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Complex Numbers, EXT2 N2 2024 HSC 9 MC

Consider the solutions of the equation  \(z^4=-9\).

What is the product of all of the solutions that have a positive principal argument?

  1. \(3\)
  2. \(-3\)
  3. \(3 i\)
  4. \(-3 i\)
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\(B\)

Show Worked Solution

\(z^4=-9\)

\(\text{Convert}\ z^4 \ \text{to Mod/Arg form:}\)

\(\left|z^4\right|=9, \ \ \arg \left(z^4\right)=\pi \ \text{(\(-9\) is on negative real axis})\)
 

\(\text{By De Moivre:}\)

   \(\abs{z}=\sqrt[4]{9}=\sqrt{3}\)

   \(\arg (z)=\dfrac{\pi}{4}\)

\(\text{Roots are} \ \ \dfrac{\pi}{2} \ \ \text{rotations of}\ \  z=\sqrt{3} \, \text{cis}\left(\dfrac{\pi}{4}\right)\)

\(z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{\pi}{4}\right), z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{3 \pi}{4}\right)\)

\(\sqrt{3}\, \text{cis}\left(\dfrac{\pi}{4}\right) \cdot \sqrt{3} \, \text{cis}\left(\dfrac{3 \pi}{4}\right)=3 \, \text{cis}(\pi)=-3\)

\(\Rightarrow B\)

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 13c

Consider the equation  `z^5+1=0`, where `z` is a complex number.

  1. Solve the equation  `z^5+1=0`  by finding the 5th roots of `-1`.  (2 marks)

  2. Show that if `z` is a solution of  `z^5+1=0`  and  `z !=-1`, then  `u=z+(1)/(z)`  is a solution of  `u^2-u-1=0`.  (2 marks)

  3. Hence find the exact value of  `cos\ (3pi)/(5)`.  (3 marks)

Show Answers Only
  1. `z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  2. `text{Proof (See Worked Solutions)}`
  3. `(1-sqrt5)/4`
Show Worked Solution

i.   `z^5+1=0\ \ =>\ \ z^5=-1`

`z=e^(i((2k+1)/5)),\ \ kin{0,1,-1,2,-2}`

`:.z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  

ii.   `z^5+1=(z+1)(z^4-z^3+z^2-z+1)`

`text{Given}\ \ z!=-1,`

`z^4-z^3+z^2-z+1=0`
 

`text{Divide by}\ z^2\ \ (z!=0)`

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

 
`text{Let}\ \ u=z+1/z:`

`:.u^2-u-1=0`
 


Mean mark (ii) 53%.

iii.  `u^2-u-1=0`

`text{By quadratic formula:}`

`u` `=(1+-sqrt(1-4xx1xx(-1)))/(2)`  
  `=(1+-sqrt5)/2`  

♦ Mean mark (iii) 43%.
`z+1/z` `=(1+-sqrt5)/2`  
`e^(i(3pi)/5)+e^(-i(3pi)/5)` `=(1-sqrt5)/2,\ \ (cos\ (3pi)/5 <0)`  
`2cos((3pi)/5)` `=(1-sqrt5)/2`  
`cos((3pi)/5)` `=(1-sqrt5)/4`  

Complex Numbers, EXT2 N2 SM-Bank 1

Find all solutions for `z`, in exponential form, given  `z^4 = -2 sqrt3 - 2 i`.   (3 marks)

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`sqrt2 e^(-frac{i 17 pi}{24}) \ , \  sqrt2 e^(-frac{i 5 pi}{24}) \ , \  sqrt2 e^(frac{i 7 pi}{24}) \ , \  sqrt2 e^(frac{i 19 pi}{24})`

Show Worked Solution

`text{Convert} \ \ z^4\ \ text{to Mod/Arg:}`

`| z^4 | = sqrt{(2 sqrt3)^2 + 2^2} = 4`
 

 

`tan \ theta` `= frac{2 sqrt3}{2} = sqrt3`
`theta` `= frac{pi}{3}`

`text{arg} (z^4) = – (frac{pi}{2} + frac{pi}{3}) = – frac{5 pi}{6}`
 

`text{By De Moivre:}`

`| z | = root4 (4) = sqrt2`

`text{arg}(z) = -frac{5 pi}{24} + frac{ k pi}{2} \ , \ k = 0 , ± 1 , ± 2`

`k = 0:\ text{arg} (z)=-frac{5 pi}{24}`

`k = 1:\ text{arg} (z)= -frac{5 pi}{24} + frac{pi}{2} = frac{7 pi}{24}`

`k = – 1:\ text{arg} (z)= -frac{5 pi}{24} – frac{pi}{2} = frac{-17 pi}{24}`

`k = 2:\ text{arg} (z)= -frac{5 pi}{24} + pi = frac{19 pi}{24}`

 
`therefore \ text{Solutions are:} \ sqrt2 e^(- frac{i 17 pi}{24}) \ , \  sqrt2 e^(frac{ -i 5 pi}{24}) \ , \  sqrt2 e^(frac{i 7 pi}{24}) \ , \  sqrt2 e^(frac{i 19 pi}{24})`

Complex Numbers, EXT2 N2 EQ-Bank 1

`z = sqrt2 e^((ipi)/15)`  is a root of the equation  `z^5 = alpha(1 + isqrt3), \ alpha ∈ R`.

