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Complex Numbers, EXT2 N2 2025 HSC 15c

  1. Show that
  2.     \(\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.\)   (3 marks)

  3. Use De Moivre's theorem to show that the sixth roots of \(-1\) are given by 
  4.    \(\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)\)  for  \(k=0,1,2,3,4,5\).   (2 marks)

  5. Hence, or otherwise, show the solutions to  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)  are 
  6. \(z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)\), and \(i \cot \left(\dfrac{11 \pi}{12}\right)\).   (2 marks)

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i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

 

Show Worked Solution

i.    \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\dfrac{\theta}{2}\right)\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 

ii.    \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

 
\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\dfrac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\dfrac{\pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\dfrac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\dfrac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\dfrac{\pi}{12}\right), \, i \cot \left(\dfrac{3 \pi}{12}\right), \, i \cot \left(\dfrac{5 \pi}{12}\right), \ldots, i \cot \left(\dfrac{11 \pi}{12}\right)\)

Complex Numbers, EXT2 N2 2021 HSC 14c

  1. Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that
  2.    `cos5theta = 16cos^5theta - 20cos^3 theta + 5cos theta`.  (3 marks)

  3. By using part (i), or otherwise, show that  `text(Re)(e^((ipi)/10)) = sqrt((5 + sqrt5)/8)`.  (3 marks)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `(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`

 

ii.   `text(Re)(e^((ipi)/10)) = cos\ pi/10`

♦♦ Mean mark 31%.

`text(Using part a:)`

`16 cos^5(pi/10) – 20cos^3(pi/10) + 5cos(pi/10) = cos((5pi)/10) = cos(pi/2) = 0`

`text(Solutions to)\ cos(5theta) = 0`

`5theta` `= pi/2, (3pi)/2, (5pi)/2, …`
`theta` `= pi/10, (3pi)/10, (5pi)/10, …`

  
`text(Let)\ cos(pi/10) = x`

`16x^5 – 20x^3 + 5x` `= 0`
`16x^4 – 20x^2 + 5` `= 0`
`x^2` `= (20 ± sqrt(20^2 – 4 · 16 · 5))/(2 xx 16)`
  `= (20 ± sqrt80)/32`
  `= (5 ±sqrt5)/8`
`x` `= ±sqrt((5 ± sqrt5)/8)`

 
`cos\ pi/10 > 0 \ => \ x = sqrt((5 ± sqrt5)/8)`

`1 > cos\ pi/10 > cos\ (3pi)/10 > 0`

`:.\ text(Re)\ (e^((ipi)/10)) = cos\ pi/10 = sqrt((5 + sqrt5)/8)`

Complex Numbers, EXT2 N2 2020 HSC 14a

Let `z_1` be a complex number and let  `z_2 = e^(frac{i pi}{3}) z_1`

The diagram shows points `A` and `B` which represent `z_1` and `z_2`, respectively, in the Argand plane.
 


 

  1. Explain why triangle  `OAB`  is an equilateral triangle.   (2 marks)

  2. Prove that  `z_1 ^2 + z_2^2 = z_1 z_2`.  (3 marks)

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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
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i.

 

`text{Let}`     `z_1` `= r(cos theta + i sin theta)`
  `z_2` `= e^(i frac{pi}{3}) z_1`
    `= r (cos ( theta + frac{pi}{3} ) + i sin (theta + frac{pi}{3}))`

 

`| z_1 | = | z_2 | => OA = OB`

` angle AOB = frac{pi}{3} \ ( z_2 \ text{is a} \ frac{pi}{3} \ text{anti-clockwise rotation of} \ z_1 )`

`=> angle OBA = angle BAO = pi/3\ \ \ text{(angles opposite equal sides)}`

`therefore \ OAB \ text{is equilateral}`

 

ii.     `z_1` `= z_2 e^(i frac{pi}{3})`
  `frac{z_1}{z_2}` `= e^(i frac{pi}{3})`
  `(frac{z_1}{z_2})^3` `= e^((3 xx i frac{pi}{3})) \ \ (text{by De Moivre})`
  `frac{z_1^3}{z_2^3}` `= e^(i pi)`
  `z_1^3` `= -z_2^3`
  `z_1^3 + z_2^3` `= 0`

COMMENT: The identity to prove suggests the addition of 2 cubes is a possible strategy.
`(z_1 + z_2)(z_1^2 – z_1 z_2 + z_2^2)` `= 0`
`z_1^2 – z_1 z_2 + z_2^2` `= 0`
`z_1^2 + z_2^2` `= z_1 z_2`

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)

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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
     
    `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)

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  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 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)

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  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 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)

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  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 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)

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  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{)}`