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Complex Numbers, SPEC2 2022 VCAA 6 MC

Given  `z=x+yi`, where `x, y \in R` and `z \in C`, an equation that has a graph that has two points of intersection with the graph given by  `|z-5|=2`  is

  1. `\text{Arg}(z-3)=\frac{\pi}{2}`
  2. `|z-1|=2`
  3. `\text{Im}(z)=2`
  4. `\text{Re}(z)+\text{Im}(z)=2`
  5. `|z-5-5 i|=4`
Show Answers Only

`E`

Show Worked Solution

`text{Using CAS, sketch all graphs together with the circle}\ |z-5|=2`

`|z-5-5 i|=4\ \ text{has 2 points of intersection.}`

`=>E`

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 + 1`.

  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.  
     

     
    1.  

       
    2. `(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%.

 

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 2021 VCAA 5 MC

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


 

For points on this circle, the maximum value of  `|z|`  is

  1. `sqrt3 + 1`
  2. `3`
  3. `sqrt13`
  4. `sqrt7 + 1`
  5. `8`
Show Answers Only

`D`

Show Worked Solution
♦ Mean mark part 42%.

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

`text(Radius = 1)`

`text(By Pythagoras, line from)`

`text(origin to centre of circle)`

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

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

`=>\ D`

Complex Numbers, SPEC2 2020 VCAA 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. Write down the function that describes the ray  `text(Arg)(z - u) = pi/4`, giving the rule in cartesian form.  (1 mark)
  6. The points representing `u` and `v` and  `−5i`  lie on the circle given by  `|z - z_c| = r`, where `z_c` is the centre of the circle and `r` is the radius.
  7. Find `z_c` in the form  `a + ib`, where  `a, b ∈ R`, and find the radius `r`.  (3 marks)
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`
  6. `z_c = −5/3 – 10/3 i and r = (5sqrt2)/3`
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.   

♦♦ Mean mark (d)(ii) 25%.

 

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

`:. f: (−2, ∞) -> RR, \ f(x) = x + 1`

 

e.   `|z_c – u| = |z_c – v| = |z_c – (−5i)|`

♦ Mean mark (e) 40%.
`z_c – u` `= (a + 2) + (b + 1)i`
`z_c – v` `= (a + 4) + (b + 3)i`
`z_c +5i` `= a + (b + 5)i`

 
`a^2 + (b + 5)^2 = (a + 2)^2 + (b + 1)^2\ …\ (1)`

`a^2 + (b + 5)^2 = (a + 4)^2 + (b + 3)^2\ …\ (2)`

`a = −5/3, b = −10/3\ \ text{(by CAS)}`
 

`:.z_c = −5/3 – 10/3 i`

`:.r` `= |z_c – (−5i)|`
  `= sqrt((−5/3)^2 + (−10/3 + 5)^2)`
  `= (5sqrt2)/3`

Complex Numbers, SPEC2-NHT 2019 VCAA 5 MC

The circle defined by  `|z + a| = 3 |z + i|`, where  `a ∈ R`, has a centre and radius respectively given by

  1. `(a/8, −9/8), \ 3/8sqrt(a^2 + 1)`
  2. `(a/8, −9/8), \ (9a^2 + 9)/64`
  3. `(a/8, −9/8), \ 1/8sqrt(153 - 7a^2)`
  4. `(−a/8, 9/8), \ (9a^2 + 9)/64`
  5. `(−a/8, 9/8), \ 3/8sqrt(a^2 + 1)`
Show Answers Only

`A`

Show Worked Solution
`|z + a|` `= 3|z + i|`
`|(x + a) + yi|` `= 3|x + (y + 1)i|`
`sqrt((x + a)^2 + y^2)` `= 3sqrt(x^2 + (y + 1)^2)`
`x^2 + 2ax + a^2 + y^2` `= 9(x^2 + y^2 + 2y + 1)`
`8x^2 – 2ax + 8y^2 + 18y` `= a^2 – 9`
`8(x^2 – a/4x + (a^2)/64) + 8(y^2 + 9/4y + 81/64)` `= a^2 – 9 + (a^2)/8 + 81/8`
`(x – a/8)^2 + (y + 9/8)^2` `= (9a^2)/64 + 9/64`
  `= 9/64(a^2 + 1)`

