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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, SPEC1 2023 VCAA 2

Consider the complex number  \(z=(b-i)^3\), where  \(b \in R^{+}\).

Find \(b\) given that  \(\arg (z)=-\dfrac{\pi}{2}\).   (3 marks)

Show Answers Only

\(b=\sqrt{3} \)

Show Worked Solution

\(-\dfrac{\pi}{2} = \arg (z) = \arg[(b-i)^3]\)

\(b-i\ \Rightarrow \ \text{complex number that lives on}\ \ y=-i \)

\(3 \times \arg(b-i) = -\dfrac{\pi}{2} \)

\(\arg(b-i) = -\dfrac{\pi}{6} \)

\(\tan^{-1}\Big{(}\dfrac{-1}{b} \Big{)} \) \(= -\dfrac{\pi}{6} \)  
\(-\dfrac{1}{b}\) \(=-\dfrac{1}{\sqrt{3}}\)  
\(\therefore b\) \(=\sqrt{3}\)  

Complex Numbers, SPEC2 2021 VCAA 4 MC

For  `z ∈ C`, if  `text(Im)(z) > 0`, then  `text(Arg)((zbarz)/(z - barz))` is

  1. `-pi/2`
  2. `0`
  3. `pi/4`
  4. `pi/2`
  5. `pi`
Show Answers Only

`A`

Show Worked Solution

`text(Let)\ \ z=x+iy \ => \ barz=x-iy`

♦♦ Mean mark 35%.
`text(Arg)((zbarz)/(z – barz))` `= text(Arg)(zbarz) – text(Arg)(z – barz)`
  `= text(Arg)(x^2 + y^2) – text(Arg)(2yi)`
  `= 0 – text(Arg)(2yi),\ \ text(where)\ y > 0`
  `= -pi/2`

`=>\ A`

Complex Numbers, SPEC1 2019 VCAA 7

  1. Show that  `3-sqrt3 i = 2sqrt3 text(cis)(-pi/6)`.   (1 mark)

  2. Find  `(3-sqrt3 i)^3`, expressing your answer in the form  `x + iy`, where  `x`,  `y ∈ R`.   (2 marks)

  3. Find the integer values of  `n`  for which  `(3-sqrt3 i)^n`  is real.   (1 mark)

  4. Find the integer values of  `n`  for which  `(3-sqrt3 i)^n = ai`, where  `a`  is a real number.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0-i 24sqrt3`
  3. `n = 6k\ \ (k ∈ ZZ)`
  4. `n = 3 + 6k\ \ (k ∈ ZZ)`
Show Worked Solution

a.  `|3-sqrt3 i|= sqrt(3^2 + (-sqrt3)^2)= sqrt12= 2sqrt3`

`text(Arg)(3-sqrt3 i)` `= tan^(-1)(-(sqrt3)/3)= -pi/6`
   

`:. 3-sqrt3 i = 2sqrt3\ text(cis)(-pi/6)`

b.    `(3-sqrt3 i)^3` `= (2sqrt3)^3\ text(cis)(3 xx-pi/6)`
    `= 24sqrt3\ text(cis)(-pi/2)`
    `= 24sqrt3(cos(-pi/2) + isin(-pi/2))`
    `= 0-i 24sqrt3`

 

c.   `(3-sqrt3 i)^n = (2sqrt3)^n\ text(cis)((-npi)/6)`

`text(Real when)\ \ sin(-(npi)/6) = -sin((npi)/6) = 0`

`(npi)/6 = 0, pi, 2pi, …, kpi\ \ (k ∈ ZZ)`

`:. n = 6k\ \ (k ∈ ZZ)`

 

d.  `(3-sqrt3 i)^n = ai\ \ text(when)\ \ cos(-(npi)/6) = cos((npi)/6) = 0`

`(npi)/6 = pi/2, (3pi)/2, …, pi/2 + kpi\ \ (k ∈ ZZ)`

`:. n = 3 + 6k\ \ (k ∈ ZZ)`

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 2013 VCAA 8 MC

The principal arguments of the solutions to the equation  `z^2 = 1 + i`  are

  1. `pi/8`  and  `(9pi)/8`
  2. `−pi/8`  and  `(7pi)/8`
  3. `−(7pi)/8`  and  `pi/8`
  4. `(7pi)/8`  and  `(15pi)/8`
  5. `−(3pi)/4`  and  `pi/4`
Show Answers Only

