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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 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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a.    `frac{1+sqrt3}{2} + i (frac{sqrt3 – 1}{2})`

b.    `2 \ text{cis} (frac{pi}{3})`

c.    `sqrt2 \ text{cis} (frac{pi}{12})`

d.    `frac{sqrt6-sqrt2}{4}`

Show Worked Solution
a.     `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})`

 

b.     `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})`

 

c.     `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})`

 

d.    `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 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)

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a.    `frac{1 + sqrt3}{2}-i ( frac{1-sqrt3}{2} )`

b.    `sqrt2 (cos (frac{pi}{12}) + i sin (frac{pi}{12}))`

c.    `frac{sqrt2 + sqrt6}{4}`

Show Worked Solution
a.      `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} )`

 

b.    `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}) )`

 

c.    `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}`

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 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`

Show Worked Solution
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 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)

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a.    `sqrt2\ text(cis)(-pi/4)`

b.    `−4/5-2/5i`

Show Worked Solution

a.    `z + w= −2-2i + 3 + i= 1-i`

  `|\ z+w\ |` `= sqrt2`
  `text(arg)\ (z+w)` `=- pi/4`
  `:. z+w`   `= sqrt2\ text(cis)(-pi/4)`

 

b.     `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`