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Complex Numbers, EXT2 N2 2021 HSC 14c*

Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that

      `cos5theta = 16cos^5theta-20cos^3 theta + 5cos theta`.   (3 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution

`(cos theta + i sin theta)^5 = cos5theta + i sin 5theta\ \ text{(by De Moivre)}`

`text(Using binomial expansion:)`

`(cos theta + i sin theta)^5`

`= cos^5theta + 5cos^4theta · isin theta + 10cos^3theta · i^2sin^2theta + 10 cos^2theta · i^3sin^3theta`

`+ 5costheta · i^4sin^4theta + i^5sin^5theta`

`= cos^5theta-10cos^3thetasin^2theta + 5costhetasin^4theta + i\ \ text{(imaginary part)}`
 

`text(Equating real parts:)`

`cos5theta` `= cos^5theta-10cos^3thetasin^2theta + 5costhetasin^4theta`
  `= cos^5theta-10cos^3theta(1-cos^2theta) + 5costheta(1-cos^2theta)sin^2theta`
  `= cos^5theta-10cos^3theta + 10cos^5theta + (5costheta-5cos^3theta)(1-cos^2theta)`
  `= 11cos^5theta-10cos^3theta + 5costheta-5cos^3theta-5cos^3theta + 5cos^5theta`
  `= 16cos^5theta-20cos^3theta + 5costheta`

Complex Numbers, EXT2 N2 2025 SPEC1 8

Consider the function with rule  \(f(z)=z^4+6 z^2+25\), where \(z \in C\).

  1. Consider  \(z_1=1+2 i\).
  2. Plot and label \(z_1\) and \(\overline{z}_1\) on the Argand plane below.   (1 mark)  

     
  3. Given that  \(1+2 i\)  is a solution of  \(f(z)=0\), find a quadratic factor of \(f(z)\).   (2 marks)  

  4. Hence, find all remaining solutions of  \(f(z)=0\).   (2 marks)

Show Answers Only

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(z^2-2 z+5\)

c.    \(z=-1+2 i, z=-1-2 i\)

Show Worked Solution

a.    \(z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i\)
 

b.    \(\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:\)

\(\Rightarrow 1-2 i \ \ \text{is also a solution (conjugate factor theorem).}\)

\(\text {Express as a quadratic factor:}\)

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

 

c.    \(f(z)=z^4+6 z^2+25, \quad z \in C\)

\(\text{Since \(\ z^2-2 z+5\ \) is a factor:}\)

\(f(z)\) \(=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)\)
  \(=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)\)
  \(=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b\)
  \(=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b\)
Mean mark (c) 53%.

\(\text{Equating co-efficients:}\)

\(a-2=0 \ \Rightarrow \ a=2\)

\(5 b=25 \ \Rightarrow \ b=5\)

\(f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)\)
 

\(\text{Solve:} \ \ z^2+2 z+5=0\)

\(z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i\)

\(\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i\)

Complex Numbers, EXT2 N2 2025 HSC 14c

Let \(w\) be a complex number such that  \(1+w+w^2+\cdots+w^6=0\).

  1. Show that \(w\) is a 7th root of unity.   (1 mark)

The complex number  \(\alpha=w+w^2+w^4\)  is a root of the equation  \(x^2+b x+c=0\), where \(b\) and \(c\) are real and \(\alpha\) is not real.

  1. Find the other root of  \(x^2+b x+c=0\)  in terms of positive powers of \(w\).  (2 marks)

  2. Find the numerical value of \(c\).  (1 mark)

Show Answers Only

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \text{(given)}\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7=1\) \(=1\)

ii.   \(w^6+w^5+w^3\)

iii.  \(2\)

Show Worked Solution

i.    \(\text{If \(w\) is a \(7^{\text{th}}\) root of \(1 \ \Rightarrow \ w^7=1\)}\)

\(1+w+w^2+\ldots+w^6=0\ \ \text{(given,}\ w\neq 1)\)

\((1-w)\left(1+w+w^2+\cdots+w^6\right)\) \(=0\)
\(1-w^7\) \(=0\)
\(w^7\) \(=1\)
♦♦ Mean mark (i) 35%.

ii.    \(\text {Find the other root of:} \ \ x^2+b x+c=0\)

\(\text{Since \(b, c\) are real (given),}\)

\(\text{Using conjugate root theory, other root}\ =\bar{\alpha}\)

\(\bar{\alpha}\) \(=\overline{w+w^2+w^4}\)
  \(=\overline{w}+\overline{w^2}+\overline{w^4}\)
  \(=\dfrac{1}{w}+\dfrac{1}{w^2}+\dfrac{1}{w^4} \quad\left( \bar{w}=\dfrac{1}{w} \ \text{since} \ \ \abs{w}=1\right)\)
  \(=\dfrac{w^7}{w}+\dfrac{w^7}{w^2}+\dfrac{w^7}{w^4}\)
  \(=w^6+w^5+w^3\)

 

iii.    \(\text{Product of roots}=\dfrac{c}{a}=c\)

\(c\) \(=\left(w+w^2+w^4\right)\left(w^6+w^5+w^3\right)\)
  \(=w^7+w^6+w^4+w^8+w^7+w^5+w^{10}+w^9+w^7\)
  \(=1+w^6+w^4+\left(w^7 \cdot w\right)+1+w^5+\left(w^7 \cdot w^3\right)+\left(w^7 \cdot w^2\right)+1\)
  \(=2+\underbrace{1+w+w^2+w^3+w^4+w^5+w^6}_{=0}\)
  \(=2\)
♦♦ Mean mark (iii) 32%.

Complex Numbers, EXT2 N2 2024 HSC 9 MC

Consider the solutions of the equation  \(z^4=-9\).

What is the product of all of the solutions that have a positive principal argument?

  1. \(3\)
  2. \(-3\)
  3. \(3 i\)
  4. \(-3 i\)
Show Answers Only

\(B\)

Show Worked Solution

\(z^4=-9\)

\(\text{Convert}\ z^4 \ \text{to Mod/Arg form:}\)

\(\left|z^4\right|=9, \ \ \arg \left(z^4\right)=\pi \ \text{(\(-9\) is on negative real axis})\)

Mean mark 57%.

\(\text{By De Moivre:}\)

   \(\abs{z}=\sqrt[4]{9}=\sqrt{3}\)

   \(\arg (z)=\dfrac{\pi}{4}\)

\(\text{Roots are} \ \ \dfrac{\pi}{2} \ \ \text{rotations of}\ \  z=\sqrt{3} \, \text{cis}\left(\dfrac{\pi}{4}\right)\)

\(z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{\pi}{4}\right), z=\sqrt{3} \, \text{cis}\left( \pm \dfrac{3 \pi}{4}\right)\)

\(\sqrt{3}\, \text{cis}\left(\dfrac{\pi}{4}\right) \cdot \sqrt{3} \, \text{cis}\left(\dfrac{3 \pi}{4}\right)=3 \, \text{cis}(\pi)=-3\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N2 2024 HSC 4 MC

A monic polynomial, \(f(x)\), of degree 3 with real coefficients has \(3\) and  \(2+i\)  as two of its roots.

