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Complex Numbers, SPEC2 2024 VCAA 2

  1. Express the relation  \(\abs{z-z_1}=\abs{z-z_2}\)  in the form  \(y=m x+c\), where  \(x, y, m, c \in R\), \(z=x+i y, \ z_1=1+2 i\)  and  \(z_2=4\).   (2 marks)

  2. The line segment from  \(z_1=1+2 i\)  to  \(z_2=4\)  is the diameter of a circle.
  3. Find the equation of this circle in the form \(\abs{z-z_c}=r\), where \(z_c\) is the centre of the circle and \(r\) is the radius.   (2 marks)

  4. A second circle is given by  \(\abs{z-(1+2 i)}=2\).
  5. Sketch this circle on the Argand diagram below, labelling the imaginary axis intercepts with their values.  (2 marks)
     

     

  1. A ray originating at the point  \(z=2-i\)  passes through the point  \(z=-2+3 i\),  cutting the second circle into two segments.
    1. Sketch the ray on the Argand diagram provided in part c.  (1 mark)

    2. Find the equation of the ray in the form \(\operatorname{Arg}\left(z-z_0\right)=\theta\)  where  \(z_0 \in C\) and \(\theta\) is measured in radians in terms of \(\pi\).  (1 mark)

  3. Find the area of the minor segment formed by the intersection of the ray and the circle.  (2 marks)

Show Answers Only

a.    \(y=\dfrac{3}{2} x-\dfrac{11}{4}\)

b.    \(\abs{z-\left(\dfrac{5}{2}+i\right)} = \dfrac{\sqrt{13}}{2}\)

c.    

d.i.    \(\text{See image above}\)

d.ii.   \(\operatorname{Arg}(z-2+i)=\dfrac{3 \pi}{4}\)

e.   \(A=\pi-2\)

Show Worked Solution

a.    \(\text{Method 1}\)

\(z_1=1+2i, \quad z_2=4\)

\(\abs{z-z_1}=\abs{z-z_2}\)

\(\text{Equation can be written:}\)

\((x-1)^2+(y-2)^2\) \(=(x-4)^2+y^2\)
\(-2x-4y+5\) \(=-8x+16 \ \ \ \text{(all squares cancel)}\)
\(6x+4y\) \(=11\)
\(y\) \(=\dfrac{3}{2} x-\dfrac{11}{4}\)

 
\(\text{Method 2}\)

\(\text {Find line of points equidistant from \(\ z_1\)  and  \(z_2\)}\)

\(m_{\text{line}\ z_1 z_2}=\dfrac{0-2}{4-1}=\dfrac{-2}{3}\)

\(m_{\perp}=\dfrac{3}{2}\)

\(\text{Mid point} \  z_1 z_2 =\dfrac{5+2i}{2}=\left(\dfrac{5}{2},1\right)\)

\(\perp \ \text{bisector:} \ m=\dfrac{3}{2}, \ \text{passes through} \ \left(\dfrac{5}{2}, 1\right)\)

\(y-1\) \(=\dfrac{3}{2}\left(x-\dfrac{5}{2}\right)\)
\(y\) \(=\dfrac{3}{2} x-\dfrac{11}{4}\)

 

b.    \(\text{Centre of circle }=\text {midpoint}\left(z_1 z_2\right)\)

\(z_c=\dfrac{z_1+z_2}{2}=\dfrac{5+2 i}{2}=\dfrac{5}{2}+i\)
 

\(\text{Radius}(r)=\dfrac{1}{2} \times \text{diameter}\)

\(r=\dfrac{1}{2} \sqrt{(4-1)^2+(0-2)^2}=\dfrac{\sqrt{13}}{2}\)
 

\(\text{Circle equation:}\)

\(\abs{z-\left(\dfrac{5}{2}+i\right)} = \dfrac{\sqrt{13}}{2}\)
 

c.    \(\text{Find \(y\)-axis intercepts \((z=y i)\):}\)