 
`:. text(Centre)\ (a/8, −9/8),\ text(radius) = 3/8sqrt(a^2 + 1)`

`=>\ A`

Complex Numbers, SPEC2 2019 VCAA 2

  1. Show that the solutions of  `2z^2 + 4z + 5 = 0`, where  `z ∈ C`, are  `z = −1 ± sqrt6/2 i`.  (1 mark)
  2. Plot the solutions of  `2z^2 + 4z + 5 = 0`  on the Argand diagram below.  (1 mark)
     

     

Let  `|z + m| = n`, where  `m, n ∈ R`, represent the circle of minimum radius that passes through the solutions of  `2z^2 + 4z + 5 = 0`.

    1. Find  `m`  and  `n`.  (2 marks)
    2. Find the cartesian equation of the circle  `|z + m| = n`.  (1 mark)
    3. Sketch the circle on the Argand diagram in part a.ii. Intercepts with the coordinate axes do not need to be calculated or labelled.  (1 mark)
  2. Find all values of  `d`, where  `d ∈ R`, for which the solutions of  `2z^2 + 4z + d = 0`  satisfy the relation  `|z + m| <= n`.  (2 marks)
  3. All complex solutions of  `az^2 + bz + c = 0`  have non-zero real and imaginary parts.

     

    Let  `|z + p| = q`  represent the circle of minimum radius in the complex plane that passes through these solutions, where  `a, b, c, p, q ∈ R`.

     

    Find  `p`  and  `q`  in terms of  `a, b`  and  `c`.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2.   
  3. `m = 1, n = sqrt6/2`
  4. `(x + 1)^2 + y^2 = 3/2`
  5. `−1 <= d <= 5\ \ (text(by CAS))`
  6. `p = b/(2a), \ q = |sqrt(b^2 – 4ac)/(2a)|`
Show Worked Solution
a.i.    `z` `= (−b ± sqrt(b^2 – 4ac))/(2a)`
    `= (−4 ± sqrt(16 – 4 · 2 · 5))/(4)`
    `= (−4 ± 2sqrt6 i)/(4)`
    `= −1 ± sqrt6/2 i\ \ …\ text(as required)`

 

a.ii.   

 

b.i.   `text(Radius of circle = )sqrt6/2`

 `text(Centre) = (0, −1)`

`:. m = 1, \ n = sqrt6/2`

 

b.ii.    `|z + 1|` `= sqrt6/2`
  `|x + iy + 1|` `= sqrt6/2`
  `(x + 1)^2 + y^2` `= 3/2`

 

b.iii.   

 

c.   `text(Solve:)\ 2z^2 + 4z + d = 0`

`z = −1 ± sqrt(4 – 2d)/2 = −1 ± sqrt((2 – d)/2)`

`z + 1 = ± sqrt((2 – d)/2)`
 

`text(Solve for)\ d\ text(such that:)`

`|sqrt((2 – d)/2)| <= sqrt6/2`

`−1 <= d <= 5\ \ (text(by CAS))`

 

d.   `z = (−b ± sqrt(b^2 – 4ac))/(2a) = (−b)/(2a) ± sqrt(b^2 – 4ac)/(2a)`

`z + b/(2a)` `= ± sqrt(b^2 – 4ac)/(2a)`
`|z + b/(2a)|` `= |sqrt(b^2 – 4ac)/(2a)|`

 
`:. p = b/(2a), \ q = |sqrt(b^2 – 4ac)/(2a)|`

Complex Numbers, SPEC2 2011 VCAA 7 MC

In the complex plane, the circle with equation  `|\ z - (2 + 3i)\ | = 1`  is intersected exactly twice by the curve with equation

