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Complex Numbers, EXT2 N2 2023 HSC 16a

Let \(w\) be the complex number  \(z=e^{\small{\dfrac{2i \pi}{3}}} \).

  1. Show that  \(1+w+w^2=0\).   (2 marks)

The vertices of a triangle can be labelled \(A, B\) and \(C\) in anticlockwise or clockwise direction, as shown.
 

Three complex numbers \(a, b\) and \(c\) are represented in the complex plane by points \(A, B\) and \(C\) respectively.

  1. Show that if triangle \(A B C\) is anticlockwise and equilateral, then  \(a+b w+c w^2=0\).   (2 marks)

  2. It can be shown that if triangle \(A B C\) is clockwise and equilateral, then \(a+b w^2+c w=0\). (Do NOT prove this.)
  3. Show that if \(A B C\) is an equilateral triangle, then

\(a^2+b^2+c^2=a b+b c+c a .\)  (2 marks)

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  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution

i.    \(w=e^{\small{\dfrac{2i \pi}{3}}} \)

\(w^2=e^{\small{-\dfrac{2i \pi}{3}}} = \overline w \)

\(\text{Re}(w)=\cos \Big{(}\dfrac{2\pi}{3} \Big{)}=-\dfrac{1}{2} \)

\(1+w+w^2\) \(=1+w+\overline w \)  
  \(=1+ 2 \times \text{Re}(w) \)  
  \(=1+2 \times -\dfrac{1}{2} \)  
  \(=0\)  

 
ii.    \(\text{Rotate}\ \ b-c\ \ \text{anticlockwise by}\ \ \dfrac{2\pi}{3} \)

\(\Rightarrow (b-c)w = c-a\)

\(\therefore bw-cw-c+a = a+bw-c(1+w) = 0 \)

\(\text{Substitute}\ \ 1+w=-w^2\ \ \text{(using part (i))} \)

\(a+bw-c(-w^2) \) \(=0\)  
\(a+bw+cw^2\) \(=0\ \ …\ \text{as required} \)  
♦♦♦ Mean mark (ii) 18%.

iii.   \(a+b w^2+c w=0\ \ \text{(if}\ \Delta ABC\ \text{is clockwise – given)} \)

\(a+bw+cw^2=0\ \ \text{(if}\ \Delta ABC\ \text{is anticlockwise)} \)

\(\Rightarrow (a+b w^2+c w)(a+bw+cw^2) = 0\ …\ (1)\)
 

\(\text{Expand (1) using part (i):}\ \ w+w^2=-1\ \ \text{and}\ \ w^3 = z^3 = 1 \)

\(a^2+abw^2+acw+bwa+b^2w^3+bcw^2+cw^2a+cbw^4+c^2w^3\) \(=0\)  
\(a^2+b^2+c^2+ab(w+w^2)+bc(w+w^2)+ca(w+w^2)\) \(=0\)  
\(a^2+b^2+c^2+ab(-1)+bc(-1)+ca(-1) \) \(=0\)  

 
\(\therefore a^2+b^2+c^2 = ab+bc+ca\ …\ \text{as required} \)

♦♦♦ Mean mark (iii) 11%.

Complex Numbers, EXT2 N2 2020 HSC 14a

Let `z_1` be a complex number and let  `z_2 = e^(frac{i pi}{3}) z_1`

The diagram shows points `A` and `B` which represent `z_1` and `z_2`, respectively, in the Argand plane.
 


 

  1. Explain why triangle  `OAB`  is an equilateral triangle.   (2 marks)

  2. Prove that  `z_1 ^2 + z_2^2 = z_1 z_2`.  (3 marks)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
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i.

