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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)

Show Answers Only
  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 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`