A.   `|\ z - 3i\ | = 1`

B.   `|\ z + 3\ | = |\ z - 3i\ |`

C.   `|\ z - 3\ | = |\ z - 3i\ |`

D.   `text(Im)(z) = 4`

E.   `text(Re)(z) = 3`

Show Answers Only

`C`

Show Worked Solution

`text(Consider option C:)`

`|z – 3| = |z – 3i|`

`=> y = x\ text{(perpendicular bisector}`

`text{between (3, 0) and (0, 3)}`
 

`=> C`

Complex Numbers, SPEC2 2013 VCAA 5 MC

The region in the complex plane that is outside the circle of radius `b` centred at the origin is given by the set of points `z`, where  `z ∈ C`, such that

A.   `|\ z\ | < b`

B.   `|\ z\ | > b`

C.   `|\ z\ | > b^2`

D.   `|\ z\ | = b`

E.   `|\ z\ | < b^2`

Show Answers Only

`B`

Show Worked Solution

`text(Circle:)\ |z| = b`

`text(Outside:)\ |z| > b`

`=> B`

Complex Numbers, SPEC2-NHT 2018 VCAA 2

In the complex plane, `L` is the line given by  `|z + 1| = |z + 1/2 - sqrt 3/2 i|`.

  1. Show that the cartesian equation of `L` is given by  `y = -1/sqrt 3 x`.  (2 marks)
  2.  Find the point(s) of intersection of `L` and the graph of the relation  `z bar z = 4`  in cartesian form.  (2 marks)
  3. Sketch `L` and the graph of the relation  `z bar z = 4`  on the Argand diagram below.  (2 marks)

 

 

The part of the line `L` in the fourth quadrant can be expressed in the form  `text(Arg)(z) = a`.

  1. State the value of `a`.  (1 mark)
  2. Find the area enclosed by `L` and the graphs of the relations  `z bar z = 4, \ text(Arg)(z) = pi/3`  and  `text(Re)(z) = sqrt 3`.  (2 marks)
  3. The straight line `L` can be written in the form `z = k bar z`, where  `k in C`.

     

    Find `k` in the form  `r text(cis)(theta)`, where  `theta`  is the principal argument of `k`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `z = sqrt 3 – i, quad -sqrt 3 + i`
  3. `text(See Worked Solutions)`
  4. `a = -pi/6`
  5. `pi/3 + sqrt 3`
  6. `k = cis (-pi/3)`
Show Worked Solution

a.   `(x + 1)^2 + y^2 = (x + 1/2)^2 + (y – sqrt 3/2)^2`

`x^2 + 2x + 1 + y^2 = x^2 + x + 1/4 + y^2 – y sqrt 3 + 3/4`

`0` `=-x – y sqrt 3`
`y sqrt 3` `= -x`
`:. y` `= -1/sqrt 3 x`

 

b.   `(x + iy) (x – iy)` `=4`
  `x^2 – i^2 y^2` `=4`
  `x^2 + y^2` `=4`

 

`text(Substitute)\ \ y = -1/sqrt 3 x\ \ text(into)\ \  x^2 + y^2 = 4:`

`x^2 + 1/3 x ^2` `=4`
`4/3 x^2` `=4`
`x^2` `=3`
`x` `=+- sqrt3`
`=>y` `= +- 1`

 

`:.\ text(Intersection at):\  (sqrt 3, -1), quad (-sqrt 3, 1)`

 

c.   

 

d.   `alpha = tan^(-1) (-1/sqrt 3), quad alpha in (-pi/2, 0)`

`alpha = -pi/6`
 

e.   

