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