Which of the following could be \(f(x)\) ?

  1. \(f(x)=x^3-7 x^2-17 x+15\)
  2. \(f(x)=x^3-7 x^2+17 x-15\)
  3. \(f(x)=x^3+7 x^2-17 x+15\)
  4. \(f(x)=x^3+7 x^2+17 x-15\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Since coefficients are real, roots are:}\ \ 3, 2+i, 2-i\)

\[ \sum \text{roots} = 7 = \dfrac{-b}{1}\ \ \Rightarrow \ b=-7\]

\(\text{Product of roots}\ = 3(2+i)(2-i)=15=\dfrac{-d}{1}\ \ \Rightarrow \ d=-15\)

\(\Rightarrow B\)

Complex Numbers, EXT2 N2 2022 SPEC1 1

Consider the equation  `p(z)=z^2 + 6iz-25`, `z ∈ C`.

  1. Express `p(z)` in the form  `p(z) = (z+ai)^2 + b`  where  `a`, ` b  ∈ R`.   (1 mark)
  2. Hence, or otherwise, find the solutions of the equation  `p(z) = 0`.   (2 marks)
Show Answers Only
  1. `p(z) = (z + 3i)^2-16`
  2. `z =  ± 4-3i`
Show Worked Solution
a.   `p(z)` `= (z + 3i)^2-(3i)^2-25`
    `=(z + 3i)^2 +9-25`
    `=(z + 3i)^2-16`

 

b.  
`(z + 3i)^2` `= 16`
  `z + 3i` `= ± 4`
  `z ` `= ± 4-3i`

Complex Numbers, EXT2 N2 2023 HSC 12e

The complex number  \(2+i\)  is a zero of the polynomial

\(P(z)=z^4-3 z^3+c z^2+d z-30\)

where \(c\) and \(d\) are real numbers.

  1. Explain why  \(2-i\)  is also a zero of the polynomial \(P(z)\).  (1 marks)

  2. Find the remaining zeros of the polynomial \(P(z)\).  (2 marks)

Show Answers Only

i.    \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)

\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)

ii.   \(\text{Remaining zeros:}\ \ -3, 2 \)

Show Worked Solution

i.    \(\text{Since all coefficients are real and given}\ P(x)\ \text{has a complex root} \)

\((2+i), \ \text{then its conjugate pair}\ (2-i)\ \text{is also a root.} \)

 

ii.    \(P(z)=z^4-3 z^3+c z^2+d z-30\)

\(\text{Let roots be:}\ \ 2+i, 2-i, \alpha, \beta \)

\( \sum\ \text{roots:}\)

\(2+i+2-i+\alpha + \beta\) \(=-\dfrac{b}{a} \)  
\(4+\alpha+\beta\) \(=3\)  
\(\alpha + \beta\) \(=-1\ \ \ …\ (1) \)  

 
\(\text{Product of roots:} \)

\((2+i)(2-i)\alpha\beta \) \(= \dfrac{e}{a} \)  
\(5\alpha\beta\) \(=-30\)  
\(\alpha \beta \) \(=-6\ \ \ …\ (2) \)  

 
\(\text{Substitute}\ \ \beta=-\alpha-1\ \ \text{into (2):} \)

\(\alpha(-\alpha-1) \) \(=-6 \)  
\(-\alpha^2-\alpha \) \(=-6\)  
\(\alpha^2+\alpha-6\) \(=0\)  
\( (\alpha+3)(\alpha-2) \) \(=0\)  

 
\(\therefore\ \text{Remaining zeros:}\ \ -3, 2 \)

Complex Numbers, EXT2 N2 2023 HSC 11a

Solve the quadratic equation

\(z^2-3 z+4=0\)

where \(z\) is a complex number. Give your answers in Cartesian form.  (2 marks)

Show Answers Only

\(\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)

Show Worked Solution

\(z^2-3 z+4=0\)

\(z\) \(=\dfrac{3 \pm \sqrt{(-3)^2-4 \times 1 \times 4}}{2} \)  
  \(=\dfrac{3 \pm \sqrt{-7}}{2} \)  
  \(=\dfrac{3+\sqrt{7}i}{2}\ \ \text{or}\ \ \dfrac{3- \sqrt{7}i}{2} \)  

Complex Numbers, EXT2 N2 2022 HSC 13c

Consider the equation  `z^5+1=0`, where `z` is a complex number.

  1. Solve the equation  `z^5+1=0`  by finding the 5th roots of `-1`.  (2 marks)

  2. Show that if `z` is a solution of  `z^5+1=0`  and  `z !=-1`, then  `u=z+(1)/(z)`  is a solution of  `u^2-u-1=0`.  (2 marks)

  3. Hence find the exact value of  `cos\ (3pi)/(5)`.  (3 marks)

Show Answers Only
  1. `z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  2. `text{Proof (See Worked Solutions)}`
  3. `(1-sqrt5)/4`
Show Worked Solution

i.   `z^5+1=0\ \ =>\ \ z^5=-1`

`z=e^(i((2k+1)/5)),\ \ kin{0,1,-1,2,-2}`

`:.z=e^(i(pi)/5), e^(i(3pi)/5), e^(-i(pi)/5), -1, e^(-i(3pi)/5)`
  

ii.   `z^5+1=(z+1)(z^4-z^3+z^2-z+1)`

`text{Given}\ \ z!=-1,`

`z^4-z^3+z^2-z+1=0`
 

`text{Divide by}\ z^2\ \ (z!=0)`

`z^2-z+1-1/z+1/z^2` `=0`  
`z^2+1/z^2-(z+1/z)+1` `=0`  
`z^2+2+1/z^2-(z+1/z)-1` `=0`  
`(z+1/z)^2-(z-1/z)-1` `=0`  

 
`text{Let}\ \ u=z+1/z:`

`:.u^2-u-1=0`
 


Mean mark (ii) 53%.

iii.  `u^2-u-1=0`

`text{By quadratic formula:}`

`u` `=(1+-sqrt(1-4xx1xx(-1)))/(2)`  
  `=(1+-sqrt5)/2`  

♦ Mean mark (iii) 43%.
`z+1/z` `=(1+-sqrt5)/2`  
`e^(i(3pi)/5)+e^(-i(3pi)/5)` `=(1-sqrt5)/2,\ \ (cos\ (3pi)/5 <0)`  
`2cos((3pi)/5)` `=(1-sqrt5)/2`  
`cos((3pi)/5)` `=(1-sqrt5)/4`  

Complex Numbers, EXT2 N2 2022 HSC 6 MC

It is known that a particular complex number `z` is NOT a real number.

