SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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.  Show that the cartesian form of the equation of  `L`  is  `y = - sqrt3 x`.  (2 marks)
  3.  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)

  4.  Find, in cartesian form, the points(s) of intersection of  `L`  and the graph of  `|z| = 4`.  (2 marks)
  5.  Sketch  `L`  and the graph of  `|z| = 4`  on the Argand diagram below.  (2 marks)
     
     
         
     
  6.  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 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 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 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, 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`