\(4=\abs{z-(1+2i)}^2=\abs{yi-(1+2i)}^2=(-1)^2+(y-2)^2\)

\(y^2-4y+1=0 \ \Rightarrow \ =2 \pm \sqrt{3}\)
 

d.i.    \(\text{See image above}\)

d.ii.   \(\operatorname{Arg}(z-2+i)=\dfrac{3 \pi}{4}\)

Mean mark (d.ii.) 56%.

e.    \(\text{Circle radius}=2\)

\(\text{Angle subtending minor segment}=\dfrac{\pi}{2}\)

\(A=\dfrac{r^2}{2}(\theta-\sin \theta)=2\left(\dfrac{\pi}{2}-1\right)=\pi-2\)

Complex Numbers, SPEC2 2024 VCAA 5 MC

If the point  \(z=1+\sqrt{3} i\)  is represented on an Argand diagram, the point representing  \(-\bar{z}\)  can be located by

  1. reflecting the point representing \(z\) in the real axis.
  2. rotating the point representing \(z\) anticlockwise about the origin by 90\(^{\circ}\).
  3. reflecting the point representing \(z\) in the imaginary axis.
  4. rotating the point representing \(z\) clockwise about the origin by 90\(^{\circ}\).
Show Answers Only

\(C\)

Show Worked Solution

\(z=1+\sqrt{3}i\ \ \Rightarrow\ \ \bar{z} = 1-\sqrt{3}i\)

\(-\bar{z} = -1+\sqrt{3}i\)

\(\therefore \text{It is a reflection of}\ z\ \text{in the imaginary axis.}\)

\(\Rightarrow C\)

Complex Numbers, SPEC2 2022 VCAA 2

Two complex numbers \(u\) and \(v\) are given by  \(u=a+i\)  and  \(v=b-\sqrt{2}i\), where \(a, b \in R\).

  1.  i. Given that  \(uv=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6})i\), show that  \(a^2+(1-\sqrt{3}) a-\sqrt{3}=0\).   (2 marks)

  2. ii. One set of possible values for \(a\) and \(b\) is  \(a=\sqrt{3}\)  and  \(b=\sqrt{2}\).
  3.     Hence, or otherwise, find the other set of possible values.   (1 mark)

  4. Plot and label the points representing  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\)  on the Argand diagram below.
     

  1. The ray given by  \(\text{Arg}(z)=\theta\)  passes through the midpoint of the line interval that joins the points  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\).
  2. Find, in radians, the value of \(\theta\) and plot this ray on the Argand diagram in part b.   (2 marks)

  3. The line interval that joins the points  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\)  cuts the circle  \(|z|=2\)  into a major and a minor segment.
  4. Find the area of the minor segment, giving your answer correct to two decimal places.   (2 marks)

Show Answers Only

a.i.  \(\text{See worked solutions.}\)

a.ii.  \(a=-1, \ b=-\sqrt{6}\)

b.   
       

c.   \(\theta=-\dfrac{\pi}{24}\)

d.    \(0.69\ \text{u}^2\)

Show Worked Solution

a.i.  \(u v=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6}) i\)

\begin{aligned}
(a+i)(b-\sqrt{2} i) & =a b-\sqrt{2} a i+b i+\sqrt{2} \\
& =a b+\sqrt{2}+(b-\sqrt{2}a) i
\end{aligned}

\(\text {Equating co-efficients: }\)

  \(ab=\sqrt{6} \ \Rightarrow \ b=\dfrac{\sqrt{6}}{a}\ \cdots\ \text {(1)}\)

  \(b-\sqrt{2} a=\sqrt{2}-\sqrt{6}\ \cdots\ \text {(2)}\)

\(\text{Substitute (1) into (2):}\)

\(\dfrac{\sqrt{6}}{a}-\sqrt{2}a=\sqrt{2}-\sqrt{6}\)

\(\sqrt{6}-\sqrt{2} a^2=(\sqrt{2}-\sqrt{6}) a\)

