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Complex Numbers, SPEC1 2018 VCAA 2

  1. Show that  `1 + i = sqrt 2\ text(cis)(pi/4)`.   (1 mark)

  2. Evaluate  `(sqrt 3-i)^10/(1 + i)^12`, giving your answer in the form  `a + bi`, where  `a, b ∈ R`.   (3 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-8-8 sqrt 3 i`

Show Worked Solution

a.   `r` `= sqrt(1^2 + 1^2)`
    `= sqrt 2`
  `theta` `= tan^(-1) (1/1) = pi/4`
     
  `:. 1 + i` `= sqrt 2\ text(cis)(pi/4)`

 

b.   
 `r_2` `= sqrt((sqrt 3)^2 + (-1)^2)`
    `= sqrt (3 + 1)`
    `= 2`

 
`theta_2 = -tan^(-1) (1/sqrt 3)=-pi/6`

`sqrt 3-i` `= 2\ text(cis)(-pi/6)`
`(sqrt 3-i)^10` `= 2^10\ text(cis) ((-10 pi)/6)`
   
`=>(1 + i)^12` `= (sqrt 2)^12text(cis)((12pi)/4)`
  `=2^6 text(cis)(3pi)`

 

`:. (sqrt 3-i)^10/(1 + i)^12` `= (2^10 text(cis)((-5pi)/3))/(2^6\text(cis)(3pi))`
  `= 2^4 text(cis)((-5pi)/3-(9pi)/3)`
  `= 16 text(cis)((-14pi)/3)`
  `= 16 text(cis) ((-2pi)/3)`
  `= 16(-1/2 + (-sqrt3)/2 i)`
  `= -8-8 sqrt 3 i`