# Question 4:

## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2 - i$ is also a root | B1 | 2.2a — Can be implied by $(z-(2+i))$ |
| $(z-(2+i))(z-(2-i))$ | M1 | 1.1 — Correctly determining the form of the quadratic in factor form. DR: Detailed reasoning required |
| $(z-2)^2 + 1$ | | |
| $z^2 - 4z + 5$ | A1 | 2.1 — Working needs to be shown (Show detailed reasoning). At least 1 step between factor form and final answer |
| $(z^2 - 4z + 5)(4z^2 + \ldots + 37)$ | M1 | 1.1 — With non-zero $z$ term. Or attempt at symbolic division resulting in at least $4z^2$. Or genuine attempt at method comparing coefficients leading to at least one useful equation (e.g. $z^4$: $A=4$). Must be a product of two quadratics. Multiplication table might be used (backwards) |
| $(z^2 - 4z + 5)(4z^2 + 4z + 37)$ | A1 [5] | 1.1 |

## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times 37}}{2 \times 4}$ soi | M1* | 1.1a — Correct method for finding a root of a quadratic ($\pm$ not necessary). Or $4((z+\frac{1}{2})^2 - \frac{1}{4}) + 37$ oe. Need to see some method |
| $\dfrac{-4 \pm 24i}{8}$ or $\dfrac{-1 \pm 6i}{2}$ etc. | A1dep(*) | 2.1 — Simplification of square root to include $i$ ($\pm$ now necessary). Must be of the form $z = a + bi$ (or $z = \frac{a+bi}{c}$). No square roots left in |
| $\dfrac{-1 \pm 6i}{2}$, $2 \pm i$ | A1(ft) dep(*) [3] | 1.1 — All four - aef. Follow through on incorrect roots from previous mark as long as method mark awarded. SC: If M0 awarded but all four correct roots given with no extra then award SC B1 |

## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r_2 = |2 \pm i| = \sqrt{5}$ | B1 | 2.2a — Need to see $|2+i|$ or $|2-i|$. Or $\sqrt{2^2+1^2} = \sqrt{5}$ etc. |
| $r_1 = \left|\dfrac{-1 \pm 6i}{2}\right| = \dfrac{\sqrt{37}}{2}$ | B1ft | 2.2a — ft the magnitude of the other conjugate pair |
| $\pi\left(\dfrac{\sqrt{37}}{2}\right)^2 - \pi(\sqrt{5})^2$ | M1 | 3.1a — Either way round, using their $r_1$ and/or $r_2$ |
| Area between two circles is $\dfrac{17}{4}\pi$ | A1ft [4] | 3.2a — $\pi(r_1^2 - 5)$ simplified. Positive area needed, must follow B1B1(ft)M1 |

## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\omega = -\dfrac{-12}{4} = 3\ldots$ | B1 | 2.2a — or $\frac{-1+6i}{2} + \frac{-1-6i}{2} + 2+i+2-i = 3$. Some justification needed, not just $\omega = 3$. Allow $3+0i$ |
| $\ldots$ and $\sqrt{5} < 3 < \dfrac{\sqrt{37}}{2}$ so $\omega$ is in $R$ | E1 [2] | 2.3 — Or $5 < 9 < 37/4$. Or $2.23\ldots < 3 < 3.04\ldots$. Both end comparisons needed. No follow through given in this part. Conclusion needed. Diagram is ok for E1 if it implies both end comparisons |

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