 

`text{Let}`     `z_1` `= r(cos theta + i sin theta)`
  `z_2` `= e^(i frac{pi}{3}) z_1`
    `= r (cos ( theta + frac{pi}{3} ) + i sin (theta + frac{pi}{3}))`

 

`| z_1 | = | z_2 | => OA = OB`

` angle AOB = frac{pi}{3} \ ( z_2 \ text{is a} \ frac{pi}{3} \ text{anti-clockwise rotation of} \ z_1 )`

`=> angle OBA = angle BAO = pi/3\ \ \ text{(angles opposite equal sides)}`

`therefore \ OAB \ text{is equilateral}`

 

ii.     `z_1` `= z_2 e^(i frac{pi}{3})`
  `frac{z_1}{z_2}` `= e^(i frac{pi}{3})`
  `(frac{z_1}{z_2})^3` `= e^((3 xx i frac{pi}{3})) \ \ (text{by De Moivre})`
  `frac{z_1^3}{z_2^3}` `= e^(i pi)`
  `z_1^3` `= -z_2^3`
  `z_1^3 + z_2^3` `= 0`

COMMENT: The identity to prove suggests the addition of 2 cubes is a possible strategy.
`(z_1 + z_2)(z_1^2 – z_1 z_2 + z_2^2)` `= 0`
`z_1^2 – z_1 z_2 + z_2^2` `= 0`
`z_1^2 + z_2^2` `= z_1 z_2`

Complex Numbers, EXT2 N2 2016 HSC 16b

  1. The complex numbers  `0, \ u`  and  `v`  form the vertices of an equilateral triangle in the Argand diagram.

     

    Show that  `u^2 + v^2 = uv.`  (2 marks)

  2. Give an example of non-zero complex numbers  `u`  and  `v`, so that  `0, \ u`  and  `v`  form the vertices of an equilateral triangle in the Argand diagram.  (1 mark)

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  1. `text(Show Worked Solutions)`
  2. `u = 1/2 + sqrt3/2 i`
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i.   `0,u,v\ text(are vertices of an equilateral triangle.)`

♦♦ Mean mark 23%.

 

ext2-2016-hsc-16b-answer

`=> u` `= v text(cis)(±pi/3)`
`u^3` `= v^3text(cis)(±pi)`
`u^3` `= − v^3`
`u^3 + v^3` `= 0`

 
`(u + v)(u^2 – uv + v^2) = 0`

`text(S)text(ince)\ u != v,`

`u^2 – uv + v^2` `= 0`
`:. u^2 + v^2` `= uv`

 

ii.   `text(Let)\ \ v = 1,`

♦♦ Mean mark 33%.
`:. u` `= text(cis)(pi/3)`
  `= cos\ pi/3 + isin\ pi/3`
  `= 1/2 + sqrt3/2 i`

Complex Numbers, EXT2 N2 2016 HSC 16a

  1. The complex numbers  `z = cos theta + i sin theta`  and  `w = cos alpha + i sin alpha`, where  `-pi < theta < pi`  and  `-pi < alpha <= pi`, satisfy

        `1 + z + w = 0.`

    By considering the real and imaginary parts of      `1 + z + w`, or otherwise, show that  `1, \ z`  and  `w` form the vertices of an equilateral triangle in the Argand diagram.  (3 marks)

  2. Hence, or otherwise, show that if the three non-zero complex numbers  `2i, \ z_1`  and  `z_2`  satisfy

     

     
        `|\ 2i\ | = |\ z_1\ | = |\ z_2\ |`  AND  `2i + z_1 + z_2 = 0.`

    then they form the vertices of an equilateral triangle in the Argand diagram.      (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
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i.    `z` `= costheta + isintheta, \ |\ z\ | = 1`
  `w` `= cosalpha + isinalpha, \ |\ w\ | = 1`

 

`text(S)text(ince)\ 1 + z + w = 0`

♦♦♦ Mean mark 25%.
STRATEGY: A clear graphical image simplifies both parts of this question significantly.
`text(Im)(1 + z + w)` `= 0`
`sintheta + sinalpha` `= 0`
`sintheta` `= −sinalpha`
`:. theta` `= −alpha\ …\ (1)`

 