`C`

Show Worked Solution
`z^2` `= 1 + i`
  `= sqrt2 text(cis)(pi/4 + 2kpi)`
`z` `= 2^(1/4) text(cis) (pi/8 + kpi)`

 
`:.\ text(Principal arguments are)\ −(7pi)/8\ text(and)\ pi/8.`

`=> C`

Complex Numbers, SPEC1 2016 VCAA 6

Write  `(1-sqrt 3 i)^4/(1 + sqrt 3 i)`  in the form  `a + bi`, where  `a`  and  `b`  are real constants.   (3 marks)

Show Answers Only

`4 + 4 sqrt 3 i`

Show Worked Solution

`(1-sqrt 3 i) \ \ => \ r_1=sqrt (1^2 + (sqrt 3)^2)=2`

`theta_1 = tan^(-1) (-sqrt 3)=- pi/3`

`(1 + sqrt 3 i) \ \ => \ r_2=2, \ theta_2 =pi/3`
 

`:. (1-sqrt 3 i)^4/(1 + sqrt 3 i)` `= (2 text(cis) (-pi/3))^4/(2 text(cis) (pi/3))`
  `= (16 text(cis) (-(4 pi)/3))/(2 text(cis) (pi/3))`
  `= 8 text(cis) (-(5 pi)/3)`
  `= 8 text(cis) (pi/3)\ \ \ (-pi<=theta<=pi)`
  `= 8(1/2 + sqrt 3/2 i)`
  `= 4 + 4 sqrt 3 i`

Complex Numbers, SPEC2 2015 VCAA 7 MC

If  `z = sqrt3 + 3i`, then  `z^63`  is

  1. real and negative
  2.  equal to a negative real multiple of `i`
  3. real and positive
  4. equal to a positive real multiple of `i`
  5. a positive real multiple of  `1 + isqrt3`
Show Answers Only

`A`

Show Worked Solution
`z` `= sqrt3 + 3i`
  `= 2sqrt3 text(cis)\ (pi/3)`

 

`:. z^63` `= (2sqrt3)^63 text(cis)\ ((63pi)/12)`
  `= (2sqrt3)^63 text(cis)\ (21pi)`
  `= (2sqrt3)^63 text(cis)\ (pi)`
  `= -(2sqrt3)^63`

 
`=> A`

Complex Numbers, SPEC2 2015 VCAA 5 MC

Given  `z = (1 + isqrt3)/(1 + i)`, the modulus and argument of the complex number  `z^5`  are respectively

  1. `2sqrt2`  and  `(5pi)/6`
  2. `4sqrt2`  and  `(5pi)/12`
  3. `4sqrt2`  and  `(7pi)/12`
  4. `2sqrt2`  and  `(5pi)/12`
  5. `4sqrt2`  and  `-pi/12`
Show Answers Only

`B`

Show Worked Solution

`z_1 = 1 + isqrt3`

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

 
`z_1 = 2\ text(cis)(pi/3)`
 

`z_2 = 1 + i`

`r` `= sqrt(1 + 1)=sqrt2`
`theta` `= tan^(−1)(1)=pi/4`

 

`z_1/z_2` `= 2/sqrt2\ text(cis)(pi/3 – pi/4)`
  `= sqrt2\ text(cis)(pi/12)`

 

`:. z^5` `= (sqrt2)^5\ text(cis)((5pi)/12)\ \ \ text{(De Moivre)}`
  `= 4sqrt2\ text(cis)((5pi)/12)`

 
`=> B`

Complex Numbers, SPEC2 2014 VCAA 8 MC

The principal argument of  `(−3sqrt2 - isqrt6)/(2 + 2i)`  is

A.   `(−13pi)/12`

B.   `(7pi)/12`

C.   `(11pi)/12`

D.   `(13pi)/12`

E.   `(−11pi)/12`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ z_1=−3sqrt2 – isqrt6`

`=>\ text(Arg)\ z_1 = pi + tan^(-1) (sqrt6/(3 sqrt2)) = (7pi)/6`

`text(Let)\ \ z_2 = 2 + 2i`

`=>\ text(Arg)\ z_2 = pi/4`
 

`:.\ text(Arg)\ (z_1/z_2)` `= text(Arg)\ ((−3sqrt2 – isqrt6)/(2 + 2i))`
  `= (7pi)/6 – pi/4`
  `= (11pi)/12`

 
`=> C`

Complex Numbers, SPEC2 2018 VCAA 5 MC

Let  `z = a + bi`, where  `a, b in R\ text(\ {0})`

If  `z + 1/z \ in R`, which one of the following must be true?

