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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, SPEC2 2023 VCAA 4 MC

If \(z=-(2 a+1)+2 a i\), where \(a\) is a non-zero real constant, then \(\dfrac{4 a}{1+\bar{z}}\) is equal to

  1. \(\sqrt{2} \text{cis}\left(\dfrac{\pi}{4}\right)\)
  2. \(\sqrt{2} \text{cis}\left(\dfrac{3 \pi}{4}\right)\)
  3. \(\text{cis}\left(\dfrac{\pi}{4}\right)\)
  4. \(\sqrt{2} \text{cis}\left(-\dfrac{3 \pi}{4}\right)\)
  5. \(\text{cis}\left(-\dfrac{\pi}{4}\right)\)
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\(B\)

Show Worked Solution

\(\text{Define}\ z\ \text{(by CAS):} \)

\(\dfrac{4 a}{1+\bar{z}} = -1+i \)

\(\Rightarrow B\)

Complex Numbers, EXT2 N1 2021 SPEC2 4 MC

For  the complex number `z `, if  `text(Im)(z) > 0`, then  `text(Arg)((zbarz)/(z - barz))` is

  1. `-pi/2`
  2. `0`
  3. `pi/4`
  4. `pi`
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`A`

Show Worked Solution

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

`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, EXT2 N1 SM-Bank 9

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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  1. `2 sqrt3 text{cis} (frac{-pi}{3})`
  2. `3`
Show Worked Solution
i.    `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})`

 

ii.   `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 2004 HSC 2b

Let  `alpha = 1 + i sqrt3`  and  `beta = 1 + i`.

  1. Find  `frac{alpha}{beta}`, in the form  `x + i y`.   (1 mark)

  2. Express `alpha` in modulus-argument form.   (3 marks)

  3. Given that `beta` has the modulus-argument form
     
         `beta = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`.
     
    find the modulus-argument form of  `frac{alpha}{beta}`.   (1 mark)

  4. Hence find the exact value of  `sin frac{pi}{12}`   (1 mark)

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  1. `frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`
  2. `2 \ text{cis} (frac{pi}{3})`
  3. `sqrt2 \ text{cis} (frac{pi}{12})`
  4. `frac{sqrt6-sqrt2}{4}`
Show Worked Solution
i.     `frac{alpha}{beta}` `= frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}`
    `= frac{(1 + i sqrt3)(1 – i)}{1^2 – i^2}`
    `= frac{1 – i + i sqrt3 – i^2 sqrt3}{2}`
    `= frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`

 

ii.   `alpha` `= 1 + i sqrt3`
  `| alpha |` `= sqrt(1^2 + (sqrt3)^2) = 2`

`text{arg} \ (alpha) = tan^-1 (frac{sqrt3}{1}) = frac{pi}{3}`

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

 

iii.   `beta` `= sqrt2 text{cis} (frac{pi}{4})`
  `frac{alpha}{beta}` `= frac{2}{sqrt2} \ text{cis} (frac{pi}{3} – frac{pi}{4})`
    `= sqrt2 text{cis}  (frac{pi}{12})`

 

iv.  `text{Equating imaginary parts of i and ii:}`

`sqrt2 \ sin \ (frac{pi}{12})` `= frac{sqrt3 – 1}{2}`
`sin (frac{pi}{12})` `= frac{sqrt3 – 1}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
  `= frac{sqrt6 – sqrt2}{4}`

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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  1. `2 \ text{cis} (-frac{pi}{3})`
  2. `32 \ text{cis} (frac{pi}{3})`
  3. `16 + i 16 sqrt3`
Show Worked Solution

i.    `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})`

 

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

 

iii.   `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 2008 HSC 2b

  1. Write  `frac{1 + i sqrt3}{1 + i}`  in the form  `x + iy`, where `x` and `y` are real.  (2 marks)

  2. By expressing both  `1 + i sqrt3`  and  `1 + i`  in  modulus-argument form, write  `frac{1 + i sqrt3}{1 + i}`  in modulus-argument form.   (3 marks)

  3. Hence find  `cos frac{pi}{12}`  in surd form.  (1 mark)

  4. By using the result of part (ii), or otherwise, calculate  `(frac{1 + i sqrt3}{1 + i})^12`.   (1 mark)

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  1. `frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`
  2. `sqrt2 (cos (frac{pi}{12}) + i sin (frac{pi}{12}))`
  3. `frac{sqrt2 + sqrt6}{4}`
  4. `-64`
Show Worked Solution
i.      `frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}` `= frac{(1 + i sqrt3)(1 – i)}{1 – i^2}`
    `= frac{1 – i + i sqrt3 – sqrt3 i^2}{2}`
    `= frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )`