`text(Total Area)`

`=\ text(Area of sector + Area of triangle)`

`=pi xx 2^2 xx ((pi/3 – pi/6)/(2 pi)) + 1/2 xx sqrt 3 xx 2`

`=4 pi (1/12) + sqrt3`

`=pi/3 + sqrt3\ \ text(u²)`

 

f.    `r\ text(cis)(theta)` `=k (r\ text(cis)(-theta))`
  `:. k` `=(r\ text(cis)(theta))/(r\ text(cis)(-theta))`
    `= text(cis)(2 xx ((-pi)/6))`
    `= text(cis)(- pi/3)`

Complex Numbers, SPEC2 2015 VCAA 8 MC

A relation that does not represent a circle in the complex plane is

  1. `zbarz = 4`
  2. `|\ z + 3i\ | = 2|\ z − i\|`
  3. `|\ z − i\ | = |\ z + 2\ |`
  4. `|\ z − 1 + i\ | = 4`
  5. `|\ z\ | + 2|\ barz\ | = 4`
Show Answers Only

`C`

Show Worked Solution

`text(Consider each option:)`

`A:` `(x + iy)(x – iy)` `= 4`
  `x^2 + y^2` `= 4\ \ \ \ (text{circle equation)}`

 

`B:\ x^2+ (y + 3)^2=4(x^2 + (y – 1)^2)`

`x^2 + y^2 + 6y + 9` `= 4x^2 + 4y^2 – 8y + 4`
`3x^2 + 3y^2 – 14y -5` `=0\ \ \ \ (text{circle equation)}`

 

`C:` `x^2 + (y – 1)^2` `= (x + 2)^2 + y^2`
  `x^2 + y^2 – 2y + 1` `= x^2 + 4x + 4 + y^2`
  `4x+2y+3` `=0 \ \ \ \ (text{not a circle equation)}`

 
`text(Similarly, D and E can be shown to represent circle equations.)`

`=> C`

Complex Numbers, SPEC2 2018 VCAA 2

  1. State the centre in the form  `(x, y)`, where  `x, y in R`, and the state the radius of the circle given by  `|z - (1 + 2i)| = 2`, where  `z in C`.  (1 mark)
  2. By expressing the circle given by  `|z + 1| = sqrt 2 |z - i|`  in cartesian form, show that this circle has the same centre and radius as the circle given by  `|z - (1 + 2i)| = 2`.  (2 marks)
  3. Graph the circle given by  `|z + 1| = sqrt 2 |z - i|`  on the Argand diagram below, labelling the intercepts with the vertical axis.  (2 marks)
     

         
     

The line given by  `|z - 1| = |z - 3|`  intersects the circle given by  `|z + 1| = sqrt 2 |z - i|`  in two places.

  1. Draw the line given by  `|z - 1| = |z - 3|`  on the Argand diagram in part c. Label the points of intersection with their coordinates..  (2 marks)
  2. Find the area of the minor segment enclosed by an arc of the circle given by  `|z + 1| = sqrt 2 |z - i|`  and part of the line given by  `|z - 1| = |z - 3|`.  (3 marks)

 

Show Answers Only
  1. `text(Centre): (1, 2), quad text(radius): 2`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `(4 pi)/3 – sqrt 3`
Show Worked Solution

a.   `(x – 1)^2 + (y – 2)^2 = 2^2`

`:.\ text(Centre)\ (1, 2), quad text(radius) = 2`
 

b.    `y^2 + (x + 1)^2` `= (sqrt 2)^2 (x^2 + (y – 1)^2)`
  `y^2 + x^2 + 2x + 1` `= 2(x^2 + y^2 – 2y + 1)`
  `y^2 + x^2 + 2x + 1` `= 2x^2 + 2y^2 – 4y + 2`
`0` `= 2x^2 + 2y^2 – 4y + 2 – y^2 – x^2 – 2x – 1`
`0` `= x^2 + y^2 – 4y – 2x + 1`
`0` `= (x^2 – 2x + 1) + (y^2 – 4y)`
`0` `= (x – 1)^2 + (y^2 – 4y + 2^2) – 4`
`0` `= (x – 1)^2 + (y – 2)^2 – 4`
`4` `= (x – 1)^2 + (y – 2)^2`

 
`:.\ text(Centre)\ (1, 2), quad text(radius) = 2`

 

c.   `ytext(-axis intercepts occur when)\ \ x=0:`

`(0 – 1)^2 + (y – 2)^2` `= 4`
`1 + (y – 2)^2` `= 4`
`(y – 2)^2` `= 3`
`y – 2` `= +- sqrt 3`
`y` `= 2 +- sqrt 3`

 

d.   `|z – 1| = |z – 3|`

`=>\ text(Graph is a line equidistant from:)\ \ 1 + 0i and 3 + 0i`

`text(Circle intercepts occur when)\ \ x=2:`

`(2- 1)^2 + (y – 2)^2` `= 4`
`(y – 2)^2` `= 3`
`y – 2` `= +- sqrt 3`
`y` `= 2 +- sqrt 3`

 

e.   