Which of the following could be true for this number `z` ?

  1. `bar(z)=iz`
  2. `bar(z)=|z^(2)|`
  3. `text{Re}(iz)= text{Im}(z)`
  4. `text{Arg}(z^(3))= text{Arg}(z)`
Show Answers Only

`A`

Show Worked Solution

`z in CC, z !in RR`

`text{Let}\ \ z=a+ib\ \ =>\ \ barz=a-ib`

`iz=i(a+ib)=-b+ia`

`text{If}\ \ barz=iz,`

`a-ib=-b+ia`

`a=-b\ \ text{(satisfies)}`

`:.∃ a,b (b!=0)\ \ text{such that}\ \ barz=iz`

`=>A`


♦♦ Mean mark 31%.

Complex Numbers, EXT2 N2 SM-Bank 10

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. State the relationship between `z_2` and `z_3`.  (1 mark)

  2. 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)

Show Answers Only
  1. `z_2 = barz_3`
  2. `alpha = -3, beta = 9, gamma = -27`
Show Worked Solution

i.   `text(By conjugate root theory)`

`z_2 = barz_3`

 

ii.   `text(Let)\ \ z_1 = a + bi, \ z_2 = a – bi`

`|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`

Complex Numbers, EXT2 N2 2021 SPEC1 8

  1. Solve  `z^2 + 2z + 2 = 0`  for `z`, where  `z ∈ C`.  (1 mark)

  2. Solve  `z^2 + 2barz + 2 = 0`  for `z`, where  `z ∈ C`.  (3 marks)

Show Answers Only
  1. `z = -1 – i\ \ text(or)\ \ -1 + i`
  2. `z = 1 ± sqrt5 i`
Show Worked Solution
a.    `z^2 + 2z + 2` `= 0`
  `z^2 + 2z + 1 + 1` `= 0`
  `(z + 1)^2 + 1` `= 0`
  `(z + 1)^2 – i^2` `= 0`
  `(z + 1 + i)(z + 1 – i)` `= 0`

 
`:. z = -1 – i\ \ \ text(or)\ \ -1 + i`

 

b.   `z = x + yi \ => \ barz = x – yi`

`z^2 + 2barz + 2` `= 0`
`(x + yi)^2 + 2(x – yi) + 2` `= 0`
`x^2 + 2xyi – y^2 + 2x – 2yi + 2` `= 0`
`x^2 – y^2 + 2x + 2 + (2xy – 2y)i` `= 0`

 

`text(Find)\ \ x, y\ text(such that)`

`x^2 – y^2 + 2x + 2` `= 0\ …\ (1)`
`2xy – 2y` `= 0\ …\ (2)`

 
`text(When)\ \ 2xy – 2y = 0`

`2y(x – 1)` `= 0`
`x` `= 1`

 
`text(Substitute)\ \ x = 1\ \ text{into (1)}`

`1 – y^2 + 2 + 2` `= 0`
`y^2` `= 5`
`y` `= ±sqrt5`

 
`:. z = 1 ± sqrt5 i`

Complex Numbers, EXT2 N2 2021 HSC 11d

  1. Find the two square roots of `−i` , giving the answers in the form  `x + iy`, where `x` and `y` are real numbers.   (2 marks)

  2. Hence, or otherwise, solve  `z^2 + 2z + 1 + i = 0`  giving your solutions in the form  `a + i b`  where `a` and `b` are real numbers.   (2 marks)

Show Answers Only

i.    `z = 1/sqrt2-1/sqrt2 i \ \ text{or} \ \ z =-1/sqrt2+ 1/sqrt2 i`

ii.   `z = -1 + 1/sqrt2-1/sqrt2 i \ \ text{or} \ \ z = -1-1/sqrt2+ 1/sqrt2 i`

Show Worked Solution

i.     `text{Solution 1}`

`z = x + iy \ , \ z^2 = -i`

`z^2 = x^2-y^2 + 2x yi = -i`

`x^2-y^2 = 0 \ …\ (1)`

`2xy = -1 \ …\ (2)`

`x= ± y \ …\ (1)′`
 

`text{Substitute} \ \ x = y \ \ text{into} \ (2):`

`2x^2 = -1 \ -> \ text{no real solutions}`

`text{Substitute} \ \ x= -y \ \ text{into} \ (2):`

`-2x^2` `= -1`  
`x` `= ± 1/sqrt2`  

 
`:. \ z = 1/sqrt2-1/sqrt2 i \ \ text{or} \ \ z =-1/sqrt2+ 1/sqrt2 i`
 

`text{Solution 2}`

`z = re^{i theta} \ , \ z^2 = -i`

`r^2 e^{i2 theta} = e^{i {3pi}/{2}} \ \ text{or} \ \ e^{-i pi/2}`

`=> \ r = 1 \ , \  theta = {3pi}/{4} \ \ text{or} \ \-pi/4`
 

`z= text{cos} {3pi}/{4} + i \ text{sin} {3pi}/{4} =-1/sqrt2 + 1/sqrt2 i`

`z= text{cos} (-pi/4) + i \ text{sin} (-pi/4) = 1/sqrt2-1/sqrt2 i`

 

ii.    `z^2 + 2z + 1 + i = 0`

`z` `= {2 ± sqrt{4-4 * 1 * (1 + i)}}/{2}`  
  `= {-2 ± sqrt{4-4-4 \ i}}/{2}`  
  `= -1 ± sqrt(-i)`  

 
`:. \ z = -1 + 1/sqrt2-1/sqrt2 i \ \ text{or} \ \ z = -1-1/sqrt2 + 1/sqrt2 i`

Complex Numbers, EXT2 N2 2021 HSC 6 MC

Which polynomial could have  `2 + i`  as a zero, given that `k` is a real number?

  1. `x^3 − 4 x^2 + k x`
  2. `x^3 − 4 x^2 + k x + 5`
  3. `x^3 − 5 x^2 + k x`
  4. `x^3 − 5 x^2 + k x + 5`
Show Answers Only

`A`

Show Worked Solution

`text{Roots:} \ 2 + i \ , \ 2- i \ text{(conjugative roots),} \ alpha\ text{(real)}`

`-b` `= 2 + i + 2 – i + 2 = 4 + alpha`  
`k` `= (2 + i)(2 – i) + alpha(2 + i) + alpha (2 – i)`  
  `= 5 + 4 alpha`  
`-d` `= (2 + i)(2 – 1) α = 5 alpha`  

 

`text{Test coefficients for each option}`

`A: \ 4 + alpha = 4 \ => \ alpha = 0 \ , \ d= 0 \ text{(correct)}` 

`B: \ 4 + alpha = 4 \ => \ alpha = 0 \ , \ d= 5 ≠ 0`

`C: \ 4 + alpha = 5 \ => \ alpha = 1 \ , \ d= 0 ≠ -5`

`D: \ 4 + alpha = 5 \ => \ alpha = 1 \ , \ d= 5 ≠ -5`
 

`=>\ A`

Complex Numbers, EXT2 N2 2020 SPEC1 3

Find the cube roots of  `1/sqrt 2 - 1/sqrt 2 i`. Express your answers in modulus-argument form.  (3 marks)

