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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 1

In the complex plane, `L` is the with equation `|z + 2| = |z-1-sqrt3 i|`.

  1.  Verify that the point (0, 0) lies on `L`.   (1 marks)

  2.  The line  `L`  can also be expressed in the form  `|z-1| = |z-z_1|`, where  `z_1 ∈ C`.

     

     Find  `z_1` in cartesian form.   (2 marks)

  3.  Find, in cartesian form, the points(s) of intersection of  `L`  and the graph of  `|z| = 4`.   (2 marks)

  4.  Sketch  `L`  and the graph of  `|z| = 4`  on the Argand diagram below.   (2 marks)

     
     
         
     

  5.  Find the area of the sector defined by the part of  `L`  where  `text(Re)(z) ≥ 0`, the graph of  `|z| = 4`  where  `text(Re)(z) ≥ 0`, and imaginary axis where  `text(Im)(z) > 0`.   (1 marks)

Show Answers Only
  1. `text(Proof(See Worked Solution))`
  2. `text(Proof(See Worked Solution))`
  3. `-(1)/(2)-(sqrt3)/(2) i`
  4. `(2,-2 sqrt3) text(and) (-2, 2 sqrt3)`
  5.  

     
  6. `(20pi)/(3)`
Show Worked Solution

a.   `text(Substitute)\ \ z = 0 + 0i\ \  text(into both sides:)`

`text(LHS) = |2| = 2`

`text(RHS) = |-1-sqrt3i| = sqrt{(-1)^2 + (-sqrt3)^2} = 2`

`:. (0,0)\ \ text(lies on both sides.)`

 

b.  `|x + yi + 2|` `= |x + yi-1-sqrt3 i|`
  `|(x + 2) + yi|` `= |(x-1) + (y-sqrt3) i|`
  `sqrt(x^2 + 4x + 4 + y^2` `= sqrt(x^2-2x + 1 + y^2 -2 sqrt3 y + 3`
  `x^2 + 4x + 4 + y^2` `= x^2-2x + 4-2 sqrt3 y + y^2`
  `6x` `= -2 sqrt3 y`
  `y` `= -(3)/(sqrt3) x`
  `y` `= -sqrt3 x`


c.

`m_text(perp) = (sqrt2)/(3) , text(through)\ (1, 0)`

`y = (sqrt3)/(3) (x-1)\ …\ L_1`

`text(Intersection) \ L\ text(and) \ L_1,`

`text(Solve:) \ (sqrt3)/3 (x-1) = -sqrt3 x\ \ \ text{(by CAS)}`

`=> x = (1)/(4) , y = -(sqrt3)/(4) \ \ \ text{(point}\ Ptext{)}`
 

`P(x_1,y_1) \ text(is midpoint of) \ \ z_1 \ text(and) \ \ (1, 0):`

`(x_1 + 1)/(2) = (1)/(4) \ => \ x_1 = -(1)/(2)`

`(y_1 + 0)/(2) = -sqrt3/(4) \ => \ y_1 = -sqrt3/(2)`

`:. \ z_1 = -(1)/(2)-(sqrt3)/(2) i`

 

d.   `|z| = 4 => \ text(circle, centre) \ (0,0), \ text(radius) = 4`

`x^2 + y^2 = 16\ …\ (1)`

`y =-sqrt3 x\ …\ (2)`

`text(Substitute)\ (2) \ text(into) \ (1)`

`x^2 + 3x^2 = 16`

`x = ±2`

`:. \ text(Intersection at)\ (2,-2 sqrt3) \ text(and) \ (-2, 2 sqrt3)`

 

e.


 

f.    

`text(Area)` `= (5)/(12) xx  pi xx 4^2`
  `= (20pi)/(3)`

Complex Numbers, SPEC2 2019 VCAA 5 MC

Let  `z = x + yi`, where  `x, y ∈ R`. The rays  `text(Arg)(z - 2)=pi/4`  and  `text(Arg)(z - (5 + i)) = (5pi)/6`, where  `z ∈ C`, intersect on the complex plane at a  point  `(a,b)`.