`text(Re)(1 + z + w)` `= 0`
`1 + costheta + cosalpha` `= 0`
`2costheta` `= −1qquad(text(S)text(ince)\ cosalpha  = cos(−theta) = costheta)`
`costheta` `= −1/2`
`:. theta` `= (2pi)/3`
`alpha` `= −(2pi)/3`

 
ext2-2016-hsc-16a-answer2 

 
`=>\ text(All points are on the unit circle)`

`text(separated by)\ (2pi)/3\ text(radians.)`

`:.\ text(They are vertices of an equilateral triangle.)`

 

ii.   `|\ 2i\ | = 2`

`text(Let)\ \ z_1` `= 2(costheta + isintheta), \  |\ z_1\ | = 2`
`z_2`  `= 2(cosalpha + isinalpha), \  |\ z_2\ | = 2`

 

♦♦♦ Mean mark 5%.
`text(Re)(2i + z_1 + z_2)` `= 0`
`2(costheta + cosalpha)` `= 0`
`:. costheta` `= −cosalpha`
`:. theta` `= pi – alpha`

 

`text(Im)(2i + z_1 + z_2)` `= 0`
`2(1 + sintheta + sinalpha)` `= 0`
`sintheta + sinalpha` `= −1`
`2sintheta` `= −1qquad(text(S)text(ince)\ sinalpha = sin(pi – theta) = sintheta)`
`sintheta` `= −1/2`
`:. theta` `= (7pi)/6`
`:. alpha` `= −pi/6`

 
ext2-2016-hsc-16a-answer3 
 

`=>\ text(All points are on the 2 unit)`

`text(circle separated by)\ (2pi)/3\ text(radians.)`

`:. 2i, z_1\ text(and)\ z_2\ text(are vertices of an)`

`text(equilateral triangle.)`

Complex Numbers, EXT2 N2 2007 HSC 2d

The points  `P,Q`  and  `R`  on the Argand diagram represent the complex numbers  `z_1, z_2`  and  `a`  respectively.

The triangles  `OPR`  and  `OQR`  are equilateral with unit sides, so  `|\ z_1\ | = |\ z_2\ | = |\ a\ | = 1.`

Let  `omega = cos­ pi/3 + i sin­ pi/3.`

  1. Explain why  `z_2 = omega a.`   (1 mark)

  2. Show that  `z_1 z_2 = a^2.`   (1 mark)

  3. Show that  `z_1` and `z_2`  are the roots of  `z^2 - az + a^2 = 0.`   (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(S) text(ince)\ \ Delta ORQ\ \ text(is equilateral, each angle is)\ \ pi/3\ \ text(radians.)`

`text(From)\ \ R(a):`

`Q(z_2)\ \ text(is an anticlockwise rotation through)\ \ pi/3.`

`:. z_2 = a(cos\ pi/3+i sin\ pi/3)=omega a.`

 

ii.  `text(Solution 1)`

`text(Similarly,)\ \ a` `=z_1 omega`
`z_1` `=a/omega`
`:z_1z_2` `=a/omega xx omega a`
  `=a^2`

 

`text(Solution 2)`

`P(z_1)\ \ text(is a clockwise rotation of)\ \ R(a)\ \ text(through)\ \ pi/3.`

`:.z_1` `= bar omega a.`
`:. z_1 z_2` `= bar omega a xx omega a`
  `=a^2(cos­ pi/3 – i sin­ pi/3) xx (cos­ pi/3 + i sin­ pi/3)`
  `=a^2(cos^2­ pi/3 + sin^2­ pi/3)`
  `= a^2`

 

iii.  `z^2-az + a^2 = 0`

`text(Let the roots be)\ \  alpha and beta.`

`alpha + beta` `=-b/a=a`
`alpha beta` `=c/a=a^2`

 

`z_1 z_2` `= a^2\ \ \ \ \ text{(part (ii))}`
`z_1 + z_2` `=bar omega a + omega a`
  `=(cos­ pi/3 + i sin­ pi/3 + cos­ pi/3-i sin­ pi/3) a`
  `=2 cos ­ pi/3 xx a`
  `=2 xx 1/2 xx a`
  `=a`

 

`:.\ z_1 and z_2\ \ text(are the roots of)\ \  z^2-az + a^2 = 0.`

Complex Numbers, EXT2 N2 2012 HSC 12d

On the Argand diagram the points  `A_1`  and  `A_2`  correspond to the distinct complex numbers  `u_1`  and  `u_2`  respectively. Let  `P`  be a point corresponding to a third complex number  `z`.

