## Question 4(i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2+3i}{5+i}\times\frac{5-i}{5-i}$ **or** $2+3i = k(1+i)(5+i) = \ldots$ | M1 | 1.1a |
| $\frac{10-2i+15i+3}{25+1}$ or $\frac{13+13i}{26}$ **or** $2+3i = k(5+i+5i-1) = \ldots$ | dM1 | 1.1b |
| $\frac{1}{2}(1+i)$ cso **or** $2+3i = k(4+6i)$ therefore $\frac{2+3i}{5+i} = k(1+i)$ where $k=\frac{1}{2}$ cso | A1 | 2.1 |

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## Question 4(i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=4$ | B1 | 2.2a |

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## Question 4(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $|z|=3$ | B1 | 1.2 |
| $\arg(z^{10})=10\arg(z)=-\frac{5\pi}{3} \Rightarrow \arg(z)=\ldots\left\{-\frac{\pi}{6}\right\}$ **or** $\arg(z^{10})=10\arg(z)=\frac{\pi}{3}\Rightarrow\arg(z)=\ldots\left\{\frac{\pi}{30}\right\}$ | M1 | 1.1b |
| $z=3\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)=\ldots$ | M1 | 2.1 |
| $z=\frac{3\sqrt{3}}{2}-\frac{3}{2}i$ or $a=\frac{3\sqrt{3}}{2}$ and $b=-\frac{3}{2}$ | A1 | 1.1b |

**Alternative:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $a^2+b^2=9$ | B1 | 1.2 |
| $10\arg z = -\frac{5\pi}{3} \Rightarrow \arg z = -\frac{5\pi}{3}\div 10$ or e.g. $10\arg(z)=\frac{\pi}{3}\Rightarrow\arg(z)=\ldots\left\{\frac{\pi}{30}\right\}$ | M1 | 1.1b |
| $\frac{b}{a}=\arctan\left(-\frac{\pi}{6}\right)\Rightarrow\frac{b}{a}=-\frac{\sqrt{3}}{3}\Rightarrow b=-a\frac{\sqrt{3}}{3}$ or $\frac{b}{a}=\arctan\frac{\pi}{30}\Rightarrow b=0.104\ldots a$ | M1 | 2.1 |
| $z=\frac{3\sqrt{3}}{2}-\frac{3}{2}i$ or $a=\frac{3\sqrt{3}}{2}$ and $b=-\frac{3}{2}$ | A1 | 1.1b |

# Question (i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Selects process $\frac{2+3\text{i}}{5+\text{i}} \times \frac{5-\text{i}}{5-\text{i}}$ | M1 | |
| Evidence of multiplying out brackets | dM1 | |
| Achieves $\frac{1}{2}(1+\text{i})$ or $\frac{13}{26}(1+\text{i})$ with no errors | A1 | Correct answer from no working scores no marks. Going from $\frac{13+13\text{i}}{26}$ and stating $k=\frac{1}{2}$ is A0 |
| **Alternative:** Multiplies across by $(5+\text{i})$ and expands | M1 | |
| Collects terms | dM1 | |
| Achieves $2+3\text{i} = k(4+6\text{i})$ concluding $\frac{2+3\text{i}}{5+\text{i}} = k(1+\text{i})$ where $k=\frac{1}{2}$ | A1 | |

# Question (i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $n=4$ only | B1 | |

# Question (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $|z|=3$, can be implied by $a^2+b^2=9$ | B1 (M1 on ePen) | |
| Uses $\arg(z_1 z_2)=\arg(z_1)+\arg(z_2)$ to find $\arg(z)=-\frac{5\pi}{3}\div 10$ or $\arg(z)=\frac{\pi}{3}\div 10$ | M1 | Send to review if finding additional solutions where arg of $z$ is $\frac{(6k-5)\pi}{30}$; correctly uses $\arg(z)=\frac{\pi}{30}$ |
| Uses $z=|z|(\cos(\text{arg})+\text{i}\sin(\text{arg}))$ to find $z$ or values of $a$ or $b$ | M1 | As long as modulus has changed |
| Correct complex number or values for $a$ and $b$ | A1 | |
| **Alternative:** $a^2+b^2=9$ | B1 | |
| Uses $\arg(z_1 z_2)=\arg(z_1)+\arg(z_2)$ to find $\arg(z)=-\frac{5\pi}{3}\div 10$ | M1 | |
| Uses argument of $z$ to find equation in $a$ and $b$, solve simultaneously | M1 | As long as $\sqrt{a^2+b^2}\neq 59049$ |
| Correct complex number or values for $a$ and $b$ | A1 | |

Correct answers include: $z_1=\frac{3\sqrt{3}}{2}-\frac{3}{2}\text{i}$, $z_2\approx 2.98+0.314\text{i}$, $z_3\approx 2.23+2.01\text{i}$, $z_4\approx 0.624+2.93\text{i}$, $z_5\approx -1.22+2.74\text{i}$, $z_6=-\frac{3\sqrt{3}}{2}+\frac{3}{2}\text{i}$, and negatives thereof.

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