The value of `b` is

  1. `−sqrt3`
  2. `2 - sqrt3`
  3. `0`
  4. `sqrt3`
  5. `2 + sqrt3`
Show Answers Only

`D`

Show Worked Solution

`text(Convert to cartesian equations)`

`text(Arg)(z – 2) = pi/4`

`y = x – 2\ …\ (1)`

`text(Arg)(z – (5 + i)) = (5pi)/6`

`y – 1 = −1/sqrt3 (x – 5)\ …\ (2)`

`text{Solve (1) and (2) simultaneously:}`

`text(Intersection)\ (a,b) :\ \ x = 2+sqrt3, \ y = sqrt3`

`:. b = sqrt3`

`=>D`

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 2012 VCAA 7 MC

The set of points in the complex plane defined by  `|\ z + 2i\ | = |\ z\ |`  corresponds to

A.   the point given by  `z = −i`

B.   the line  `text(Im)(z) = −1`

C.   the line  `text(Im)(z) = −i`

D.   the line  `text(Re)(z) = −1`

E.   the circle with centre  `−2i`  and radius `1`

Show Answers Only

`B`

Show Worked Solution

`|\ z + 2i\ | = |\ z\ |\ \ =>\ \  |\ z – (−2i)\ | = |\ z\ |`

`text{Find all points equidistant from (0, −2) and (0, 0):}`

`text(Midpoint):\ ((0 + 0)/2, (0 – 2)/2) = (0, -1)`

`m= (-2 – 0)/(0 – 0) = oo`

`:. m_(_|_) = 0\ \ text{(i.e. horizontal line)}`

`:. y = -1 \ \ or\ \ text{Im} (z) = -1`

 
`=> B`

Complex Numbers, SPEC2 2017 VCAA 4

  1. Express  `−2-2sqrt3 i`  in polar form.  (1 mark)

  2. Show that the roots of  `z^2 + 4z + 16 = 0`  are  `z = −2-sqrt3 i`  and  `z = −2 + 2sqrt3 i`.  (1 mark)

  3. Express the roots of  `z^2 + 4z + 16 = 0`  in terms of  `2-2sqrt3 i`.  (1 mark)

  4. Show that the cartesian form of the relation  `|z| = |z-(2-2sqrt3 i)|`  is  `x-sqrt3 y-4 = 0`  (2 marks)

  5. Sketch the line represented by  `x-sqrt3y -4 = 0`  and plot the roots of  `z^2 + 4z + 16 = 0`  on the Argand diagram below.  (2 marks)

     
       

  6. The equation of the line passing through the two roots of  `z^2 + 4z + 16 = 0`  can be expressed as  `|z-a| = |z-b|`, where  `a, b ∈ C`.

     

    Find `b` in terms of `a`.  (1 mark)

  7. Find the area of the major segment bounded by the line passing through the roots of  `z^2 + 4z + 16 = 0`  and the major arc of the circle given by  `|z| = 4`.  (2 marks)

Show Answers Only
  1. `4text(cis)((−2pi)/3)`
  2. `text(See Worked Solutions)`
  3. `−(2-2sqrt3 i) and-bar((2-2sqrt3 i))`
  4. `text(See Worked Solutions)`
  5.  
  6. `−4-bara`
  7. `4sqrt3 + (32pi)/3`
Show Worked Solution
a.    `r` `= sqrt((−2)^2 + (−2sqrt3)^2)=4`

 

`theta` `= −pi + tan^(−1)((2sqrt3)/2)`
  `= −pi + pi/3`
  `=(-2pi)/3`

 
`:. −2-2sqrt3 i = 4text(cis)((−2pi)/3)`
 

b.   `z^2 + 4z + z^2-4 + 16` `=0`
`(z + 2)^2 + 12` `= 0`
`(z + 2)^2` `= -12`
`(z + 2)^2` `= 12i^2`
`z + 2` `= ±sqrt12 i`
`z + 2` `= ±2sqrt3 i`
`:. z` `= -2 ± 2sqrt3 i`