Show Answers Only

`z_1 = text(cis)((7 pi)/12)`

`z_2 = text(cis)(-pi/12)`

`z_3 = text(cis)((-3 pi)/4)`

Show Worked Solution

`z^3 = 1/sqrt 2 – 1/sqrt 2 i = text(cis)(-pi/4) = text(cis)(-pi/4 + 2 pi k), k ∈ Z`
 

`text(By De Moivre):`

`z = text(cis)(-pi/12 + (2 pi k)/3)`

`text(If)\ \ k = 1, \ z_1 = text(cis)(-pi/12 + (2 pi)/3) = text(cis)((7 pi)/12)`

`text(If)\ \ k = 0, \ z_2 = text(cis)(-pi/12)`

`text(If)\ \ k = – 1, \ z_3= text(cis)(-pi/12 – (2pi)/3) = text(cis)((-3 pi)/4)`

Complex Numbers, EXT2 N2 2020 HSC 11e

Solve  `z^2 + 3 z + (3-i) = 0`, giving your answer(s) in the form  `a + bi`, where  `a`  and  `b`  are real.  (4 marks)

Show Answers Only

`z= -1 + i \ \ text{or} \ \ -2-i`

Show Worked Solution

`z^2 + 3z + (3-i) = 0`

`z` `= frac{-3 ± sqrt(9-4 · 1 (3-i))}{2}`
  `= frac{-3 ± sqrt(4i-3)}{2} `

 
`text{Consider} \ \ Delta = sqrt(4i-3) :`

`x + i y` `= sqrt(4i-3)`
`(x + iy)^2` `= 4i-3`
`x^2-y^2 + 2xyi` `= 4i-3`

 
`text{Equating real and imaginary parts:}`

`2 xy` `= 4`
`xy` `= 2\ …\ (1)`
`x^2-y^2` `= -3\ …\ (2)`

 
`=> x = 1 \ , \ y =2`

`=> \ x + iy = 1 + 2 i`
 

`therefore z` `= frac{-3 ± (1 + 2i)}{2}`
`z ` `= -1 + i \ \ text{or} \ \ -2-i`

Complex Numbers, EXT2 N2 2020 HSC 2 MC

Given that  `z = 3 + i`  is a root of  `z^2 + pz + q = 0`, where `p` and `q` are real, what are the values of `p` and `q`?

  1.  `p = -6 \ , \ q = sqrt(10)`
  2.  `p = -6 \ , \ q = 10`
  3.  `p = 6 \ , \ q = sqrt(10)`
  4.  `p = 6 \ , \ q = 10`
Show Answers Only

`B`

Show Worked Solution

`text{S}text{ince} \ \ z_1 = 3 + i \ \ text{is a root}`

`=> \ z_2 = 3 – i \ \ text{is a root}`
 

`z_1 + z_2` `= frac{-b}{a}`
`(3 + i) + (3 – i)` `= -p`
`therefore \ p` `= -6`

 

`z_1 z_2` `= frac{c}{a}`
`(3 + i) (3 – i)` `= q`
`3^2 – i^2` `= q`
`therefore \ q` `= 10`

 
`=> \ B`

Complex Numbers, EXT2 N1 SM-Bank 4

  1. Express `z` in the form  `a + bi`, given  `z = sqrt(−3 - 4i)`  (2 marks)

  2. Hence, using the quadratic formula, solve
     
         `z^2 - 7z + 13 + i = 0`  (1 mark)

Show Answers Only
  1. `z_1 = −1 + 2i`
  2. `z = 3 + i\ \ text(or)\ \ 4 – i`
Show Worked Solution
i.    `z` `= sqrt(−3 – 4i)`
  `z^2` `= −3 – 4i`
`−3 – 4i` `= (x + iy)^2`
  `= x^2 – y^2 + 2xyi`
   
`x^2 – y^2` `= −3\ …\ \ (1)`
`2xy` `= −4`
`xy` `= −2\ …\ \ (2)`

 
`x=-1,\ \ y=2`

`x=1,\ \ y=-2`
 

`:. z_1` `= −1 + 2i`
`z_2` `= 1 – 2i`


ii.   `z^2 – 7z + 13 + i = 0`

 
`text(Using general formula:)`

`z` `= (−b ± sqrt(b^2 – 4ac))/(2a)`
  `= (7 ± sqrt(49 – 4 · 1(13 + i)))/2`
  `= (7 ± sqrt(−3 – 4i))/2`

 
`text(Using)\ \ z_1 = −1 + 2i,`

COMMENT: Since  `z_1 = – z_2`, the general formula only produces 2 distinct solutions.

`z = (7 + (−1 + 2i))/2 = 3+i`
 

`text(Using)\ \ z_2 = 1 – 2i,`

`z = (7 + (1 – 2i))/2 = 4 – i`
 

`:. z = 3 + i\ \ text(or)\ \ z=4 – i`

Complex Numbers, EXT2 N2 2019 SPEC1-N 1

A cubic polynomial has the form  `p(z) = z^3 + bz^2 + cz + d, \ z ∈ C`, where  `b, c, d ∈ R`.

Given that a solution of  `p(z) = 0`  is  `z_1 = 3 - 2i`  and that  `p(–2) = 0`, find the values of  `b, c` and `d`.   (4 marks)

Show Answers Only

`b =-4 \ , \ c = 1 \ , \ d = 26`

Show Worked Solution

`text(Roots:)\  \ z_1 = 3 – 2i \ , \ z_2 = 3 + 2i \ , \ z_3 = -2`
 

`p(z)` `= (z – 3 + 2i )(z – 3 – 2i )(z + 2)`
  `= ((z-3)^2 – (2i)^2)(z+2)`
  `= (z^2 – 6z + 9 + 4)(z + 2)`
  `= (z^2 – 6z + 13)(z + 2)`
  `= z^3 + 2z^2 – 6z^2 – 12z + 13z + 26`
  `= z^3 – 4z^2 + z + 26`

 

`:. \ b =-4 \ , \ c = 1 \ , \ d = 26`

Complex Numbers, EXT2 N2 2019 HSC 16b

Let  `P(z) = z^4 - 2kz^3 + 2k^2z^2 + mz + 1`, where  `k`  and  `m`  are real numbers.

The roots of  `P(z)`  are  `alpha, bar alpha, beta, bar beta`.

It is given that  `|\ alpha\ | = 1`  and  `|\ beta\ | = 1`.

