Let `beta = 1-i sqrt3`.
- Express `beta` in modulus-argument form. (2 marks)
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- Express `beta^5` in modulus-argument form. (2 marks)
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- Hence express `beta^5` in the form `x+iy`. (1 mark)
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Show Worked Solution
a. `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})`
| b. | `beta^5` | `= 2^5 \ text{cis} (-frac{pi}{3} xx5)` |
| `= 32 \ text{cis} (-frac{5pi}{3} + 2 pi)` | ||
| `= 32 \ text{cis} (frac{pi}{3})` |
| c. | `beta^5` | `= 32 ( cos (frac{pi}{3}) + i sin (frac{pi}{3}) )` |
| `= 32 ( frac{1}{2} + i frac{sqrt3}{2})` | ||
| `= 16 + i 16 sqrt3` |