`sin theta = sqrt3/2\ \ =>\ \ theta = pi/3`

♦ Mean mark 39%.
MARKER’S COMMENT: Students who used standard formulae rather than definite integrals were more successful.

`text(Angle at centre of segment) = (2pi)/3`

`text(Height of triangle)\ (h) = 2^2 – (sqrt3)^2 =1`
 

`:.\ text(Shaded Area)`

`=\ text(Area of segment – Area of triangle)`

`=((2pi)/3)/(2pi) xx pi xx 2^2 – 1/2 xx 2sqrt3 xx 1`

`=(4pi)/3 – sqrt3`

Complex Numbers, SPEC2 2014 VCAA 9 MC

The circle  `| z - 3 - 2i | = 2`  is intersected exactly twice by the line given by

A.   `| z - i | = | z + 1 |`

B.   `| z - 3 - 2i | = | z - 5 |`

C.   `| z - 3 - 2i | = | z - 10i |`

D.   `text(Im)(z) = 0`

E.   `text(Re)(z) = 5`

Show Answers Only

`B`

Show Worked Solution

`text{Circle has centre (3,2) and radius 2.}`

`text{Consider each option (using CAS):}`

`text(Options A and C do not intersect.)`

`text(Options D and E are tangents.)`

`text(Only option B intersects twice).`

`=> B`

Complex Numbers, SPEC1-NHT 2018 VCAA 8

A circle in the complex plane is given by the relation  `|z - 1 - i| = 2, \ z in C`.

  1. Sketch the circle on the Argand diagram below.  (1 mark)

 

  1. i.  Write the equation of the circle in the form  `(x - a)^2 + (y - b)^2 = c`  and show that the gradient of a tangent to the circle can be expressed as  `(dy)/(dx) = (1 - x)/(y - 1)`.  (2 marks)
  2. ii. Find the gradient of the tangent to the circle where  `x = 2`  in the first quadrant of the complex plane.  (1 mark)
  3. Find the equations of all rays that are perpendicular to the circle in the form  `text(Arg) (z) = alpha`.  (2 marks)
Show Answers Only
  1. `text(See Worked Solutions)`
    1.  `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `(-1)/sqrt 3`
  3. `text(Arg) (z) = pi/4; quad text(Arg) (z) = (-3 pi)/4`
Show Worked Solution
a.  

 

b.i.   `(x – 1)^2 + (y – 1)^2` `= 4`
  `2(x – 1) + d/(dx) ((y – 1)^2)` `= 0`
  `2(x – 1) + 2(y-1)*(dy)/(dx)` `= 0`
  `2 (y-1)*(dy)/(dx)` `= -2(x – 1)`
  `(dy)/(dx)` `= (-(x – 1))/(y-1)`
    `= (1 – x)/(y – 1)`

 

b.ii.   `(2 – 1)^2 + (y – 1)^2` `= 4`
  `1 + (y – 1)^2` `= 4`
  `(y – 1)^2` `= 3`
  `y` `= 1 + sqrt 3\ \ \ (y > 0)`
     
  `(dy)/(dx)|_{(2, 1 + sqrt 3)}` `= (1 – 2)/(1 + sqrt 3 – 1`
    `= (-1)/sqrt 3`

 

c.   `P (0, 0) quad C(1, 1)`
  `-> y = x`
  `:. alpha = pi/4, quad (-3 pi)/4` 

 
`text(Arg) (z) = pi/4`

`text(Arg) (z) = (-3 pi)/4`