  1. Show that  `(text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`.  (3 marks)

  2. The diagram shows the position of  `alpha`.
     


 

On the diagram, accurately show all possible positions of  `beta`.  (2 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.    `P(z) = z^4 – 2kz^3 + 2k^2z^2 + mz + 1,\ \ k, m in RR`

`text(Roots):\ \ alpha, bar alpha, beta, bar beta and |\ alpha\ | = 1, |\ beta\ | = 1`

`text(Show)\ \ (text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`

♦♦ Mean mark part (i) 26%.

`alpha + bar alpha + beta + bar beta` `= 2k`
`2 text{Re} (alpha) + 2 text{Re} (beta)` `= 2k`
`text{Re} (alpha) + text{Re} (beta)` `= k`

 

`alpha bar alpha + alpha beta + alpha bar beta + bar alpha beta + bar alpha bar beta + beta bar beta` `= 2k^2`
`|\ alpha\ |^2 + alpha(beta + bar beta) + bar alpha(beta + bar beta) + |\ beta\ |^2` `= 2k^2`
`1 + (alpha + bar alpha)(beta + bar beta) + 1` `= 2k^2`
`2 + 2 text{Re} (alpha) ⋅ 2 text{Re} (beta)` `= 2 (text{Re} (alpha) + text{Re} (beta))^2`
`2 + 4 text{Re} (alpha) text{Re} (beta)` `= 2 text{Re} (alpha)^2 + 4 text{Re} (alpha) text{Re} (beta) + 2 text{Re} (beta)^2`
`2` `= 2(text{Re} (alpha)^2 + text{Re} (beta)^2)`
`:. 1` `= text{Re} (alpha)^2 + text{Re} (beta)^2`

 

ii.    `|\ alpha\ | = |\ beta\ |\ \ \ text{(given)}`
  `text{Re}(alpha)^2 + text{Re}(beta)^2 = 1\ \ \ text{(see part (i))}`
  `text{Re}(alpha)^2 + text{Im}(alpha)^2 = 1\ \ \ (|\ alpha\ | = 1)`
  `=> text{Re}(beta)^2 = text{Im} (alpha)^2`
  `\ \ \ \ \ \ text{Re}(beta) = +-text{Im}(alpha)`

 

♦♦♦ Mean mark part (ii) 10%.

Complex Numbers, EXT2 N2 2018 HSC 15b

  1. Use De Moivre's theorem and the expansion of `(costheta + isintheta)^8` to show that
     
    `sin8theta = ((8),(1)) cos^7thetasintheta - ((8),(3)) cos^5thetasin^3theta`
     
                        `+ ((8),(5)) cos^3thetasin^5theta - ((8),(7)) costhetasin^7theta`  (2 marks)

  2. Hence, show that
     
    `(sin8theta)/(sin2theta) = 4(1 - 10sin^2theta + 24sin^4theta - 16sin^6theta)`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(By De Moivre)`

`costheta + isintheta^8 = cos8theta + isin8theta\ \ …\ (text{*})`
 

`text(Using Binomial Expansion)`

`(costheta + isintheta)^8`

`= cos^8theta + ((8),(1))cos^7theta * isintheta + ((8),(2)) cos^6theta *i^2sin^2theta`

`+ ((8),(3)) cos^5theta *i^3sin^3theta + ((8),(4)) cos^4theta *i^4sin^4theta + ((8),(5)) cos^3theta *i^5sin^5theta`

`+ ((8),(6)) cos^2theta *i^6sin^6theta + ((8),(7)) costheta *i^7sin^7theta + i^8sin^8theta`

 
`text(Equating imaginary parts of the expansion equation (*)):`

`isin8theta = ((8),(1)) cos^7theta* isintheta + ((8),(3)) icos^5theta* i^3sin^3theta`

`+ ((8),(5)) cos^3theta* i ^5sintheta + ((8),(7)) costheta *i^7sin^7theta`

`:. sin8theta = ((8),(1)) cos^7theta sintheta – ((8),(3)) cos^5theta sin^3theta`

`+ ((8),(5)) cos^3theta sin^5theta – ((8),(7)) costheta sin^7theta`
 

ii.    `sin8theta` `= 8cos^7theta sintheta – 56cos^5 sin^3theta + 56cos^3theta sin^5theta – 8costheta sin^7theta`
    `= 2sinthetacostheta (4cos^6theta – 28cos^4theta sin^2theta + 28cos^2theta sin^4theta – 4sin^6theta)`

 
`:. (sin8theta)/(sin2theta)`

`= 4cos^6theta – 28cos^4theta sin^2theta + 28cos^2theta sin^4theta – 4sin^6theta`

`= 4(1 – sin^2theta)^3 – 28(1 – sin^2theta)^2 sin^2theta + 28(1 – sin^2theta) sin^4theta – 4sin^6theta`

`= 4(1 – 3sin^2theta + 3sin^4theta + sin^6theta) – 28sin^2theta (1 – 2sin^2theta + sin^4theta)`

`+ 28sin^4theta (1 – sin^2theta) – 4sin^6theta`

`= 4 – 40sin^2theta + 96sin^4theta – 56sin^6theta`

`= 4(1 – 10sin^2theta + 24sin^4theta – 16sin^6theta)`

Complex Numbers, EXT2 N2 2018 HSC 6 MC

Which complex number is a 6th root of `i`?

  1. `−1/sqrt2 + 1/sqrt2i`
  2. `−1/sqrt2 - 1/sqrt2i`
  3. `−sqrt2 + sqrt2i`
  4. `−sqrt2 - sqrt2i`
Show Answers Only

`A`

Show Worked Solution

`text(Consider option A:)`
 

`|−1/sqrt2 + 1/sqrt2i|= sqrt((−1/sqrt2)^2 + (1/sqrt2)^2) = 1`

`text(arg)(z)` `= (3pi)/4`
`z` `= 1(cos\ (3pi)/4 + i sin\ (3pi)/4)`
`z^6` `= cos\ (18pi)/4 + i sin\ (18pi)/4\ \ \ text{(De Moivre)}`
  `= i`

 
`=> A`

Complex Numbers, EXT2 N2 2017 HSC 6 MC

It is given that  `z = 2 + i`  is a root of  `z^3 + az^2 - 7z + 15 = 0`, where `a` is a real number.

What is the value of `a`?

