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Complex Numbers, EXT2 N2 2025 HSC 6 MC

The complex numbers \(z\) and \(w\) lie on the unit circle. The modulus of  \(z+w\)  is \(\dfrac{3}{2}\).

What is the modulus of  \(z-w\) ?

  1. \(\dfrac{1}{8}\)
  2. \(\dfrac{\sqrt{7}}{2}\)
  3. \(\dfrac{3}{2}\)
  4. \(\dfrac{7}{4}\)
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\(B\)

Show Worked Solution

\(|z|=|w|=1, \quad \abs{z+w}=\dfrac{3}{2} \ \text{(given)}\)

\(\text{Find}\ \ \abs{z-w}:\)

\(\abs{z+\omega}^2=(z+\omega)(\bar{z}+\bar{\omega})=z \bar{z}+z \bar{\omega}+\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z-\omega}^2=(z-\omega)(\bar{z}-\bar{\omega})=\bar{z} z-z \bar{\omega}-\omega \bar{z}+\omega \bar{\omega}\)

\(\abs{z+\omega}^2+|z-\omega|^2\) \(=2 z \bar{z}+2 \omega \bar{\omega}=2\abs{z}^2+2\abs{\omega}^2=4\)
\(\dfrac{9}{4}+\abs{z-\omega}^2\) \(=4\)
\(\abs{z-w}^2\) \(=\dfrac{7}{4}\)
\(\abs{z-\omega}\) \(=\dfrac{\sqrt{7}}{2}\)

 
\(\Rightarrow B\)

Complex Numbers, EXT2 N2 2024 HSC 7 MC

It is given that \(\abs{z-1+i}=2\).

What is the maximum possible value of \(\abs{z}\)?

  1. \(\sqrt{2}\)
  2. \(\sqrt{10}\)
  3. \(2+\sqrt{2}\)
  4. \(2-\sqrt{2}\)
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\(C\)

Show Worked Solution

\(\abs{z-1+i}=2 \ \Rightarrow \ \text{circle centre} \ \ (1,-i), \ \ \text {radius}=2\)
 
 

\(OA=\sqrt{1^2+1^2}=\sqrt{2}\)

\(AB=2 \ \ \text{(radius)}\)

\(\therefore \abs{z}_{\text{max}}=2+\sqrt{2}\)

\(\Rightarrow C\)

Complex Numbers, EXT2 N2 2023 14a

Let \(z\) be the complex number  \(z=e^{\small{\dfrac{i \pi}{6}}} \)  and \(w\) be the complex number  \(w=e^{\small{\dfrac{3 i \pi}{4}}} \).

  1. By first writing \(z\) and \(w\) in Cartesian form, or otherwise, show that
  2.    \(|z+w|^2=\dfrac{4-\sqrt{6}+\sqrt{2}}{2}\).  (3 marks)

  3. The complex numbers \(z, w\) and \(z+w\) are represented in the complex plane by the vectors \(\overrightarrow{O A},\overrightarrow{O B}\) and \(\overrightarrow{O C}\) respectively, where \(O\) is the origin.
  4. Show that  \(\angle A O C=\dfrac{7 \pi}{24}\).  (2 marks)

  5. Deduce that  \(\cos \dfrac{7 \pi}{24}=\dfrac{\sqrt{8-2 \sqrt{6}+2 \sqrt{2}}}{4}\).  (1 mark)

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

i.    \(z=e^{\small\dfrac{i \pi}{6}} = \cos\,\dfrac{\pi}{6} + i \,\sin\,\dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{1}{2}i \)

\(w=e^{\small\dfrac{3i \pi}{4}} = \cos\,\dfrac{3\pi}{4} + i \,\sin\,\dfrac{3\pi}{4} = -\dfrac{1}{\sqrt2} + \dfrac{i}{\sqrt2} \)