  1. Express  `1 + isqrt3`  in exponential form.  (2 marks)

  2. Find the value of `alpha`.  (1 mark)

  3. Find the other 4 roots of the equation in exponential form.  (3 marks)

Show Answers Only
  1. `2e^((ipi)/3)`
  2. `alpha = 2sqrt2`
  3. `e^((i11pi)/15), e^(-(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`
Show Worked Solution

i.   `beta = 1 + isqrt3`

`|beta| = sqrt(1 + (sqrt3)^2) = 2`

`text(arg)(beta) = tan^(−1) (sqrt3/1) = pi/3`

`beta = 2e^((ipi)/3)`

 

ii.    `z` `= sqrt2 e^((ipi)/15)`
  `z^5` `= (sqrt2 e^((ipi)/15))^5`
    `= (sqrt2)^5 e^((ipi)/15 xx 5)`
    `= 4sqrt2 e^((ipi)/3)`

 
`:. alpha = 2sqrt2`

 

iii.   `text(arg)(z^5) = pi/3 + 2kpi, \ \ k = 0, ±1, ±2, …`

`text(arg)(z) = pi/15 + (2kpi)/5`

`k = 1:\ text(arg)(z) = pi/15 + (2pi)/5 = (11pi)/15`

`k = text(−1):\ text(arg)(z) = pi/15 – (2pi)/5 = −pi/3`

`k = 2:\ text(arg)(z) = pi/15 + (4pi)/5 = (13pi)/15`

`k =text(−2):\ text(arg)(z) = pi/15 – (4pi)/5 = −(11pi)/15`
 

`:. 4\ text(other roots are:)`

`e^((i11pi)/15), e^(−(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)` 

Complex Numbers, EXT2 N2 2018 HSC 6 MC

Which complex number is a 6th root of `i`?

  1. `−1/sqrt2 + 1/sqrt2i`
  2. `−1/sqrt2 - 1/sqrt2i`
  3. `−sqrt2 + sqrt2i`
  4. `−sqrt2 - sqrt2i`
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`A`

Show Worked Solution

`text(Consider option A:)`
 

`|−1/sqrt2 + 1/sqrt2i|= sqrt((−1/sqrt2)^2 + (1/sqrt2)^2) = 1`

`text(arg)(z)` `= (3pi)/4`
`z` `= 1(cos\ (3pi)/4 + i sin\ (3pi)/4)`
`z^6` `= cos\ (18pi)/4 + i sin\ (18pi)/4\ \ \ text{(De Moivre)}`
  `= i`

 
`=> A`

Complex Numbers, EXT2 N2 2017 HSC 1 MC

The complex number `z` is chosen so that  `1, z, …, z^7`  form the vertices of the regular polygon shown.

Which polynomial equation has all of these complex numbers as roots?

  1. `x^7 - 1 = 0`
  2. `x^7 + 1 = 0`
  3. `x^8 - 1 = 0`
  4. `x^8 + 1 = 0`
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`C`

Show Worked Solution

`P(x)\ \ text(has 8 separate roots.)`

`:.\ text(Must be of degree at least 8.)`

`text(S) text(ince 1 is also a root,)`

`=>  C`

Complex Numbers, EXT2 N2 2009 HSC 2e

  1. Find all the 5th roots of  `–1`  in modulus-argument form.   (2 marks)

  2. Sketch the 5th roots of  `–1`  on an Argand diagram.  (1 mark)

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  1. `z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis) (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,``z_4 = text(cis) (7 pi)/5,\ \ z_5 = text(cis) (9 pi)/5`
  2.  
Show Worked Solution
i.   `z` `=cos theta+i sin theta`
  `z^5` `=cos\ 5 theta+i sin\ 5 theta=-1,\ \ \ \ text{(De Moivre)}`
  `:. cos\ 5 theta` `=-1`
  `5 theta` `=pi,\ 3pi,\ 5pi,\ 7pi,\ 9pi`
  `theta` `=pi/5,\ (3pi)/5,\ pi,\ (7pi)/5,\ (9pi)/5`

 

`:.\ text(The roots are)`

`z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis)\ (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,`

`z_4 = text(cis)\ (7 pi)/5,\ \ z_5 = text(cis)\ (9 pi)/5`

 

ii.