Points  `B_1`  and  `B_2`  are positioned so that  `ΔA_1PB_1`  and  `ΔA_2B_2P`, labelled in an anti-clockwise direction, are right-angled and isosceles with right angles at  `A_1`  and  `A_2`, respectively. The complex numbers  `w_1`  and  `w_2`  correspond to  `B_1`  and  `B_2`, respectively.
 

Complex Numbers, EXT2 2012 HSC 12d1 
 

  1. Explain why  `w_1 = u_1 + i(z − u_1)`.  (1 mark)

  2. Find the locus of the midpoint of  `B_1B_2`  as  `P`  varies.  (2 marks)

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  1. `text(See Worked Solutions.)`
  2. `(u_1 + u_2)/2 + (u_2 − u_1)/2 i`
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i. `vec (A_1P)` `= z − u_1`
  `vec (A_1B_1)`  `= w_1 − u_1` 

`B_1A_1 ⊥ A_1P\ text(and)\ |vec (A_1P)| = |vec (A_1B_1)|`

`vec (A_1B_1)\ text(is an anticlockwise rotation of)\ vec (A_1P)\ text(through)\ 90^@`

`:.w_1 − u_1 = i(z −u_1)`

`:.w_1 = u_1+ i(z −u_1)`

 

ii.   `vec (A_2B_2)` `= w_2 − u_2`
  `vec (A_2P)` `= z − u_2`

`A_2B_2 ⊥ A_2P\ text(and)\ |vec (A_2B_2)| = |vec (A_2P)|`

`vec (A_2P)\ text(is an anticlockwise rotation of)\ vec (A_2B_2)\ text(through)\ 90^@`

`z − u_2` `= i(w_2 −u_2)`
`iw_2` `= z − u_2 + iu_2`
`−w_2` `= iz − iu_2 − u_2`
`:. w_2` `= u_2 + i(u_2 − z)`

 

`:.\ text(The midpoint of)\ B_1B_2\ text(is)\ (w_1 + w_2)/2`

`= 1/2[u_1 + i(z − u_1) + u_2 + i(u_2 − z)]`
`= 1/2[u_1 + u_2 + i(u_2 − u_1)]`
`= (u_1 + u_2)/2 + (u_2 − u_1)/2 i\ \ \ \ text{(which is a fixed point)}`

Complex Numbers, EXT2 N2 2013 HSC 15a

The Argand diagram shows complex numbers  `w`  and  `z`  with arguments  `phi`  and  `theta`  respectively, where  `phi < theta`. The area of the triangle formed by  `0, w`  and  `z` is  `A`.
 


 

Show that  `z bar w - w bar z = 4iA.`  (3 marks)

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`text{(See Worked Solutions)}`

Show Worked Solution
♦♦ Mean mark 29%.

STRATEGY: The angles shown in the graphic should alert students that the mod-arg approach is likely to be easier than the `x + iy` form.
`A` `=1/2 ab sin C`
  `= 1/2 |z| |w|\ sin (theta – phi)`

 

`z bar w – w bar z` `= |z| text(cis)\ theta* |w| text(cis)(-phi) – |w| text(cis)\ phi *|z| text(cis)(-theta)`
  `= |z| |w| (text(cis)\ theta\ text(cis)(-phi) – text(cis)\ phi\ text(cis) (-theta))`
  `= |z| |w| (text(cis)(theta – phi) – text(cis)(phi – theta))`

 

`text(S)text(ince)\ \ cos(phi – theta)` `= cos(theta – phi),\ \ \ text(and)`
`sin(phi – theta)` `= -sin(theta – phi)`
`=>text(cis)(theta – phi) – text(cis)(phi – theta)` `= 2i sin(theta – phi)`

 

`:.z bar w – w bar z`  `= 2i |z| |w| sin(theta – phi)`
  `=4i(1/2 |z| |w|\ sin (theta – phi))`
  `=4iA\ \ \ \ text(… as required)`