Let `P(z) = z^4 - 2kz^3 + 2k^2z^2 + mz + 1`, where `k` and `m` are real numbers.
The roots of `P(z)` are `alpha, bar alpha, beta, bar beta`.
It is given that `|\ alpha\ | = 1` and `|\ beta\ | = 1`.
- Show that `(text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`. (3 marks)
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- The diagram shows the position of `alpha`.
On the diagram, accurately show all possible positions of `beta`. (2 marks)
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Show Worked Solution
i. `P(z) = z^4 – 2kz^3 + 2k^2z^2 + mz + 1,\ \ k, m in RR`
`text(Roots):\ \ alpha, bar alpha, beta, bar beta and |\ alpha\ | = 1, |\ beta\ | = 1`
`text(Show)\ \ (text{Re} (alpha))^2 + (text{Re} (beta))^2 = 1`♦♦ Mean mark part (i) 26%.
| `alpha + bar alpha + beta + bar beta` | `= 2k` |
| `2 text{Re} (alpha) + 2 text{Re} (beta)` | `= 2k` |
| `text{Re} (alpha) + text{Re} (beta)` | `= k` |
| `alpha bar alpha + alpha beta + alpha bar beta + bar alpha beta + bar alpha bar beta + beta bar beta` | `= 2k^2` |
| `|\ alpha\ |^2 + alpha(beta + bar beta) + bar alpha(beta + bar beta) + |\ beta\ |^2` | `= 2k^2` |
| `1 + (alpha + bar alpha)(beta + bar beta) + 1` | `= 2k^2` |
| `2 + 2 text{Re} (alpha) ⋅ 2 text{Re} (beta)` | `= 2 (text{Re} (alpha) + text{Re} (beta))^2` |
| `2 + 4 text{Re} (alpha) text{Re} (beta)` | `= 2 text{Re} (alpha)^2 + 4 text{Re} (alpha) text{Re} (beta) + 2 text{Re} (beta)^2` |
| `2` | `= 2(text{Re} (alpha)^2 + text{Re} (beta)^2)` |
| `:. 1` | `= text{Re} (alpha)^2 + text{Re} (beta)^2` |
| ii. | `|\ alpha\ | = |\ beta\ |\ \ \ text{(given)}` |
| `text{Re}(alpha)^2 + text{Re}(beta)^2 = 1\ \ \ text{(see part (i))}` | |
| `text{Re}(alpha)^2 + text{Im}(alpha)^2 = 1\ \ \ (|\ alpha\ | = 1)` | |
| `=> text{Re}(beta)^2 = text{Im} (alpha)^2` | |
| `\ \ \ \ \ \ text{Re}(beta) = +-text{Im}(alpha)` |
♦♦♦ Mean mark part (ii) 10%.
