# Question 4(a):

| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| Circle centred at $-i$ passing through origin | M1A1 | M1: a circle anywhere; A1: circle **correctly positioned** with centre $-i$ or $-1$ marked at $(0,-1)$ or $(-1,0)$ or $(0,-i)$ or $(-i,0)$ in **correct place**, passing through $(0,0)$; centre may be indicated away from sketch but sketch takes precedence; ignore shading |

# Question 4(b):

| Working/Answer | Marks | Guidance Notes |
|---|---|---|
| $z = \frac{wi+2}{3i-w}$ | M1A1 | M1: attempt to make $z$ the subject; A1: correct rearrangement oe |
| $z+i = \frac{wi+2}{3i-w}+i = \frac{wi+2-3-wi}{3i-w}$ | M1 | Applies $z+i$ and finds common denominator |
| $\left\lvert\frac{wi+2-3-wi}{3i-w}\right\rvert = 1$ | M1A1 | M1: sets $|z+i|=1$; A1: correct equation simplified or unsimplified |
| $\left\lvert\frac{-1}{3i-w}\right\rvert=1 \Rightarrow |w-3i|=1 \Rightarrow |u+iv-3i|=1$ | dM1A1 | dM1: introduces $u$ and $v$ (or $x$ and $y$) and uses Pythagoras to find Cartesian form — **dependent on all previous method marks**; A1: $u^2+(v-3)^2=1$ or equivalent (allow $u,v$ or $x,y$ or $a,b$) |

## Question (Complex Numbers - Way 2):

**Way 2:**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{wi+2}{3i-w}$ | M1A1 | M1: Attempt to make $z$ the subject; A1: Correct rearrangement |
| $z = \frac{(u+iv)i+2}{3i-(u+iv)} = \frac{(2-v)+ui}{-u+(3-v)i} \times \frac{-u-(3-v)i}{-u-(3-v)i}$ | M1 | Introduces $u+iv$ and multiplies numerator and denominator by complex conjugate of denominator |
| $z+i = \frac{u+(5v-6-u^2-v^2)i+(u^2+v^2+9-6v)i}{u^2+(3-v)^2} \left(= \frac{u+(3-v)i}{u^2+(3-v)^2}\right)$ | M1A1 | M1: Applies $z+i$ and finds common denominator; A1: Correct expression (simplified or unsimplified) with no $i$'s in denominator |
| $\|z+i\|=1 \Rightarrow \left\|\frac{u+(3-v)i}{u^2+(3-v)^2}\right\|=1 \Rightarrow \frac{\sqrt{u^2+(3-v)^2}}{u^2+(3-v)^2}=1$ | dM1A1 | dM1: Introduces $u,v$ or $x,y$, uses Pythagoras to find Cartesian form. **Dependent on all previous method marks.** A1: Correct equation |

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## Question (Complex Numbers - Way 3):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{wi+2}{3i-w}$ | M1A1 | M1: Attempt to make $z$ the subject; A1: Correct rearrangement |
| $z = \frac{(u+iv)i+2}{3i-(u+iv)} = \frac{(2-v)+ui}{-u+(3-v)i} \times \frac{-u-(3-v)i}{-u-(3-v)i}$ | M1 | Introduces $u+iv$ and multiplies by complex conjugate of denominator |
| $z = \frac{u+(-u^2-v^2+5v-6)i}{u^2+(3-v)^2} \Rightarrow x = \frac{u}{u^2+(3-v)^2},\quad y = -\frac{u^2+(v-3)^2+v-3}{u^2+(3-v)^2}$ | M1A1 | M1: Obtains $x$ and $y$ in terms of $u$ and $v$; A1: Correct equations |
| $x^2+(y+1)^2=1 \Rightarrow \frac{u^2+(v-3)^2}{\left(u^2+(v-3)^2\right)^2}=1$ | dM1A1 | dM1: Uses $\|z+i\|=1$ to find equation connecting $u$ and $v$. **Dependent on all previous method marks.** A1: Correct equation |

