpha-6\) \(=0\)  
\( (\alpha+3)(\alpha-2) \) \(=0\)  

 
\(\therefore\ \text{Remaining zeros:}\ \ -3, 2 \)

Complex Numbers, EXT2 N2 2022 HSC 6 MC

It is known that a particular complex number `z` is NOT a real number.

Which of the following could be true for this number `z` ?

  1. `bar(z)=iz`
  2. `bar(z)=|z^(2)|`
  3. `text{Re}(iz)= text{Im}(z)`
  4. `text{Arg}(z^(3))= text{Arg}(z)`
Show Answers Only

`A`

Show Worked Solution

`z in CC, z !in RR`

`text{Let}\ \ z=a+ib\ \ =>\ \ barz=a-ib`

`iz=i(a+ib)=-b+ia`

`text{If}\ \ barz=iz,`

`a-ib=-b+ia`

`a=-b\ \ text{(satisfies)}`

`:.∃ a,b (b!=0)\ \ text{such that}\ \ barz=iz`

`=>A`


♦♦ Mean mark 31%.

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)

Show Answers Only

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`
Show Answers Only

`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 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 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)

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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 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`
Show Answers Only

`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`