 

♦♦ Mean mark part (c) 31%.

c.    `z_1` `= -2 + 2sqrt3 i = -(2-2sqrt3 i)`
  `z_2` `=-2-2sqrt3 i =-bar((2-2sqrt3 i))`

 

d. `|z|` `= |z-(2-2sqrt3 i)|`
     `x^2 + y^2` `= (x-2)^2 + (y + 2sqrt3)^2`
    `x^2 + y^2` `= x^2-4x + 4+ y^2 + 4sqrt3 y + 12`
  `0` `= −4x + 4sqrt3 y + 16`
  `0` `= −x + sqrt3 y + 4`

 
`:. x-sqrt3 y-4=0`

 

e.   

 

f.   `x = − 2\ \ text(is equidistant from)\ \ z_1 = a\ \ text(and)\ \ z_2 = b`

♦♦♦ Mean mark 1%!

`=> text(Im)(a) = text(Im)(a)`
 

`text(Let)\ \ a = alpha + betai, \ b = gamma + betaj`

`(alpha + gamma)/2` `= −2`
`alpha + gamma` `= −4`
`gamma` `= -4-alpha`

  

`:. b` `= -4-alpha + betaj`
  `= -4-(alpha + betaj)`
  `= -4-bara`

 

♦♦ Mean mark 31%.

g.    `text(Area)\ DeltaOAB` `= 1/2 xx (4sqrt3 xx 2)`
    `= 4sqrt3`

 

`text(Area of sector)\ AOB` `= pi xx 4^2 xx (2 xx pi/3)/(2pi)`
  `= (16pi)/3`

 
`:.\ text(Area of major segment area)`

`=pi(4)^2-((16pi)/3-4sqrt3)`

`= 4sqrt3 + (32pi)/3`

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-NHT 2017 VCAA 6 MC


 

The relation that defines the line `S` above is

A.   `|z + 2| = |z + 2i|`

B.   `text(Arg)(z) = (3 pi)/4`

C.   `|z - 2| = |z + 2i|`

D.   `text(Im)(z) = text(Arg) ((3 pi)/4) + text(Arg)(-pi/4)`

E.   `|z - 2| = |z - 2i|`

Show Answers Only

`C`

Show Worked Solution

`text(Consider the options:)`

`A:\ text(Need)\ z\ text(to be equidistant from)`

`x = -2 \ and\  y = -2\ \ =>\ text(not correct)`
 

`B: text(Arg)(z) = (3 pi)/4\ \ text(is a ray originating at the)`

`text(origin)\ =>\ text(not correct)`

 

`C: text(Need)\ z\ text(to be equidistant from)`

`x = 2 \ and\  y = -2`
 

`text(Midpoint): ((2 +0)/2, (0 + (-2))/2) ≡(1,-1)`

`m = (-2 – 0)/(0 – 2) = (-2)/(-2) = 1`

`m_⊥ = (-1)/m_1 = -1`
 

`:.\ text(Option)\ C\ text(describes the correct relation.)`

`=>   C`

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, SPEC2 2017 VCAA 5 MC

 On an Argand diagram, a point that lies on the path defined by  `|\ z - 2 + i\ | = |\ z - 4\ |`  is

  1.  `(3, −1/2)`
  2.  `(−3, −1/2)`
  3.  `(−3, 3/2)`
  4.  `(3, 1/2)`
  5.  `(3, −3/2)`
Show Answers Only

`A`

Show Worked Solution

`text(Equidistant from)\ \ z = 4\ \ text(and)\ \ z= 2 + i`

`text(Midpoint:)\ ((2 + 4)/2, (−1 + 0)/2)`

`= (3, −1/2)`

`=>A`