# Question 7:

## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $z = iy \Rightarrow w = \frac{(1+i)iy + 2(1-i)}{iy - i} = \frac{-y+2+i(y-2)}{y-1}$ | M1 | Correct method to find image line; substitute $z = iy$, rearrange to Cartesian form |
| $\Rightarrow u = v$ or $\operatorname{Im} w = \operatorname{Re} w$ | A1 | Accept $x = y$ if they set $w = x + iy$ |
| | **(2)** | |

## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $w = \frac{(1+i)z + 2(1-i)}{z-i} \Rightarrow z = \frac{2(1-i)+iw}{w-1-i} = \frac{2-v+i(u-2)}{u-1+i(v-1)}$ | M1 | Makes $z$ subject, substitutes $w = u+iv$ |
| Multiplies by complex conjugate of denominator and extracts imaginary part | M1 | Sets imaginary part to zero; do not be concerned about denominator |
| $\Rightarrow (u-1)(u-2)-(2-v)(v-1)=0 \Rightarrow u^2-3u+2+v^2-3v+2=0$ | A1 | Correct equation in $u$ and $v$ |
| $\Rightarrow \left(u-\frac{3}{2}\right)^2 + \left(v-\frac{3}{2}\right)^2 = \frac{1}{2}$ | M1 | Completes the square |
| Centre is $\frac{3}{2}+\frac{3}{2}i$ and radius is $\frac{\sqrt{2}}{2}$ | A1A1 | Accept $\left(\frac{3}{2},\frac{3}{2}\right)$ for centre; allow if only error is in denominator (M1M0A0M1A1A1 possible) |
| | **(6)** | |

**Alt 1 (real axis substitution):**
| Working | Mark | Guidance |
|---------|------|----------|
| Set $z = x$, multiply by complex conjugate $\frac{x+i}{x+i}$ | M1 | |
| Eliminate $x$ from $u$ and $v$ equations | M1 A1 | Correct equation in $u$ and $v$ |
| Complete the square | M1 | |
| Correct centre and radius | A1A1 | |

**Alt 2 (inverse points):**
| Working | Mark | Guidance |
|---------|------|----------|
| $i \to \infty$, $-i \to \frac{-i+1+2-2i}{-2i} = \frac{3}{2}+\frac{3}{2}i$; centre is $\frac{3}{2}+\frac{3}{2}i$ | M1 M1 A1 M1 | |
| $0 \to \frac{2-2i}{-i} = 2+2i$; radius $= \left|\frac{3}{2}+\frac{3}{2}i - 2-2i\right| = \frac{\sqrt{2}}{2}$ | A1 A1 | |

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