\(\sqrt{2} a^2+(\sqrt{2}-\sqrt{6}) a-\sqrt{6}\) \(=0\)  
\(a^2+(1-\sqrt{3}) a-\sqrt{3}\) \(=0\)  
Mean mark (a) 54%.

 
a.ii. \(\text{Solve } a^2+(1-\sqrt{3}) a-\sqrt{3}=0 \ \ \text{for}\  a :\)

\(a=\sqrt{3} \ \text{ or } -1\)

\(\therefore \text{ Other solution: } a=-1, \ b=-\sqrt{6}\)
 

b.   
       


c.    		 \begin{aligned}
\theta & =\frac{\operatorname{Arg}(u)-\operatorname{Arg}(v)}{2} \\
& =\frac{1}{2}\left(\frac{\pi}{6}-\frac{\pi}{4}\right) \\
& =-\frac{\pi}{24}
\end{aligned}
♦ Mean mark (c) 39%.

d.   \(\text {Sector angle (centre) }=\dfrac{\pi}{6}+\dfrac{\pi}{4}=\dfrac{5 \pi}{12}\)

\(\text {Area of sector }=\dfrac{\frac{5 \pi}{12}}{2 \pi} \times \pi \times 2^2=\dfrac{5 \pi}{6}\)

\begin{aligned}
\text {Area of segment } & =\frac{5 \pi}{6}-\dfrac{1}{2} \times 2 \times 2 \times \sin \left(\dfrac{5 \pi}{6}\right) \\
& \approx 0.69\ \text{u} ^2
\end{aligned}

♦ Mean mark (d) 40%.

Complex Numbers, SPEC2 2021 VCAA 2

The polynomial  `p(z) = z^3 + alpha z^2 + beta z + gamma`, where  `z ∈ C`  and  `alpha, beta, gamma ∈ R`, can also be written as  `p(z) = (z-z_1)(z-z_2)(z-z_3)`, where  `z_1 ∈ R`  and  `z_2, z_3 ∈ C`.

  1.  i. State the relationship between `z_2` and `z_3`.  (1 mark)

  2. ii. Determine the values of `alpha, beta` and `gamma`, given that  `p(2) = -13, |z_2 + z_3| = 0`  and  `|z_2-z_3| = 6`.  (3 marks)

Consider the point  `z_4 = sqrt3 + i`.

  1. Sketch the ray given by  `text(Arg)(z-z_4) = (5pi)/6`  on the Argand diagram below.  (2 marks)
     
  2. The ray  `text(Arg)(z-z_4) = (5pi)/6`  intersects the circle  `|z-3i| = 1`, dividing it into a major and a minor segment.
  3.  i. Sketch the circle  `|z-3i| = 1` on the Argand diagram in part b.  (1 mark)

  4. ii. Find the area of the minor segment.  (2 marks)

Show Answers Only
    1. `z_2 = barz_3`
    2. `alpha = -3, beta = 9, gamma = -27`
  2.  
     

     

c.i.   

c.ii.   `(2pi-3sqrt3)/12\ text(u²)`

Show Worked Solution

a.i.   `text(By conjugate root theory)`

`z_2 = barz_3`

 

a.ii.   `text(Let)\ \ z_1 = a + bi, \ z_2 = a-bi`

♦♦ Mean mark part (a.ii.) 35%.

`|z_2 + z_3| = |2a| = 0 \ => \ a = 0`

`|z_2-z_3| = |2b| = 6 \ => \ b = ±3`
 

`text(Using)\ \ p(2) = -13`

`(2-z_1)(2-3i)(2 + 3i)` `= -13`
`(2-z_1)(4 + 9)` `= -13`
`2-z_1` `= -1`
`z_1` `= 3`

 

`p(z)` `= (z-3)(z-3i)(z + 3i)`
  `= (z-3)(z^2 + 9)`
  `= z^3-3z^2 + 9z-27`

 
`:. alpha = –3, \ beta = 9, \ gamma = –27`

♦ Mean mark part (b) 45%.
MARKER’S COMMENT: Point of emanation is not part of required ray (see open circle).