  1. `−1`
  2. `1`
  3. `7`
  4. `−7`
Show Answers Only

`A`

Show Worked Solution

`text(S)text(ince)\ \ z_1 = 2 + i\ \ text(is a root.)`

`=> z_2 = 2 – i\ \ text(is a root.)`

 

`text(Roots are)\ \ z_1, z_2, alpha.`

`text(Using product of roots:)`

`z_1z_2alpha` `= −15`
`(2 + i)(2 – i)alpha` `= −15`
`5alpha` `= −15`
`a` `= −3`

 

`text(Using sum of roots:)`

`z_1 + z_2 + alpha` `=-a`
`2 + i + 2 – i + −3` `= −a`
`a` `= −1`

`=> A`

Complex Numbers, EXT2 N2 2017 HSC 12b

Solve the quadratic equation  `z^2 + (2 + 3i)z + (1 + 3i) = 0`, giving your answers in the form  `a + bi`, where `a` and `b` are real numbers.  (3 marks)

Show Answers Only

`-1 or -1 – 3i`

Show Worked Solution

`z^2 + (2 + 3i)z + (1 + 3i) = 0`

 

`z`

 `= (-(2 + 3i) +- sqrt((2 + 3i)^2 – 4 · 1(1 + 3i)))/2`
  `= (-(2 + 3i) +- sqrt (4 + 12i + 9i^2 – 4 – 12i))/2`
  `= (-(2 + 3i) +- sqrt(-9))/2`
  `= ((-2 – 3i) +- 3i)/2`
   
`:. z` `= (-2)/2 or (-2 – 6i)/2`
  `= -1 or -1 – 3i`

Complex Numbers, EXT2 N2 2017 HSC 1 MC

The complex number `z` is chosen so that  `1, z, …, z^7`  form the vertices of the regular polygon shown.

Which polynomial equation has all of these complex numbers as roots?

  1. `x^7 - 1 = 0`
  2. `x^7 + 1 = 0`
  3. `x^8 - 1 = 0`
  4. `x^8 + 1 = 0`
Show Answers Only

`C`

Show Worked Solution

`P(x)\ \ text(has 8 separate roots.)`

`:.\ text(Must be of degree at least 8.)`

`text(S) text(ince 1 is also a root,)`

`=>  C`

Complex Numbers, EXT2 N2 2016 HSC 12c

Let  `z = cos theta + i sin theta.`

  1. By considering the real part of  `z^4`, show that  `cos 4 theta`  is
     
    `qquad cos^4 theta - 6 cos^2 theta sin^2 theta + sin^4 theta.`  (2 marks)

     

  2. Hence, or otherwise, find an expression for  `cos 4 theta`  involving only powers of `cos theta.`  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `8cos^4theta – 8cos^2theta + 1`
Show Worked Solution

i.   `z = costheta + isintheta`

`z^4` `= (costheta + isintheta)^4`
 

`= cos^4theta + 4cos^3theta*(isintheta) + 6cos^2theta*(isintheta)^2 +`

`4costheta*(isintheta)^3 + (isintheta)^4`
 

`= cos^4theta + 4icos^3thetasintheta – 6cos^2thetasin^2theta -`

`4icosthetasin^3theta + sin^4theta`

 

`z^4 = cos4theta + isin4theta\ \ text{(by De Moivre)}`
 

`text(Equating real parts:)`

`cos4theta = cos^4theta – 6cos^2thetasin^2theta + sin^4theta`

`…\ text(as required)`

 

ii.    `cos4theta` `= cos^4theta – 6cos^2theta(1 – cos^2theta) + (1 – cos^2theta)^2`
    `= cos^4theta – 6cos^2theta + 6cos^4theta + 1 – 2cos^2theta + cos^4theta`
    `= 8cos^4theta – 8cos^2theta + 1`

Complex Numbers, EXT2 N2 2016 HSC 10 MC

Suppose that  `x + 1/x = -1.`

What is the value of  `x^2016 + 1/x^2016?`

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

`=> B`

Show Worked Solution

`x + 1/x = −1`

♦ Mean mark 43%.

`x^2 + x + 1 = 0`

`:. x` `= (−1 ± sqrt(1 – 4))/2`
  `= −1/2 ± sqrt3/2 i`

 

`|\ x\ | = sqrt((−1/2)^2 + (sqrt3/2)^2) = 1`

`:. x` `= cos\ (2pi)/3 + isin\ (2pi)/3`
  `= text(cis)(2pi)/3`

`text(or)`

`x` `= cos\ (2pi)/3 – isin\ (2pi)/3`
  `= text(cis)(−(2pi)/3)`

 

`x^2016` `= text(cis)(±1344pi)quad(text(De Moivre))`
  `= text(cis)0`
  `= 1`

 

`:. x^2016 + 1/(x^2016) = 2`

`=> B`

Complex Numbers, EXT2 N2 2006 HSC 2c

Find, in modulus-argument form, all solutions of  `z^3 = -1.`   (2 marks)

Show Answers Only

`text(cis)\ pi/3,\ \ \ -1,\ \ \ text(cis)\ -pi/3`

Show Worked Solution
`z^3` `=-1`
`-1` `=\ text(cis)\ (pi+2k pi)`
`:.z` `=\ text(cis)\ (pi+2k pi)/3\ \ \ text{(De Moivre)}`

 

MARKER’S COMMENT: Another successful solution strategy was graphical, using the unit circle and three equal angles generated from pi.

`text(When)\ k=0`

`z=\ text(cis)\ pi/3`

`text(When)\ k=1`

`z=\ text(cis)\ pi = -1`

`text(When)\ k=-1`

`z=\ text(cis)\ -pi/3`
 

`:.\ text(The 3 solutions to)\ \ z^3=-1\ \ text(are)`

`z=\ text(cis)\ pi/3,\ \ \ -1,\ \ \ text(cis)\ -pi/3`

Complex Numbers, EXT2 N2 2007 HSC 8b

  1. Let  `n`  be a positive integer. Show that if  `z^2 != 1`  then
     
        `1 + z^2 + z^4 + … + z^(2n - 2) = ((z^n - z^-n)/(z - z^-1)) z^(n - 1)`.  (2 marks)

  2. By substituting  `z = cos theta + i sin theta`  where  `sin theta != 0`, into part (i), show that
     
        `1 + cos 2 theta + … + cos (2n - 2) theta + i[sin 2 theta + … + sin (2n - 2) theta]`
     
            `= (sin n theta)/(sin theta) [cos (n - 1) theta + i sin (n - 1) theta].`   (3 marks)

  3. Suppose  `theta = pi/(2n)`.  Using part (ii), show that
     
        `sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n - 1) pi)/n = cot\ pi/(2n).`   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `1 + z^2 + z^4 + … + z^(2n – 2),\ z^2 != 1`

`text(GP where)\ a = 1,\ \ r = z^2,\ \ n\ text(terms):`

`S_n` `=(1((z^2)^n – 1))/(z^2 – 1)`
  `=(z^(2n) – 1)/(z^2 – 1)`
  `=((z^n – z^-n))/(z – z^-1) xx z^n/z`
  `=((z^n – z^-n)/(z – z^-1))z^(n – 1)`

 

ii.    `z` `= cos theta + i sin theta`
  `z^n` `= cos n theta + i sin n theta\ \ …\ text(etc)\ \ \ \ text{(De Moivre)}` 
  `z^-n` `= cos( -n theta) + i sin (-n theta)`
    `= cos n theta – i sin n theta`

 