 

ii.   `z_1 = 1 +  i sqrt3`

`| z_1 | = sqrt(1 + ( sqrt3)^2) = 2`

`text{arg} (z_1) = tan^-1 (sqrt3) = frac{pi}{3}`
  

`z_1 = 2 (cos frac{pi}{3} + i sin frac{pi}{3})`
 
`z_2 = 1 + i`

`| z_2 | = sqrt(1^2 + 1^2) = sqrt2`

`text{arg} (z_2) = tan^-1 (1) = frac{pi}{4}`

`z_2 = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`
 

`frac{1 + i sqrt3}{1 + i}` `= frac{z_1}{z_2}`
  `= frac{2}{sqrt2} ( cos ( frac{pi}{3} – frac{pi}{4} ) + i sin ( frac{pi}{3} – frac{pi}{4} ) )`
  `= sqrt2 ( cos (frac{pi}{12}) + i sin (frac{pi}{12}) )`

 

iii.  `text{Equating real parts of i and ii:}`

`sqrt2 cos (frac{pi}{12})` `= frac{1 + sqrt3}{2}`
`cos(frac{pi}{12})` `= frac{1 + sqrt3}{2 sqrt2} xx frac{sqrt2}{sqrt2}`
  `= frac{sqrt2 + sqrt6}{4}`

 

iv.     `(frac{1 + i sqrt2}{1 + i})^12` `= (sqrt2)^12 (cos (frac{pi}{12} xx 12) + i sin (frac{pi}{12} xx 12))`
    `= 64 (cos pi + i sin pi)`
    `= – 64`

Complex Numbers, EXT2 N1 2018 HSC 13b

Let   `z = 1 - cos2theta + isin2theta`, where   `0 < theta <= pi`.

  1.  Show that  `|\ z\ | = 2sintheta`.  (2 marks)

  2.  Show that  `text(arg)(z) = pi/2 - theta`.  (2 marks)

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

i.   `z = 1 – cos2theta + isin2theta`

`|\ z\ |` `= sqrt((1 – cos2theta)^2 + sin^2 2theta)`
  `= sqrt(1 – 2cos2theta + cos^2 2theta + sin^2 2theta)`
  `= sqrt(2 – 2cos2theta)`
  `= sqrt(2(1 – cos2theta))`
  `= sqrt(2(2sin^2theta))`
  `= 2sintheta\ \ \ text(… as required)`

 

ii.  `z= 1 – cos2theta + isin2theta`

`text(arg)(z)` `= tan^(−1)((sin2theta)/(1 – cos2theta))`
  `= tan^(−1)((2sinthetacostheta)/(2sin^2theta))`
  `= tan^(−1)(cottheta)`
  `= tan^(−1)(tan(pi/2 – theta))`
  `= pi/2 – theta`

 
`(text(S)text(ince)\ \ 0 < theta < pi => \ −pi/2 <= pi/2 – theta < pi/2)`

Complex Numbers, EXT2 N1 2017 HSC 11a

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

  1. Find the exact value of the argument of `z`.  (1 mark)
  2. Find the exact value of the argument of  `z/w`.  (2 marks)  
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  1. `-pi/3`
  2. `-(7 pi)/12`
Show Worked Solution

i.  `z = 1 – i sqrt 3`

`text(arg)\ z = – pi/3`

 

ii.  `w = 1 + i`

`text(arg)\ w = pi/4`

`text(arg)\ (z/w)` `= text(arg)\ (z) – text(arg)\ (w)`
  `= – pi/3 – pi/4`
  `= -(7 pi)/12`

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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  1. `sqrt 2 ( cos­ pi/4 + i sin­ pi/4)`
  2. `256 + 256i`
Show Worked Solution
i.  
`|\ 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)`

 

ii.   `(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))`
Show Answers Only

`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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  1. `2 text(cis) (−pi/6)`
  2. `2^7 text(cis) ((5 pi)/6)`
  3. `64 (−sqrt 3 + i)`
Show Worked Solution
i.  
`|\ 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)`

 

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

 

iii.  `(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 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)

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  1. `2text(cis)(-(5pi)/6)`
  2. `-64`
Show Worked Solution
i.    `|-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)`

 

ii.   `(-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 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)

Show Answers Only
  1. `2\ text(cis)(-pi/6)`
  2. `i512`
Show Worked Solution
i. `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)`  

 

ii.   `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)
Show Answers Only
  1. `3 – i\ 2 sqrt 3`
  2. `2 text(cis) pi/3`
  3. `2^24`
Show Worked Solution

i.    `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`

 

ii.   `|\ 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.

 

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

Complex Numbers, EXT2 N1 2014 HSC 11a

Consider the complex numbers  `z = -2- 2i`  and  `w = 3 + i`.

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

  2. Express `z/w` in the form  `x + iy`, where  `x`  and  `y`  are real numbers.  (2 marks)

Show Answers Only
  1. `sqrt2\ text(cis)(-pi/4)`
  2. `−4/5 −2/5i`
Show Worked Solution
i.   `z + w` `= −2 − 2i + 3 + i`
     `= 1 − i`
  `|\ z+w\ |` `= sqrt2`
  `text(arg)\ (z+w)` `=- pi/4`
  `:. z+w`   `= sqrt2\ text(cis)(-pi/4)`

 

ii.  `z/w` `= (−2 − 2i)/(3 + i)`
    `= ((−2 − 2i)(3 − i))/((3 + i)(3 − i))`
    `= (−6 − 6i + 2i − 2)/(9 + 1)`
    `= −8/10 −4/10i`
    `= −4/5 − 2/5i`