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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)

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`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 N1 2024 HSC 11e

  1. Write the number  \(\sqrt{3}+i\)  in modulus-argument form.   (2 marks)

  2. Hence, or otherwise, write  \((\sqrt{3}+i)^7\)  in exact Cartesian form.   (2 marks)

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i.     \(2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)

ii.    \(-64 \sqrt{3}-64 i\)

Show Worked Solution

i.     \(z=\sqrt{3}+i\)

\(|z|=\sqrt{3+1}=2\)

\(\arg (z)=\tan ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=\dfrac{\pi}{6}\)

\(z=2 \text{cis}\left(\dfrac{\pi}{6}\right)=2\left(\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right)\)
 

ii.     \((\sqrt{3}+i)^7\) \(=2^7\left(\cos \left(\dfrac{7 \pi}{6}\right)+i \sin \left(\dfrac{7 \pi}{6}\right)\right)\)
    \(=128\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2} i\right)\)
    \(=-64 \sqrt{3}-64 i\)

Complex Numbers, EXT2 N1 EQ-Bank 14

Let  `z = sqrt3-3 i`

  1. Express `z` in modulus-argument form.   (2 marks)

  2. Find the smallest integer `n`, such that  `z^n + (overset_z)^n = 0`.   (3 marks)

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a.    `2 sqrt3 text{cis} (frac{-pi}{3})`

b.    `3`

Show Worked Solution
a.     `z` `= sqrt3-3 i`
  `|z|` `= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3`

 

`tan theta= frac{3}{sqrt3}=sqrt3\ \ =>\ \ `theta= frac{pi}{3}`

`text{arg} (z)=-frac{pi}{3}`

`therefore  z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`
 

b.    `z^n + (overset_z)^n = 0`

`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3})-i sin (frac{-pi}{3}) ]^n = 0`

`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3})-i sin (frac{-n pi}{3}) = 0`

`2 \ cos (frac{-n pi}{3})` `= 0`
`cos (frac{n pi}{3})` `= 0`
`frac{n pi}{3}` `= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …`
`frac{n}{3}` `= frac{(2k + 1)}{2}`
`n` `= frac{3 (2k + 1)}{2}`

 
`text{Numerator will always be odd  ⇒  no solution exists}`

Complex Numbers, EXT2 N1 2005 HSC 2b

Let  `beta = 1-i sqrt3`.

  1. Express  `beta`  in modulus-argument form.    (2 marks)

  2. Express  `beta^5`  in modulus-argument form.    (2 marks)

  3. Hence express  `beta^5`  in the form  `x+iy`.    (1 mark)

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a.    `2 \ text{cis} (-frac{pi}{3})`

b.    `32 \ text{cis} (frac{pi}{3})`

c.    `16 + i 16 sqrt3`

Show Worked Solution

a.    `beta = 1-i sqrt3`
 

 
`| beta | = sqrt(1^2 + (sqrt3)^2) = 2`

`tan theta` `= frac{sqrt3}{1} = sqrt3`
`theta` `= frac{pi}{3}`
`text{arg} (beta)` `= -frac{pi}{3}`

`therefore \ beta = 2 \ text{cis} (-frac{pi}{3})`

 

b.     `beta^5` `= 2^5 \ text{cis} (-frac{pi}{3} xx5)`
    `= 32 \ text{cis} (-frac{5pi}{3} + 2 pi)`
    `= 32 \ text{cis} (frac{pi}{3})`

 

c.     `beta^5` `= 32 ( cos (frac{pi}{3}) + i sin (frac{pi}{3}) )`
    `= 32 ( frac{1}{2} + i  frac{sqrt3}{2})`
    `= 16 + i 16 sqrt3`

Complex Numbers, EXT2 N1 2019 HSC 11e

Let  `z = -1 + i sqrt 3`.