`text(LHS)` `= 1 + (cos 2 theta + i sin 2 theta) + (cos 4 theta + i sin 4 theta) + `
  `… + (cos(2n – 2) theta + i sin (2n – 2) theta)`
  `= 1 + cos 2 theta + cos 4 theta + … + cos (2n – 2) theta + `
  `i (sin 2 theta + sin 4 theta + … + sin (2n – 2) theta)`
   

`text{Using part (i):}`

`text(LHS)` `=((cos n theta + i sin n theta – cos n theta + i sin n theta))/(cos theta + i sin theta – cos theta + i sin theta) xx`
  `[cos (n – 1) theta + i sin (n – 1) theta]`
  `=(2 i sin n theta)/(2 i sin theta) [cos (n – 1) theta + i sin (n – 1) theta]`
  `=(sin n theta)/(sin theta) [cos (n – 1) theta + i sin (n – 1) theta]\ \ text(… as required.)`

 

iii.  `text{Equating the imaginary parts in part (ii):}`

`sin 2 theta + sin 4 theta + … + sin 2 (n – 1) theta = (sin n theta sin (n – 1) theta)/(sin theta)`

`text(When)\ \ theta = pi/(2n),`

`sin\ (2 pi)/(2n) + sin\ (4 pi)/(2n) + … + sin\ (2(n – 1) pi)/(2 n) = (sin\ (n pi)/(2n) sin\ ((n – 1) pi)/(2n))/(sin\ pi/(2n))`

`:. sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n – 1) pi)/n`

`=(sin\ pi/2)/(sin\ pi/(2n)) xx sin\ ((n – 1) pi)/(2n)`

`=1/(sin\ pi/(2n)) sin (pi/2 – pi/(2n))`

`=(cos\ pi/(2n))/(sin\ pi/(2n))`

`=cot\ pi/(2n)`

Complex Numbers, EXT2 N2 2007 HSC 2d

The points  `P,Q`  and  `R`  on the Argand diagram represent the complex numbers  `z_1, z_2`  and  `a`  respectively.

The triangles  `OPR`  and  `OQR`  are equilateral with unit sides, so  `|\ z_1\ | = |\ z_2\ | = |\ a\ | = 1.`

Let  `omega = cos­ pi/3 + i sin­ pi/3.`

  1. Explain why  `z_2 = omega a.`   (1 mark)

  2. Show that  `z_1 z_2 = a^2.`   (1 mark)

  3. Show that  `z_1` and `z_2`  are the roots of  `z^2 - az + a^2 = 0.`   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(S) text(ince)\ \ Delta ORQ\ \ text(is equilateral, each angle is)\ \ pi/3\ \ text(radians.)`

`text(From)\ \ R(a):`

`Q(z_2)\ \ text(is an anticlockwise rotation through)\ \ pi/3.`

`:. z_2 = a(cos\ pi/3+i sin\ pi/3)=omega a.`

 

ii.  `text(Solution 1)`

`text(Similarly,)\ \ a` `=z_1 omega`
`z_1` `=a/omega`
`:z_1z_2` `=a/omega xx omega a`
  `=a^2`

 

`text(Solution 2)`

`P(z_1)\ \ text(is a clockwise rotation of)\ \ R(a)\ \ text(through)\ \ pi/3.`

`:.z_1` `= bar omega a.`
`:. z_1 z_2` `= bar omega a xx omega a`
  `=a^2(cos­ pi/3 – i sin­ pi/3) xx (cos­ pi/3 + i sin­ pi/3)`
  `=a^2(cos^2­ pi/3 + sin^2­ pi/3)`
  `= a^2`

 

iii.  `z^2-az + a^2 = 0`

`text(Let the roots be)\ \  alpha and beta.`

`alpha + beta` `=-b/a=a`
`alpha beta` `=c/a=a^2`

 

`z_1 z_2` `= a^2\ \ \ \ \ text{(part (ii))}`
`z_1 + z_2` `=bar omega a + omega a`
  `=(cos­ pi/3 + i sin­ pi/3 + cos­ pi/3-i sin­ pi/3) a`
  `=2 cos ­ pi/3 xx a`
  `=2 xx 1/2 xx a`
  `=a`

 

`:.\ z_1 and z_2\ \ text(are the roots of)\ \  z^2-az + a^2 = 0.`

Complex Numbers, EXT2 N2 2009 HSC 7b

Let  `z = cos theta + i sin theta.`

  1. Show that  `z^n + z^-n = 2 cos n theta`, where  `n`  is a positive integer.  (2 marks)

  2. Let  `m`  be a positive integer. Show that
     
    `(2 cos theta)^(2m) = 2 [cos 2 m theta + ((2m), (1)) cos (2m - 2) theta + ((2m), (2)) cos (2m - 4) theta`
     
        `+ … + ((2m), (m - 1)) cos 2 theta] + ((2m), (m)).`  (3 marks)

  3. Hence, or otherwise, prove that
     
        `int_0^(pi/2) cos^(2m) theta\ d theta = pi/(2^(2m + 1)) ((2m), (m))`
     
    where  `m`  is a positive integer.  (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.    `z` `= cos theta + i sin theta`
  `z^n` `= cos n theta + i sin n theta\ \ \ \ text{(De Moivre)}`
  `z^-n` `= cos (-n theta) + i sin (-n theta)\ \ \ \ text{(De Moivre)}`
    `= cos n theta – i sin n theta`
  `z^n + z^-n` `= cos n theta + i sin n theta + cos n theta – i sin n theta`
    `= 2 cos n theta,\ \ \ \ n > 0`

 

 

ii.  `z + z^-1 = 2 cos theta`

`:.(2 cos theta)^(2m)`

`=(z + z^-1)^(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 1) z^-1 + ((2m), (2)) z^(2m – 2) z^-2+`

` … + ((2m), (2m – 1)) z^1 z^-(2m – 1) + z^-(2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4)+`

` … + ((2m), (2m – 1)) z^-(2m – 2) + z^(-2m)`

`=z^(2m) + ((2m), (1)) z^(2m – 2) + ((2m), (2)) z^(2m – 4) + … + ((2m), (m)) z^(2m-2m) …`

`+ ((2m), (2)) z^-(2m – 4) + ((2m), (1)) z^-(2m – 2) + z^(-2m)`

`=(z^(2m) + z^(-2m)) + ((2m), (1)) (z^(2m – 2) + z^-(2m – 2)) + ((2m), (2))`

`(z^(2m – 4) + z^-(2m – 4)) + … + ((2m), (m – 1)) (z + z^-1) + ((2m), (m))`

`=2 [cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2)) cos (2m – 4) theta`

`+ … + ((2m), (m – 1)) cos 2 theta] + ((2m), (m))`

 

iii.  `int_0^(pi/2) cos^(2m) d theta`

`=1/(2^(2m))  int_0^(pi/2) (2 cos theta)^(2m)`

`=1/(2^(2m)) int_0^(pi/2)[2(cos 2 m theta + ((2m), (1)) cos (2m – 2) theta + ((2m), (2))`