\(|z+w|^2\) \(=\Bigg{|} \dfrac{\sqrt3}{2}+\dfrac{1}{2}i-\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2} \Bigg{|}\)  
  \(=\Bigg{|} \Bigg{(}\dfrac{\sqrt3}{2}-\dfrac{1}{\sqrt2} \Bigg{)} +\Bigg{(}\dfrac{1}{2}+\dfrac{1}{\sqrt2}\Bigg{)}\,i \Bigg{|}\)  
  \(=\Bigg{|} \dfrac{\sqrt6-2}{2\sqrt2}+\dfrac{\sqrt2+2}{2\sqrt2}\,i \Bigg{|}\)  
  \(= \dfrac{(\sqrt6-2)^2+(\sqrt2+2)^2}{(2\sqrt2)^2}\)  
  \(= \dfrac{6-4\sqrt6+4+2+4\sqrt2+4}{8}\)  
  \(=\dfrac{16-4\sqrt6+4\sqrt2}{8} \)  
  \(=\dfrac{4-\sqrt6+\sqrt2}{2} \)  

 
ii.   

\(\angle AOB= \arg(w)-\arg(z)=\dfrac{3\pi}{4}-\dfrac{\pi}{6}=\dfrac{7\pi}{12} \)

\( |z|=|w|=1\ \Rightarrow AOBC\ \text{is a rhombus.} \)

\(\overrightarrow{OC}\ \text{is a diagonal of rhombus}\ AOBC \)

\(\Rightarrow \overrightarrow{OC}\ \text{bisects}\ \angle AOB \)

\(\therefore \angle AOC= \dfrac{1}{2} \times \dfrac{7\pi}{12}=\dfrac{7\pi}{24} \)
  

iii.   \(\text{In}\ \triangle AOC: \)

\( \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA} = \overrightarrow{OB} \)

\(\Rightarrow \overrightarrow{OB}\ \text{is represented by}\ w. \)
 

\(\text{Using the cos rule in}\ \triangle AOC: \)

\(\cos\,\dfrac{7\pi}{24}\) \(=\dfrac{|z|^2+|z+w|^2-|w|^2}{2|z||z+w|}\)  
  \(=\dfrac{ 1+\frac{4-\sqrt6+\sqrt2}{2}-1}{2 \times 1  \sqrt{\frac{4-\sqrt6+\sqrt2}{2}}} \)  
  \(=\dfrac{\sqrt{\frac{4-\sqrt6+\sqrt2}{2}} \times 2} {2 \times 2} \)  
  \(=\dfrac{\sqrt{4( \frac{4-\sqrt6+\sqrt2}{2})}} {4} \)  
  \(=\dfrac{8-2\sqrt6+2\sqrt2}{4} \)  
♦♦ Mean mark (iii) 26%.

Vectors, EXT2 V1 2022 HSC 9 MC

Let \(A\) and \(B\) be two distinct points in three-dimensional space. Let \(M\) be the midpoint of \(A B\).

Let \(S_1\) be the set of all points \(P\) such that  \(\overrightarrow{AP} \cdot \overrightarrow{BP}=0\). 

Let \(S_2\) be the set of all points \(N\) such that \(\Big|\overrightarrow{AN}\Big|=\Big| \overrightarrow{MN} \Big| \).

The intersection of \(S_1\) and \(S_2\) is the circle \(S\).

What is the radius of the circle \(S\) ?

  1. \(\dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)
  2. \(\dfrac{\Big| \overrightarrow{AB} \Big|}{4} \)
  3. \(\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{2} \)
  4. \(\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{4} \)
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\(D\)

Show Worked Solution

`text{Diagram below is a 2-D sliced image of the 3-D geometry:}`

\(\overrightarrow{AP} \cdot \overrightarrow{BP}=0\). 

\(r=\dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)

\(S_2\ \text{includes}\ N\ \text{where}\ \Big| \overrightarrow{AN} \Big|=\Big| \overrightarrow{MN} \Big| = \dfrac{r}{2} \)  

\(\text{Let}\ r_s=\ \text{radius of}\ S\)

\(\text{Point}\ P\ \text{is intersection of}\ S_1\ \text{and}\ S_2 \)

\(\text{By Pythagoras (see diagram):}\)

\(r_s^2\) \(=r^2-(\dfrac{r}{2})^2 \)  
  \(=\dfrac{3r^2}{4}\)  
\(r_s\) \(=\dfrac{\sqrt3}{2} \times r\)  
  \(=\dfrac{\sqrt3}{2} \cdot \dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)  
  \(=\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{4} \)  

 
\(=>D\)


♦♦♦ Mean mark 20%.

