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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 2023 HSC 12d

Find the cube roots of  \(2-2 i\). Give your answer in exponential form.  (3 marks)

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\(\sqrt[3]{2-2i}= \sqrt2 e^{- \frac{i\pi}{12}}, \sqrt2 e^{\frac{i7\pi}{12}}, \sqrt2 e^{- \frac{i3\pi}{4}} \)

Show Worked Solution

\(\text{Express}\ \ 2-2i\ \ \text{in exponential form:}\)
 

\(\big{|}2-2i\big{|}=2\sqrt2 \)

\(\arg(2-2i) = – \dfrac{\pi}{4} \)

\(2-2i=2\sqrt2 e^{-\frac{i\pi}{4}} \)
 

\(\text{Find}\ \ z=re^{i\theta},\ \ \text{where}\ \ z^3=r^3e^{i 3\theta}=2\sqrt2 e^{-\frac{i\pi}{4}} \)

\(r^3=2\sqrt2\ \ \Rightarrow\ \ r=\sqrt2 \)

\( i3\theta\) \(= -\dfrac{i\pi}{4} \)  
\(\theta_1\) \(=-\dfrac{\pi}{12} \)  

 
\(\text{Since roots are found symmetrically around Argand diagram:}\)

\(\theta_2=-\dfrac{\pi}{12}+\dfrac{2\pi}{3}=\dfrac{7\pi}{12} \)

\(\theta_3=-\dfrac{\pi}{12}-\dfrac{2\pi}{3}=-\dfrac{3\pi}{4} \)

\(\therefore \sqrt[3]{2-2i}= \sqrt2 e^{- \frac{i\pi}{12}}, \sqrt2 e^{\frac{i7\pi}{12}}, \sqrt2 e^{- \frac{i3\pi}{4}} \)

Complex Numbers, EXT2 N2 2022 HSC 16d

Find all the complex numbers `z_1, z_2, z_3` that satisfy the following three conditions simultaneously.  (3 marks)

`{[|z_(1)|=|z_(2)|=|z_(3)|],[z_(1)+z_(2)+z_(3)=1],[z_(1)z_(2)z_(3)=1]:}`

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`z_1,z_2,z_3=1,i,-i \ \ text{(in any order)}`

Show Worked Solution

`z_1z_2z_3=1\ \ =>\ \ abs(z_1) abs(z_2) abs(z_3)=1`

`text{Given}\ \ abs(z_1) = abs(z_2) = abs(z_3)`

`=>abs (z_1)^3=1 \ \ => \ \ abs(z_1)=1`

`:.abs(z_1) = abs(z_2) = abs(z_3)=1`
 


♦♦♦ Mean mark 12%.

`text{Consider}\ \ z_1=e^(itheta):`

`1/z_1=e^(-itheta)=bar(e^(itheta))=bar(z)_1`

`text{Similarly,}`

`1/z_2=bar(z)_2, \ 1/z_3=bar(z)_3`

`=>z_1z_2z_3=1`

  
`text{Consider}\ \ z_1z_2:`

`z_1z_2=1/z_3=barz_3`

`text{Similarly,}`

`z_2z_3=1/z_1=barz_1, \ z_1z_3=1/z_2=barz_2`

`z_1z_2+z_1z_3+z_2z_3` `=barz_3+barz_2+barz_1`  
  `=bar(z_1+z_2+z_3)`  
  `=1`  

 
`z_1, z_2,z_3\ \ text{are zeros of polynomial:}`

`z^3-z^2+z-1` `=0`  
`(z-1)(z^2+1)` `=0`  

 
`z_1,z_2,z_3=1,i,-i \ \ text{(in any order)}`

Complex Numbers, EXT2 N2 SM-Bank 10

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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  1. `z_2 = barz_3`
  2. `alpha = -3, beta = 9, gamma = -27`
Show Worked Solution

i.   `text(By conjugate root theory)`

`z_2 = barz_3`

 

ii.   `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 SM-Bank 8

If  `(x + iy)^3 = e^( - frac{i pi}{2}),\ \ x, y ∈ R`, find a solution in the form  `x + i y, \ x, y ≠ 0`.   (2 marks)

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`frac{sqrt3}{2} – i frac{1}{2}`

Show Worked Solution
`text{Let}`   `(x + i y)` `= cos theta + i sin theta`
  `(x + i y)^3` `= cos (3 theta) + isin (3 theta)`

 
`e^(- frac{i pi}{2}) = cos (frac{-pi}{2}) + i sin (frac{-pi}{2})`

COMMENT: Read the question fully and save time – only one solution is required here.

 

`text{Equating real parts:}`

`3 theta` `= frac{-pi}{2}`
`theta` `= frac{-pi}{6}`

 

`therefore \ x + i y` `= cos (frac{-pi}{6}) + i sin (frac{-pi}{6})`
  `= frac{sqrt3}{2} – i frac{1}{2}`

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`

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`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)`