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Complex Numbers, EXT2 N2 2025 HSC 9 MC

The points \(U, V, W\) and \(Z\) represent the complex numbers \(u, v, w\) and \(z\) respectively. It is given that  \(v+z=u+w\)  and  \(u+k i z=w+k i v\)  where  \(k \in \mathbb{R} , k>1\).

Which quadrilateral best describes \(UVWZ\) ?

  1. Parallelogram
  2. Rectangle
  3. Rhombus
  4. Square
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\(C\)

Show Worked Solution

\(\text{Quadrilateral}\ UVWZ\ \ \Rightarrow\ \ \text{Diagonals are \(UW\) and \(VZ\)} \).

\(\text{Given}\ \ v+z=u+w\ \ \Rightarrow\ \ \dfrac{v+z}{2}=\dfrac{u+w}{2}\)

\(\text{Mid-points of diagonals are equal (diagonals bisect).}\)

\(u+kiz\) \(=w+kiv\)  
\(u-w\) \(=ki(v-z)\)  

 
\(\therefore UW\ \text{and}\ VZ\ \text{are perpendicular.}\)

\(\Rightarrow C\)

♦♦ Mean mark 36%.

Complex Numbers, EXT2 N2 2022 HSC 16a

A square in the Argand plane has vertices

        `5+5i,quad5-5i,quad-5-5i`  and  `-5+5i`.

The complex numbers `z_A=5+i, z_B` and `z_C` lie on the square and form the vertices of an equilateral triangle, as shown in the diagram.
 
 
                 
 
Find the exact value of the complex number `z_B`.   (4 marks)

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`(5-16/sqrt3)+5i`

Show Worked Solution

`z_A=5+i, \ \ z_B=b+5i, \ \ z_C=c-5i`

`z_B-z_A=(b-5)+4i`

`z_C-z_A=(c-5)-6i`
 


♦♦♦ Mean mark 21%.

`text{Internal angles of equilateral triangle} = pi/3:`

`=>\ (z_B-z_A)\ text{is an anti-clockwise rotation of}\ (z_C-z_A)\ text{by}\ pi/3`
 

`e^(i pi/3)(z_B-z_A)=z_C-z_A`

`(1/2+i sqrt3/2)((b-5)+4i)=(c-5)-6i`

`(b-5)/2+2i+((b-5)sqrt3)/2 i-2sqrt3` `=(c-5)-6i`  
`(b-5-4sqrt3)/2 + i((4+(b-5)sqrt3)/2)` `=(c-5)-6i`  

 
`text{Equating imaginary parts:}`

`(4+(b-5)sqrt3)/2` `=-6`  
`(b-5)sqrt3` `=-16`  
`b-5` `=-16/sqrt3`  
`b` `=5-16/sqrt3`  

 
`:.z_B=(5-16/sqrt3)+5i`

Complex Numbers, EXT2 N2 2018 HSC 11d

The points `A`, `B` and `C` on the Argand diagram represent the complex numbers `u`, `v` and `w` respectively.

The points `O`, `A`, `B` and `C` form a square as shown on the diagram.
 

It is given that  `u = 5 + 2i`.

  1.  Find  `w`.   (1 mark)

  2.  Find  `v`.   (1 mark)

  3.  Find  `text(arg)(w/v)`.   (1 mark)

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i.    `−2 + 5i`

ii.   `3 + 7i`

iii.  `pi/4`

Show Worked Solution

i.    `w= iu= i(5 + 2i)= -2 + 5i`
 

ii.    `v` `= u + w`
    `= 5 + 2i + (-2 + 5i)`
    `= 3 + 7i`

 

iii.    `text(arg)(w/v)` `= text(arg)(w)-text(arg)(v)`
    `= pi/4\ \ (text(diagonal of square bisects corner))`

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.
  2. Show that  `u^2 + v^2 = uv.`   (2 marks)
  3. 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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a.    `text(Show Worked Solutions)`

b.    `u = 1/2 + sqrt3/2 i`

Show Worked Solution

a.    `0,u,v\ text(are vertices of an equilateral triangle.)`

♦♦ Mean mark (a) 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`

 

b.    `text(Let)\ \ v = 1,`

♦♦ Mean mark (b) 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
  2.    `1 + z + w = 0.`
  3. 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)