Complex Numbers, EXT2 N2 2021 SPEC2 5

The graph of the circle given by  `|z - 2 - sqrt3i| = 1`, where  `z ∈ C`, is shown below.
 


 

For points on this circle, find the maximum value of  `|z|`.   (2 marks)

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`sqrt7+1`

Show Worked Solution

`text(Centre of circle at)\ (2, sqrt3)`

`text(Radius = 1)`

`text(By Pythagoras, line from origin to the centre of the circle)`

`d = sqrt(2^2 + sqrt(3)^2) = sqrt7`

`:. |z|_text(max) = sqrt7 + 1`

Complex Numbers, EXT2 N2 2021 HSC 10 MC

Consider the two non-zero complex numbers  `z`  and  `w`  as vectors.

Which of the following expressions is the projection of  `z`  onto  `w` ? 

  1.  `{text{Re} (zw)}/{|w|} w`
  2.  `|z/w| w`
  3.  `text{Re} (z/w) w`
  4. `{text{Re}(z)}/{|w|} w`
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`C`

Show Worked Solution

`text{Let} \ \ z = a + i b \ \ =>\ underset~z = ((a),(b))`

♦♦♦ Mean mark 30%.

`text{Let} \ \ w = c + i d \ \ =>\ underset~w = ((c),(d))`

`underset~z * underset~w = ac + bd`

`|underset~w|^2 = c^2 + d^2`

`text{proj}_(underset~w) underset~z = (underset~z*underset~w)/|underset~w|^2 *underset~w= (ac+bd)/(c^2+d^2)*underset~w`

`z/w` `=(a+ib)/(c+id) xx (c-id)/(c-id)`  
  `=(ac+bd + i(bc-ad))/(c^2+d^2)`  

 
`text{Re}(z/w) =(ac+bd)/(c^2+d^2)`

`:.\ text{proj}_(underset~w) underset~z = text{Re} (z/w) w`
 

`=> C`

Complex Numbers, EXT2 N2 2020 SPEC2 2

Two complex numbers, `u` and `v`, are defined as  `u = −2 - i`  and  `v = −4 - 3i`.

  1. Express the relation  `|z - u| = |z - v|`  in the cartesian form  `y = mx + c`, where  `m, c ∈ R`.  (3 marks)

  2. Plot the points that represent `u` and `v` and the relation `|z - u| = |z - v|` on the Argand diagram below.  (2 marks)
     
         
     
  3. State a geometrical interpretation of the graph of  `|z - u| = |z - v|`  in relation to the points that represent `u` and `v`.  (1 mark)

  4.  i. Sketch the ray given by  `text(Arg)(z - u) = pi/4`  on the Argand diagram in part b.  (1 mark)

  5. ii. In Cartesian form, write down the function that describes the ray  `text(Arg)(z - u) = pi/4`.  (1 mark)

Show Answers Only
  1. `y = −x – 5`
  2.  
  3. `|z – u| = |z – v|\ text(is the perpendicular bisector of the line joining)\ u and v.`
  4.  i.
  5. ii. `f: (−2, ∞) ->, f(x) = x + 1`
Show Worked Solution

a.   `text(Let)\ \ z = x + iy`

`z – u = x + 2 + iy + i`

`z – v = x + 4 + iy + 3i`

`|z – u| = |z – v|`

`(x + 2)^2 + (y + 1)^2` `= (x + 4)^2 + (y + 3)^2`
`x^2 + 4x + 4 + y^2 + 2y + 1` `= x^2 + 8x + 16 + y^2 + 6y + 9`
`-4y` `= 4x + 20`
`y` `= −x – 5`

 

b.   

 

c.   `|z – u| = |z – v|\ text(is the graph of the perpendicular bisector of the)`

`text(line joining)\ u and v.`

 

d.i.   