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## Question (Complex Numbers - Way 4):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = \frac{3iz-2}{z+i}$ | — | Starting form |
| $z = \frac{wi+2}{3i-w}$ | M1A1 | M1: Attempt to make $z$ the subject; A1: Correct rearrangement |
| $z+i = \frac{wi+2}{3i-w}+i = \frac{wi+2-3-wi}{3i-w}$ | M1 | Applies $z+i$ and finds common denominator |
| $z+i = \frac{-1}{3i-u-iv} \times \frac{u-(v-3)i}{u-(v-3)i} = \frac{u-(v-3)i}{u^2+(v-3)^2}$ | M1A1 | M1: Multiplies by complex conjugate and sets $=1$; A1: Correct equation with no $i$'s in denominator |
| $\frac{\sqrt{u^2+(3-v)^2}}{u^2+(3-v)^2}=1$ | dM1A1 | dM1: Introduces $u,v$ or $x,y$, uses Pythagoras. **Dependent on all previous method marks.** A1: Correct equation |

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## Question (Complex Numbers - Way 5):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = \frac{3iz-2}{z+i}$ | — | Starting form |
| $u+iv = \frac{3i(x+iy)-2}{x+iy+i} = \frac{(3ix-3y-2)(x-(y+1)i)}{x^2+(y+1)^2}$ | M1A1 | M1: Substitutes for $z$ and multiplies by $\frac{x-(y+1)i}{x-(y+1)i}$; A1: Correct expression |
| $= \frac{x+\left(3(x^2+(y+1)^2)-y-1\right)i}{x^2+(y+1)^2}$ | M1 | Express rhs in terms of $x^2+(y+1)^2$ |
| $x^2+(y+1)^2=1 \Rightarrow w = x+(2-y)i$ | M1A1 | M1: Use of $\|z+i\|=1$; A1: $w=x+(2-y)i$ |
| $x^2+(y+1)^2=1 \Rightarrow u^2+(v-3)^2=1$ | dM1A1 | dM1: Attempts equation connecting $u$ and $v$. **Dependent on all previous method marks.** A1: $u^2+(v-3)^2=1$ |

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## Question (Complex Numbers - Way 6):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $w = \frac{3iz-2}{z+i} = \frac{3i(z+i)+1}{z+i} = 3i + \frac{1}{z+i}$ | M1A1 | M1: Attempt rhs in terms of $z+i$; A1: Correct rearrangement |
| $w - 3i = \frac{1}{z+i}$ | M1 | Isolates $z+i$ |
| $\|w-3i\| = \left\|\frac{1}{z+i}\right\| = \frac{1}{\|z+i\|} = 1$ | M1A1 | M1: Applies $\|z+i\|=1$; A1: Correct equation |
| $\|w-3i\|=1 \Rightarrow u^2+(v-3)^2=1$ | dM1A1 | dM1: Introduces $u,v$ or $x,y$, uses Pythagoras. **Dependent on all previous method marks.** A1: $u^2+(v-3)^2=1$ |

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## Question (Complex Numbers - Way 7):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = \frac{wi+2}{3i-w}$ | M1A1 | M1: Attempt to make $z$ the subject; A1: Correct rearrangement |
| $\|w\| = \left\|\frac{3iz-2}{z+i}\right\| = \|3iz-2\|$ | M1 | Uses $\|w\|=\left\|\frac{3iz-2}{z+i}\right\|$ and $\|z+i\|=1$ |
| $\|w\| = \left\|3i\left(\frac{wi+2}{3i-w}\right)-2\right\| = \frac{|-3w+6i-6i+2w|}{|3i-w|}$ | M1A1 | M1: Attempts common denominator; A1: Correct equation |
| $\|w-3i\|=1 \Rightarrow u^2+(v-3)^2=1$ | dM1A1 | dM1: Introduces $u,v$ or $x,y$, uses Pythagoras. **Dependent on all previous method marks.** A1: $u^2+(v-3)^2=1$ |

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