  1. Write  `z`  in modulus-argument form.   (2 marks)

  2. Find  `z^3`, giving your answer in the form  `x + iy`, where `x` and `y` are real numbers.   (2 marks)

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i.    `z = 2 text(cis) (2 pi)/3`

ii.   `8 + 0i`

Show Worked Solution

i.    `|\ z\ |= -1 + i sqrt 3= sqrt((-1)^2 + (sqrt 3)^2)= 2`

  `tan theta` `= -sqrt 3`
  `text(arg)(z)` `= (2 pi)/3`
  `:. z` `= 2 text(cis) (2 pi)/3`

 

ii.   `z^3 = 2^3 [cos(3 xx (2 pi)/3) + i sin (3 xx (2 pi)/3)]\ \ \ text{(by De Moivre)}`

`= 8(cos 2 pi + i sin 2 pi)`

`= 8(1 + 0i)`

`= 8 + 0i`

Complex Numbers, EXT2 N2 2018 HSC 15b

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

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

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  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 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
  2. `qquad cos^4 theta-6 cos^2 theta sin^2 theta + sin^4 theta.`   (2 marks)
  3. Hence, or otherwise, find an expression for  `cos 4 theta`  involving only powers of `cos theta.`   (1 mark)

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i.    `text(See Worked Solutions)`

ii.   `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 2007 HSC 8b

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

  3. By substituting  `z = cos theta + i sin theta`  where  `sin theta != 0`, into part (a), show that
  4. `1 + cos 2 theta + … + cos (2n-2) theta + i[sin 2 theta + … + sin (2n-2) theta]`
  5. `= (sin n theta)/(sin theta) [cos (n-1) theta + i sin (n-1) theta].`   (3 marks)

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

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.    `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)`

 

b.     `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 (a):}`

`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.)`

 

c.    `text{Equating the imaginary parts in part (b):}`

`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 N1 2007 HSC 2b

  1. Write  ` 1 + i`  in the form `r (cos theta + i sin theta).`   (2 marks)

  2. Hence, or otherwise, find `(1 + i)^17` in the form  `a + ib`, where `a` and `b` are integers.   (3 marks)

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a.    `sqrt 2 ( cos­ pi/4 + i sin­ pi/4)`

b.    `256 + 256i`

Show Worked Solution
a.    
`|\ 1+i\ |` `=sqrt(1^2+1^2)=sqrt2`
`text(arg)(1+i)` `=pi/4`
`:. 1 + i =` `sqrt 2 (cos­ pi/4 + i sin­ pi/4)`

 

b.    `(1 + i)^17` `=(sqrt 2)^17 (cos\ pi/4 + i sin\ pi/4)^17`
  `=2^8 sqrt 2 (cos­ (17 pi)/4 + i sin­ (17 pi)/4)\ \ \ \ text{(De Moivre)}`
  `=2^8 sqrt 2 (cos­ pi/4 + i sin­ pi/4)`
  `=2^8 sqrt2(1/sqrt2 + 1/sqrt2 i)`
  `=2^8 (1 + i)`
  `=256 + 256 i`

Complex Numbers, EXT2 N1 2015 HSC 5 MC

Given that  `z = 1 -i`, which expression is equal to  `z^3 ?`

  1. `sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
  2. `2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))`
  3. `sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
  4. `2 sqrt 2 (cos((3 pi)/4) + i sin((3 pi)/4))`
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`B`

Show Worked Solution

 HSC 2015 5MC

`z` `=1-i`
`|\ 1-i\ |` `=sqrt2`
`text{arg}(z)` `=-pi/4`
`z` `=sqrt 2 (cos(-pi/4) + i sin(-pi/4))`
`:.z^3` `=2 sqrt 2 (cos((-3 pi)/4) + i sin((-3 pi)/4))\ \ \ \ text{(De Moivre)}`

 
`=>  B`

Complex Numbers, EXT2 N1 2006 HSC 2b

  1. Express  `sqrt 3-i`  in modulus-argument form.   (2 marks)

  2. Express `(sqrt 3-i)^7` in modulus-argument form.   (2 marks)

  3. Hence express `(sqrt 3-i)^7` in the form  `x + iy.`   (1 mark)

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a.    `2 text(cis) (−pi/6)`

b.    `2^7 text(cis) ((5 pi)/6)`

c.    `64 (−sqrt 3 + i)`

Show Worked Solution
a.    