  4. Hence, or otherwise, show that if the three non-zero complex numbers `2i, z_1` and `z_2` satisfy
  5.    `|\ 2i\ | = |\ z_1\ | = |\ z_2\ |`  AND  `2i + z_1 + z_2 = 0.`
  6. then they form the vertices of an equilateral triangle in the Argand diagram.   (2 marks)

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i.    `text(See Worked Solutions)`

ii.   `text(See Worked Solutions)`

Show Worked Solution

i.    `z= costheta + isintheta, \ |\ z\ | = 1`

`w= cosalpha + isinalpha, \ |\ w\ | = 1`

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

♦♦♦ Mean mark (i) 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 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 (ii) 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 circle separated by)\ (2pi)/3\ text(radians.)`

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

Complex Numbers, EXT2 N2 2016 HSC 5 MC

Multiplying a non-zero complex number by  `(1-i)/(1 + i)`  results in a rotation about the origin on an Argand diagram.

What is the rotation?

  1. Clockwise by  `pi/4`
  2. Clockwise by  `pi/2`
  3. Anticlockwise by  `pi/4`
  4. Anticlockwise by  `pi/2`
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`B`

Show Worked Solution
`(1-i)/(1 + i)` `= ((1-i)^2)/((1 + i)(1-i))`
  `= (−2i)/2`
  `= −i`

 
`:. text(Clockwise rotation by)\ \ pi/2.`

`=> B`

Complex Numbers, EXT2 N1 2016 HSC 4 MC

The Argand diagram shows the complex numbers `z` and `w`, where `z` lies in the first quadrant and `w` lies in the second quadrant.
  

ext2-hsc-2016-4mc

Which complex number could lie in the 3rd quadrant?

  1. `-w`
  2. `2 iz`
  3. `bar z`
  4. `w - z`
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`=> D`

Show Worked Solution

`text(Using the parallelogram method:)`
 

ext2-hsc-2016-4mc-answer1
 

`text(From the graph above,)`

`w-z\ \ text(could lie in 3rd quadrant.)`

`=> D`

Complex Numbers, EXT2 N1 2009 HSC 2c

The points `P` and `Q` on the Argand diagram represent the complex numbers `z` and `w` respectively.
 

 

 

 
 

On the diagram, mark the following points:

  1. the point `R` representing `iz.`   (1 mark)

  2. the point `S` representing `bar w.`   (1 mark)

  3. the point `T` representing  `z + w.`   (1 mark)

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a, b, and c.

Show Worked Solution

a, b, and c.

Complex Numbers, EXT2 N2 2011 HSC 2b

On the Argand diagram, the complex numbers  `0, 1 + i sqrt 3 , sqrt 3 + i`  and  `z` form a rhombus.
 

  1. Find `z` in the form  `a + ib`, where `a` and `b` are real numbers.   (1 mark)

  2. An interior angle, `theta`, of the rhombus is marked on the diagram.

     

    Find the value of `theta.`   (2 marks)

Show Answers Only

a.    `(1 + sqrt 3) + i(1 + sqrt 3)`

b.    `(5 pi)/6`

Show Worked Solution
a.    `z` `= 1 + i sqrt 3 + sqrt 3 + i`
  `= (1 + sqrt 3) + i (1 + sqrt 3)`

 

b.    `text(arg)\ z = tan^-1 ((1 + sqrt 3)/(1 + sqrt 3)) = pi/4`

`text(arg)\ (sqrt 3 + i) = tan^-1 (1/sqrt 3) = pi/6`

`text(Difference) = pi/4-pi/6 = pi/12`
 

`=>\ text(Opposite angles of a rhombus are equal)`

`=>\ text(The diagonals of a rhombus bisect the angles)`

`:.theta= pi-2 xx pi/12= (5 pi)/6\ \ text{(angle sum of triangle)`

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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a.    `text(See Worked Solutions.)`

b.    `(u_1 + u_2)/2 + (u_2-u_1)/2 i`

Show Worked Solution
a.     `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)`

 

b.       `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(zvu_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)}`