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Complex Numbers, SPEC2 2023 VCAA 2

Let \(w=\text{cis}\left(\dfrac{2 \pi}{7}\right)\).

  1. Verify that \(w\) is a root of  \(z^7-1=0\).  (1 marks)
  1. List the other roots of  \(z^7-1=0\)  in polar form.  (1 mark)
  1. On the Argand diagram below, plot and label the points that represent all the roots of  \(z^7-1=0\).  (2 marks)
     

  1.  
    1. On the Argand diagram below, sketch the ray that originates at the real root of  \(z^7-1=0\)  and passes through the point represented by \( \text{cis}\left(\dfrac{2 \pi}{7} \right)\).  (1 mark)
       

    1. Find the equation of this ray in the form \(\text{Arg}\left(z-z_0\right)=\theta\), where \(z_0 \in C\), and \(\theta\) is measured in radians in terms of \(\pi\).  (1 mark)
  1. Verify that the equation  \(z^7-1=0\)  can be expressed in the form  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\).
  2.  
    1. Express  \(\text{cis}\left(\dfrac{2 \pi}{7}\right)+\operatorname{cis}\left(\dfrac{12 \pi}{7}\right)\) in the form \(A \cos (B \pi)\), where \(A, B \in R^{+}\).
    2. Given that  \(w=\operatorname{cis}\left(\dfrac{2 \pi}{7}\right)\)  satisfies  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\), use De Moivre's theorem to show that \(\cos \left(\dfrac{2 \pi}{7}\right)+\cos \left(\dfrac{4 \pi}{7}\right)+\cos \left(\dfrac{6 \pi}{7}\right)=-\dfrac{1}{2}\).

Show Answers Only

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Show Worked Solution

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Complex Numbers, SPEC2 2023 VCAA 5 MC

Let \(z\) be a complex number where  \(\operatorname{Re}(z)>0\)  and  \(\operatorname{Im}(z)>0\).

Given  \(|\bar{z}|=4\)  and  \(\arg \left(z^3\right)=-\pi\), then \(z^2\) is equivalent to

  1. \( {4z} \)
  2. \( -2 \bar{z} \)
  3. \( 3z \)
  4. \(\bar{z}^2\)
  5. \(-4 \bar{z}\)
Show Answers Only

\(E\)

Show Worked Solution

\(\arg(z^3) = -\pi=\pi\ \ \Rightarrow \ \arg(z)=\dfrac{\pi}{3}\)

\(z\) \(=4\text{cis}\Big{(}\dfrac{\pi}{3}\Big{)}\ \ (\abs{z}=\abs{\bar z})\)  
\(z^2\) \(=16\text{cis}\Big{(}\dfrac{2\pi}{3}\Big{)} \)  
  \(=-4 \times 4\text{cis}\Big{(}-\dfrac{\pi}{3}\Big{)} \)  
  \(=-4\,\bar z\)  

 
\(\Rightarrow E\)

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 6 MC

If  `z ∈ C, z != 0`  and  `z^2 ∈ R`, then the possible values of  `text(arg)(z)`  are

  1. `(kpi)/2, k ∈ Z`
  2. `kpi, k ∈ Z`
  3. `((2k + 1)pi)/2, k ∈ Z`
  4. `((4k + 1)pi)/2, k ∈ Z`
  5. `((4k - 1)pi)/2, k ∈ Z`
Show Answers Only

`A`

Show Worked Solution
♦♦♦ Mean mark 23%.
`text(Let)\ \ z` `= r text(cis) theta`
`z^2` `= r^2 text(cis)(2theta)`

 
`text(If)\ \ z^2 ∈ R,`

`2theta` `= kpi`
`theta` `= (kpi)/2`
`text(arg)(z)` `= (kpi)/2`

 
`=>\ A`

Complex Numbers, SPEC2 2019 VCAA 6 MC

Let  `z, w ∈ C`, where  `text(Arg)(z) = pi/2`  and  `text(Arg)(w) = pi/4`.

The value of  `text(Arg)((z^5)/(w^4))`  is

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

`A`

Show Worked Solution
`text(Arg)((z^5)/(w^4))` `= text(Arg)(z^5) – text(Arg)(w^4)`
  `= 5text(Arg)(z) – 4text(Arg)(w)`
  `= (5pi)/2 – (4pi)/4`
  `= (3pi)/2`

 
`:. text(Arg)((z^5)/(w^4)) = −pi/2`

`=>A`

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 2012 VCAA 5 MC

If  `z = sqrt 2 text(cis)(-(4 pi)/5)`  and  `w = z^9`, then

A.   `w = 16 sqrt 2 text(cis)((36 pi)/5)`

B.   `w = 16 sqrt 2 text(cis)(−pi/5)`

C.   `w = 16 sqrt 2 text(cis)((4pi)/5)`

D.   `w = 9 sqrt 2 text(cis)(-pi/5)`

E.   `w = 9 sqrt 2 text(cis)((4pi)/5)`

Show Answers Only

`C`

Show Worked Solution

`z = sqrt2 text(cis) (−(4pi)/5)`

`text(By De Moivre:)`

`w` `= z^9`
  `= (sqrt 2)^9 text(cis)(-(4pi)/5 xx 9)`
  `= 16 sqrt2 text(cis)(-(36pi)/5)`
  `= 16 sqrt 2 text(cis)((4pi)/5 – 8pi)`
  `= 16 sqrt 2 text(cis)((4pi)/5)`

 
`=> C`

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, SPEC2 2013 VCAA 7 MC

If  `z = r text(cis)(theta)`, then  `(z^2)/barz`  is equivalent to

A.   `r^3text(cis)(3theta)`

B.   `r^3text(cis)(−theta)`

C.   `2text(cis)(3theta)`

D.   `r^3text(cis)(theta)`

E.   `rtext(cis)(3theta)`

Show Answers Only

`E`

Show Worked Solution
`z^2` `= r^2text(cis)(2theta)`
`barz` `= rtext(cis)(−theta)`
`(z^2)/z` `= (r^2text(cis)(2theta))/(rtext(cis)(−theta))`
  `= rtext(cis)(2theta – −theta)`
  `= rtext(cis)(3theta)`

 
`=> E`

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, 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`