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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 \(VZ\)} \).

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

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

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

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

\(\Rightarrow C\)

Complex Numbers, EXT2 N2 2024 HSC 14c

For the complex numbers \(z\) and \(w\), it is known that  \(\arg \left(\dfrac{z}{w}\right)=-\dfrac{\pi}{2}\).  

Find \(\left|\dfrac{z-w}{z+w}\right|\).   (2 marks)

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\(\abs{\dfrac{z-w}{z+w}}=1\)

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\(\arg \left(\dfrac{z}{w}\right)=\arg \, z-\arg \, w=-\dfrac{\pi}{2}\)

\(\text{Graphically, \(\arg \, z\) is a \(90^{\circ}\) anticlockwise rotation from \(\arg \, w\).}\)

♦ Mean mark 49%.

\(\abs{z-w}=\abs{z+w} \ \text{(diagonals of rectangle)}\)

\(\therefore \abs{\dfrac{z-w}{z+w}}=1\)

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 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`

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`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 EQ-Bank 1 MC

The Argand diagram shows the complex number `e^(i theta)`.
 

Which of the following could be the complex number `i e^(-i theta)`?

    

 

  

 
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`B`

Show Worked Solution

`e^(i theta) = cos theta + i sin theta`

`e^(-i theta) = cos (-theta) + i sin (-theta) = cos theta-i sin theta\ \ \ text{(conjugate)}`

`e^(i theta) \ => \ P^{′}`

`ie^(-i theta) \ => \ text(Rotate)\ e^(-i theta)\ (P^{′})\ text(90° anticlockwise.)`

`=> B`

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 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`