  1. `text(Arg)(z) = pi/4`
  2. `a = -b`
  3. `a = b`
  4. `|z| = 1`
  5. `z^2 = 1` 
Show Answers Only

`D`

Show Worked Solution
`z + 1/z` `= a + bi + 1/(a + bi)`
  `= a + bi + (a – bi)/{(a + bi)(a – bi)}`
  `= a + bi + (a – bi)/(a^2 + b^2)`
  `= {(a + bi)(a^2 + b^2) + a – bi}/(a^2 + b^2)`
  `= {a(a^2 + b^2) + a}/(a^2 + b^2) + {b (a^2 + b^2) – b}/(a^2 + b^2)\ i`

 
`text(If)\ \ z + 1/z in R\ \ =>\  {b (a^2 + b^2) – b}/(a^2 + b^2) = 0`

 

`b(a^2 + b^2) – b = 0`

`b(a^2 + b^2 – 1) = 0`

`a^2 + b^2 = 1,\ \ (b !=0)`

`a^2 + b^2 = |z|^2 = 1`

`:. |z| = 1`

`=>  D`

Complex Numbers, SPEC1 2018 VCAA 2

  1. Show that  `1 + i = sqrt 2\ text(cis)(pi/4)`.   (1 mark)

  2. Evaluate  `(sqrt 3-i)^10/(1 + i)^12`, giving your answer in the form  `a + bi`, where  `a, b ∈ R`.   (3 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-8-8 sqrt 3 i`

Show Worked Solution

a.   `r` `= sqrt(1^2 + 1^2)`
    `= sqrt 2`
  `theta` `= tan^(-1) (1/1) = pi/4`
     
  `:. 1 + i` `= sqrt 2\ text(cis)(pi/4)`

 

b.   
 `r_2` `= sqrt((sqrt 3)^2 + (-1)^2)`
    `= sqrt (3 + 1)`
    `= 2`

 
`theta_2 = -tan^(-1) (1/sqrt 3)=-pi/6`

`sqrt 3-i` `= 2\ text(cis)(-pi/6)`
`(sqrt 3-i)^10` `= 2^10\ text(cis) ((-10 pi)/6)`
   
`=>(1 + i)^12` `= (sqrt 2)^12text(cis)((12pi)/4)`
  `=2^6 text(cis)(3pi)`

 

`:. (sqrt 3-i)^10/(1 + i)^12` `= (2^10 text(cis)((-5pi)/3))/(2^6\text(cis)(3pi))`
  `= 2^4 text(cis)((-5pi)/3-(9pi)/3)`
  `= 16 text(cis)((-14pi)/3)`
  `= 16 text(cis) ((-2pi)/3)`
  `= 16(-1/2 + (-sqrt3)/2 i)`
  `= -8-8 sqrt 3 i`

Complex Numbers, SPEC1 2011 VCAA 4

Consider  `z = (1-sqrt 3 i)/(-1 + i),\ \ z in C.`

Find the principal argument of `z` in the form  `k pi,\ k in R.`   (3 marks)

Show Answers Only

`text(Arg)(z) = 11/12 pi`

Show Worked Solution
`z_1` `= 1-sqrt3 i`
  `= sqrt(1^2 + 3)\ text(cis)(−tan^(−1)(sqrt3))`
  `= 2text(cis)(−pi/3)`

 

`z_2` `= −1 + i`
  `= sqrt(1^2 + 1^2)\ text(cis)(pi-tan^(−1)(1))`
  `= sqrt2 text(cis)(pi-pi/4)`
  `= sqrt2 text(cis)((3pi)/4)`

 

`text(Arg)((1-sqrt3 i)/(−1 + i))` `= text(Arg)(1-sqrt3 i)-text(Arg)(−1 + i)`
  `= -pi/3-(3pi)/4`
  `= -(13pi)/12`

 

`-(13pi)/12 + 2pi` `= (-13pi+ 24pi)/12`  
  `= (11pi)/12`