 

b. 

 

c.i.  

 

c.ii.   `text(Arg)(z-z_4) = (5pi)/6\ \ text(cuts)\ \ ytext(-axis at angle)\ pi/3`

♦♦ Mean mark part (c.ii.) 30%.

`=>\ text(angle at centre of segment) = pi/3\ (text(equilateral triangle))`

`text(Area)` `= (pi/3)/(2pi) xx pi xx 1^2-1/2 xx 1 xx 1 xx sin (pi/3)`
  `= pi/6-sqrt3/4`
  `= (2pi-3sqrt3)/12\ text(u²)`

Complex Numbers, SPEC2 2012 VCAA 2

  1.  Given that  `cos(pi/12) = (sqrt (sqrt 3 + 2))/2`, show that  `sin(pi/12) = (sqrt (2-sqrt 3))/2`.   (2 marks)

  2. Express  `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2-sqrt3))/2`  in polar form.   (1 mark)

  3. i.  Write down  `z_1^4`  in polar form.   (1 mark)

  4. ii. On the Argand diagram below, shade the region defined by

`{z: text(Arg)(z_1) <= text(Arg)(z) <= text(Arg)(z_1^4)} ∩ {z: 1 <= |\ z\ | <= 2}, z ∈ C`.   (2 marks)


 

  1.  Find the area of the shaded region in part c.   (2 marks) 
  2. i.  Find the value(s) of `n` such that  `text(Re)(z_1^n) = 0`, where  `z_1 = (sqrt(sqrt3 + 2))/2 + i(sqrt(2-sqrt3))/2`.   (3 marks)

  3. ii. Find  `z_1^n`  for the value(s) of `n` found in part i.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. i.  `z_1 = text(cis) pi/12`

     

    ii.  `z_1^4 = text(cis) pi/3`

  3. `text(See Worked Solutions.)`
  4. `(3pi)/8`
  5. i.  `n = (2k + 1)6`

     

    ii.  `z_1^n = ±i\ text(for)\ k ∈ Z`

Show Worked Solution
a.    `cos^2(pi/12) + sin^2(pi/12)` `= 1`
  `sin^2(pi/12)` `= 1-(sqrt 3 + 2)/4`
    `= (2-sqrt 3)/4`

 
`:. sin (pi/12) = sqrt(2-sqrt 3)/2,\ \ \ (sin (pi/12) > 0)`

 

b.i.    `z_1` `= cos(pi/12) + i sin (pi/12)`
    `= text(cis)(pi/12)`

 

b.ii.    `z_1^4` `= 1^4 text(cis) ((4 pi)/12)`
    `= text(cis)(pi/3)`

 

c.   

 

d.  `text(Area of large sector)`

Mean mark 51%.

`=theta/(2pi) xx pi r^2`

`=(pi/3-pi/12)/(2pi) xx pi xx 2^2`

`=pi/2`
  

`text(Area of small sector)`

`=1/2 xx pi/4 xx 1`

`=pi/8`

 
`:.\ text(Area shaded)`

`= pi/2-pi/8`

`= (3 pi)/8`
 

♦ Mean mark (e)(i) 34%.

e.i.    `z_1^n` `= 1^n text(cis) ((n pi)/12)`
  `text(Re)(z_1^n)` `= cos((n pi)/12) = 0`
  `(n pi)/12` `=cos^(-1)0 +2pik,\ \ \ k in ZZ`
  `(n pi)/12` `= pi/2 + k pi`
  `n` `= 6 + 12k,\ \ \ k in ZZ`

 

e.ii.  `text(When)\ \ n=(6 + 12k):`

♦ Mean mark (e)(ii) 36%.