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

`­theta=tan^-1(- 1/sqrt3)=- pi/6`

 `:. sqrt 3-i = 2 text(cis) (- pi/6)`

 

b.   `(sqrt 3-i)^7 =` `2^7 text(cis) (-(7 pi)/6)\ \ \ \ text{(De Moivre)}`
`­=` `128 text(cis) ((5 pi)/6)`

 

c.  `(sqrt 3-i)^7` `=128 (cos\ (5pi)/6 + i sin\ (5pi)/6)`
  `=128 (- sqrt 3/2 + i/2)`
  `=-64 sqrt 3 + 64i`

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
  3. `(2 cos theta)^(2m) = 2 [cos (2m theta) + ((2m), (1)) cos (2m-2) theta + ((2m), (2)) cos (2m-4) theta`
  4.           `+ … + ((2m), (m-1)) cos 2 theta] + ((2m), (m)).`   (3 marks)

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

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
a.     `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`

 

 

b.    `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))`

 

c.    `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 N1 2010 HSC 2b

  1. Express  `-sqrt3-i`  in modulus–argument form.   (2 marks)

  2. Show that  `(-sqrt3-i)^6`  is a real number.   (2 marks)

Show Answers Only

a.    `2text(cis)(-(5pi)/6)`

b.    `-64`

Show Worked Solution

a.    `|-sqrt3-i\ |=sqrt((-sqrt3)^2+sqrt((-1)^2))=2`
 

Complex Numbers, EXT2 2010 HSC 2b 

`text(From the graph)`

`text{arg}(-sqrt3-i)=- (5pi)/6\ \ \ \ text{(for}\  –pi<theta<pi text{)}`

`:.-sqrt3-i= 2text(cis)(-(5pi)/6)`

 

`text{Alternative Solution (to find the argument)}`

`-sqrt3-i= 2(- sqrt3/2-1/2 i)=2text(cis)(-(5pi)/6)`
 

b.     `(-sqrt3-i)^6` `= [2text(cis)(-(5pi)/6)]^6`
    `=2^6[cos((-5pi)/6 xx6) +i sin((-5 pi)/6 xx6)]\ \ \ \ text{(De Moivre)}`
    `= 2^6[cos(-5pi) + i sin(-5pi)]`
    `= 64(-1 + 0i)`
    `= -64`

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 (a) to prove that
  3.      `cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.`   (3 marks)

  4. Hence, or otherwise, find the smallest positive solution of
  5. `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

a.    `(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`

 

b.     `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)`

 

c.    `text(If)\ \ \ 4 cos^3 theta-3 cos theta = 1`

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

`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 N1 2012 HSC 11d

  1. Write  `z = sqrt3-i`  in modulus-argument form.  (2 marks)

  2. Hence express  `z^9`  in the form  `x + iy`, where `x` and `y` are real.  (1 mark)

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a.    `2\ text(cis)(-pi/6)`

b.    `i512`

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a.     `z` `=sqrt3-i`
  `|\ z\ |` `=sqrt((sqrt3)^2+1^2)=2`

 

`:.z = sqrt3 − i` `= 2(sqrt3/2 − 1/2i)`
  `= 2(cos\ (-pi/6) + i\ sin\ (-pi/6))`  
  `= 2\ text(cis)(-pi/6)`  

 

b.     `z^9` `= 2^9\ (cos\ (-pi/6) + i\ sin\ (-pi/6))^9`
    `= 2^9\ text(cis)(-(9pi)/6)\ \ \ \ text{(by De Moivre)}`
    `=512\ text(cis)(-(3pi)/2)`
    `= 512(0 + i)`
    `=i512`

Complex Numbers, EXT2 N1 2013 HSC 11a

Let  `z = 2-i sqrt 3`  and  `w = 1 + i sqrt 3.`

  1. Find  `z + bar w.`   (1 mark)

  2. Express `w` in modulus–argument form.   (2 marks)

  3. Write `w^24` in its simplest form.   (2 marks)

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a.    `3-i\ 2 sqrt 3`

b.    `2 text(cis) pi/3`

c.    `2^24`

Show Worked Solution

a.    `z = 2-i sqrt 3\ ,\ \ w = 1 + i sqrt 3`

`bar w = 1-i sqrt 3`

`z + bar w` `= 2-i sqrt 3 + 1-i sqrt 3`
  `= 3-i\ 2 sqrt 3`

 
b.    `|\ w\ |=sqrt(1^2 + (sqrt3)^2)=2`

 `:.w` `= 2 (1/2 + i sqrt 3/2)`
  `=2(cos\ pi/3 + i sin\ pi/3)`
  `= 2 text(cis) pi/3`
MARKER’S COMMENT: The directive “in its simplest form” required students to convert `text(cis)\ 8pi` to 1.

 

c.    `w^24` `= 2^24 text(cis)\ (24 xx pi/3)`
  `= 2^24\ text(cis)(8 pi)`
  `= 2^24`