 

d.ii.   `text(Arg)(z – u) = pi/4 =>\ text(gradient) = 1, ytext(-intercept at)\ (0, 1)`

`:. f: (−2, ∞) -> RR, \ f(x) = x + 1`

Complex Numbers, EXT2 N2 2018 HSC 7 MC

Which diagram best represents the solutions to the equation  `text(arg)(z) = text(arg)(z + 1 - i)`?

A. B.
C. D.
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`text(D)`

Show Worked Solution
`text(arg)(z)` `= text(arg)(z + 1 – i)`
  `=text(arg)(z – (−1 + i))`

 
`=>\ text(arg)(z – (−1 + i))\ \ text(is the argument of)\ \ z\ \ text(from)\ \ (-1+i).`

 
`text(Plot)\ \ (-1 + i)\ \ text(on the argand diagram and then test different)`

`text(positions of)\ \ z\ \ text(along the solutions for each option.)`

`=>\ text(D)`

Complex Numbers, EXT2 N2 2007 HSC 2c

The point  `P`  on the Argand diagram represents the complex number `z`, where  `z`  satisfies

    `1/z + 1/bar z = 1.`

Give a geometrical description of the locus of  `P`  as  `z`  varies.   (3 marks)

Show Answers Only

`text(Locus is a circle, centre)\ (1, 0), text(radius 1,)`

`text(excluding the point)\ (0, 0).`

Show Worked Solution
`1/z + 1/bar z` `=1/(x + iy) +1/(x – iy)`
  `=(x – iy + x+iy)/(x^2+y^2)`
  `=(2x)/(x^2+y^2)`

 

`(2x)/(x^2+y^2)` `=1`
`x^2 + y^2` `=2x`
`x^2 – 2x + 1 + y^2` `=1`
`(x – 1)^2 + y^2` `=1`

 

`:.\ text(Locus is a circle, centre)\ (1, 0), text(radius 1,)`

`text(excluding the point)\ (0, 0).`

Complex Numbers, EXT2 N2 2015 HSC 9 MC

The complex number  `z`  satisfies  `| z - 1 | = 1.`

What is the greatest distance that  `z`  can be from the point  `i`  on the Argand diagram?

  1. `1`
  2. `sqrt 5`
  3. `2 sqrt 2`
  4. `sqrt 2 + 1`
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`D`

Show Worked Solution

HSC 2015 9MC

`| z – 1 | = 1\ \ text{is a circle, centre (1,0), radius 1.}`

`text(Consider the graph:)`

`text(Distance from)\ \(0, i)\ \ text(to the centre)=sqrt2`

`:.\ text(Greatest distance)=sqrt2 + text(radius)=sqrt2+1`

`=>  D`

Complex Numbers, EXT2 N2 2011 HSC 4a

Let  `a`  and  `b`  be real numbers with  `a != b`. Let  `z = x + iy`  be a complex number such that

    `|\ z - a\ |^2 - |\ z - b\ |^2 = 1.` 

  1. Prove that  `x = (a + b)/2 + 1/(2 (b - a)).`  (2 marks)

  2. Hence, describe the locus of all complex numbers  `z`  such that  `|\ z - a\ |^2 - |\ z - b\ |^2 = 1.`  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)`
  2. `x = (a + b)/2 + 1/(2(b – a))`
Show Worked Solution
i.    `|\ z – a\ |^2 – |\ z – b\ |^2` `= 1`
  `|\ (x-a)+iy\ |^2-|\ (x-b)+iy\ |^2` `=1`
  `(x – a)^2 + y^2 – ((x – b)^2 + y^2)` `=1`
  `(x – a)^2 – (x – b)^2` `=1`
  `(x – a – (x – b)) (x – a + x – b)` `=1`
  `(b – a) (2x – a – b)` `=1`
`2x – a – b` `= 1/(b – a)`
`2x` `= a + b + 1/(b – a)`
`:. x` `= (a + b)/2 + 1/(2(b – a))`

  

♦ Mean mark part (ii) 44%.

ii.  `text(The locus is the vertical line)`

`x = (a + b)/2 + 1/(2(b – a)).`