  `z_1^n` `= 1^(6 + 12k) text(cis) (((6 + 12k)pi)/12), quad k in ZZ`
    `= i xx sin (((6 + 12k)pi)/12)`
    `= i xx sin (pi/2 + k pi)`

 
`text(If)\ \ k=0  or text(even,)\ \ z_1^n=i`

`text(If)\ \ k\ text(is odd,)\ \ z_1^n=-i`

`:. z_1^n = +-i,\ \ \ k in ZZ`

Complex Numbers, SPEC2 2012 VCAA 6 MC

For any complex number `z`, the location on an Argand diagram of the complex number  `u = i^3 bar z`  can be found by

A.   rotating `z` through `(3 pi)/2` in an anticlockwise direction about the origin

B.   reflecting `z` about the `x`-axis and then reflecting about the `y`-axis

C.   reflecting `z` about the `y`-axis and then rotating anticlockwise through `pi/2` about the origin

D.   reflecting `z` about the `x`-axis and then rotating anticlockwise through `pi/2` about the origin

E.   rotating `z` through `(3 pi)/2` in a clockwise direction about the origin

Show Answers Only

`C`

Show Worked Solution

`z -> bar z:\ text(reflect in)\ x text(-axis)`

`x+iy\ \ =>\ \ x-iy`

`bar z -> i^3 bar z:\ text(rotate)\ 90^@ xx 3\ text(anticlockwise, or)\ 90^@\ text(clockwise)`

`i^3 bar z` `= i^3 (x – iy)`
  `= -i(x – iy)`
  `= -y – ix `

 
`text(Consider option)\ C:`

♦♦ Mean mark 37%.

`z\ text(reflected about)\ y text(-axis)`

`-x + iy`

`text(then rotated)\ pi/2\ text(anticlockwise)`

`i(-x + iy)` `= -ix + i^2y`  
  `= -y – ix`  ✔  

 
`=> C`

Complex Numbers, SPEC2 2015 VCAA 9 MC

Let  `z_1 = r_1 text(cis)(theta_1)`  and  `z_2 = r_2 text(cis)(theta_2)`, where  `z_1`  and  `z_1z_2`  are shown in the Argand diagram below;  `theta_1`  and  `theta_2`  are acute angles.
 

SPEC2 2015 VCAA 9 MC
 

A statement that is necessarily true is

A.   `r_2 > 1`

B.   `theta_1 < theta_2`

C.   `|\ (z_1)/(z_2)\ | > r_1`

D.   `theta_1 = theta_2`

E.   `r_1 > 1`

Show Answers Only

`C`

Show Worked Solution

`text(Consider each option:)`

♦ Mean mark 47%.

`A:\ \ r_1r_2 < r_1\ \ =>\ \ r_2 < 1`

`B:`

`alpha + pi/2 – theta_1 = theta_2 ~~ theta_1\ \ text(not necessarily true)`

`C:\ \ (r_1)/(r_2) = (|\ z_1\ |)/(|\ z_2\ |) = |(z_1)/(z_2)|`

`text(S)text(ince)\ \ r_2<1\ \ =>\ \ |(z_1)/(z_2)| > r_1`

`D:\ \ theta_1 = theta_2\ \ text(possible but no scale given)`

`E:\ \ r_1>1\ \ text(possible but no scale given.)`

 

`=> C`

Complex Numbers, SPEC2-NHT 2018 VCAA 4 MC

On the Argand diagram shown above, `4 text(cis)(-(2pi)/3)`  is represented by the point

  1. `P`
  2. `Q`
  3. `R`
  4. `S`
  5. `T`
Show Answers Only

`D`

Show Worked Solution

`text(Point is in 3rd quadrant, and)\ \ r = 4.`

`text(Consider:)\  R, S, T`

`r_R` `~~ sqrt((-1)^2 + (-1.75)^2)`
  `~~ 2.02 < 4`

 
`r_T > sqrt((-2)^2 + (-4)^2)`

`r_T > sqrt 20> 4`

 
`text(By elimination, point must be)\ \ S.` 

`=>   D`