`cos (2m – 4) theta + … + ((2m), (m – 1)) cos 2 theta) + ((2m), (m))] d theta`

`=1/(2^(2m)) [2((sin 2 m theta)/(2m) + ((2m), (1)) (sin (2m – 2) theta)/(2m – 2)`

`+ … + ((2m), (m – 1)) (sin 2 theta)/2) + ((2m), (m)) theta]_0^(pi/2)`

`=1/(2^(2m)) [2(0 + 0 + … + 0) + ((2m), (m)) pi/2 – (0)]`

`=pi/(2^(2m + 1)) ((2m), (m))`

Complex Numbers, EXT2 N2 2009 HSC 2f

  1. Find the square roots of  `3 +4i.`  (3 marks)

  2. Hence, or otherwise, solve the equation

     

        `z^2 + iz - 1 - i = 0.`  (2 marks)

Show Answers Only
  1. `+-(2 + i)`
  2. `1 or -1-i`
Show Worked Solution
i.    `z` `=sqrt(3+4i)`
  `z^2` `=3+4i`
  `(x + iy)^2` `= 3 + 4i`
  `x^2  – y^2+ 2xyi` `= 3 + 4i`
`=>x^2 – y^2 = 3,\ \ \ xy = 2`

 
`text(By inspection,)`

`text(If)\ \ x=2,\ \ \ y=1`

`text(If)\ \ x=-2,\ \ \ y=-1` 

`:.z= 2 + i,\ \ text(or)\ \ -(2+i)`

 

ii.   `z^2 + iz – 1 – i = 0`

`:.z =` `(-i +- sqrt (-1 + 4 (1 + i)))/2`
`=` `\ \ (-i +- sqrt(3 + 4i))/2`
`=` `\ \ (-i +- (2+i))/2`
`=` `\ \ 1\ \ \ text(or)\ \ \ -1-i`

Complex Numbers, EXT2 N2 2009 HSC 2e

  1. Find all the 5th roots of  `–1`  in modulus-argument form.   (2 marks)

  2. Sketch the 5th roots of  `–1`  on an Argand diagram.  (1 mark)

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  1. `z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis) (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,``z_4 = text(cis) (7 pi)/5,\ \ z_5 = text(cis) (9 pi)/5`
  2.  
Show Worked Solution
i.   `z` `=cos theta+i sin theta`
  `z^5` `=cos\ 5 theta+i sin\ 5 theta=-1,\ \ \ \ text{(De Moivre)}`
  `:. cos\ 5 theta` `=-1`
  `5 theta` `=pi,\ 3pi,\ 5pi,\ 7pi,\ 9pi`
  `theta` `=pi/5,\ (3pi)/5,\ pi,\ (7pi)/5,\ (9pi)/5`

 

`:.\ text(The roots are)`

`z_1 = text(cis)\ pi/5,\ \ \ z_2 = text(cis)\ (3 pi)/5,\ \ \ z_3 = text(cis) pi=-1,`

`z_4 = text(cis)\ (7 pi)/5,\ \ z_5 = text(cis)\ (9 pi)/5`

 

ii.  

Complex Numbers, EXT2 N2 2011 HSC 2d

  1. Use the binomial theorem to expand  `(cos theta + i sin theta)^3.`  (1 mark)

  2. Use de Moivre’s theorem and your result from part (i) to prove that
     
        `cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.`  (3 marks)

  3. Hence, or otherwise, find the smallest positive solution of
     
        `4 cos^3 theta - 3 cos theta = 1.`  (2 marks)

Show Answers Only
  1. `cos^3 theta + 3 i cos^2 theta sin theta – 3 cos theta sin^2 theta – i sin^3 theta`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `theta = (2 pi)/3`
Show Worked Solution

i.   `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

`= cos^3 theta + 3 cos^2 theta (i sin theta)+ 3 cos theta (i sin theta)^2 + (i sin theta)^3`

`= cos^3 theta + 3 i cos^2 theta sin theta- 3 cos theta sin^2 theta – i sin^3 theta`

 

ii.  `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`
 

`text(Equate real parts)`

`cos 3 theta` `= cos^3 theta – 3 cos theta sin^2 theta`
`cos 3 theta` `= cos^3 theta – 3 cos theta (1 – cos^2 theta)`
`cos 3 theta` `= 4 cos^3 theta – 3 cos theta`
`4 cos^3 theta`  `=cos 3 theta+3cos theta`
`:.cos^3 theta` `= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

 

iii.   `text(If)\ \ \ 4 cos^3 theta – 3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (ii))}`

`3 theta` `= 2 k pi`
`:. theta` `= (2 k pi)/3`

 

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`

Complex Numbers, EXT2 N2 2011 HSC 2c

Find, in modulus-argument form, all solutions of  `z^3 = 8.`  (2 marks)

Show Answers Only

`2text(cis)\ 0,\ 2text(cis)\ (2pi)/3,\ 2text(cis)((-2pi)/3)`

Show Worked Solution

`z^3 = 8`

`z^3 = 8 [cos (2 k pi) + i sin (2 k pi)],\ \ k=0, +-1, +-2`

`z = 2[cos ((2 k pi)/3) + i sin ((2 k pi)/3)]\ \ \ text{(De Moivre)}`
 

`text(When)\ \ k = 0,`

`z` `= 2 (cos 0 + i sin 0)`
  `= 2text(cis)\ 0`

 

`text(When)\ \ k = 1,`

`z` `= 2 [cos ((2 pi)/3) + i sin((2 pi)/3)]`
  `=2text(cis)\ (2pi)/3`

 
`text(When)\ \ k = -1,`

`z` `= 2 [cos ((-2 pi)/3) + i sin ((-2 pi)/3)]`
  `=2 text(cis)((-2 pi)/3)`

Complex Numbers, EXT2 N1 2013 HSC 11c

Factorise  `z^2 + 4iz + 5.`  (2 marks)

Show Answers Only

`(z – i) (z + 5i)`

Show Worked Solution

`text(Solution 1)`

`alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2`

`:. z^2 + 4iz + 5 = (z – i) (z + 5i)`

`(alpha + beta = 4i,\ \ \ alpha beta = 5 = -5i^2)`

MARKER’S COMMENT: Best practice includes stating the general quadratic formula before substituting in values.

 
`text(Solution 2)`

`z` `=(-b+- sqrt(b^2 – 4ac))/(2a)`
  `=(-4i +- sqrt((4i)^2-4 xx 5))/2`
  `=(-4i+-sqrt(-16-20))/2`
  `=(-4i +- 6i)/2`
  `=i\ \ text(or)\ \ -5i`

 
`:.z^2 + 4iz + 5 = (z – i) (z + 5i)`