# Question 6:

## Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{z-i}{z+1} \Rightarrow z = \frac{w+i}{1-w}$ | B1 | Correct rearrangement as shown or equivalent. |
| $z = \frac{w+i}{1-w} = \frac{u+iv+i}{1-u-iv} \times \frac{1-u+iv}{1-u+iv}$ | M1 | Introduces $w = u+iv$ and multiplies numerator and denominator by conjugate of denominator. |
| $z = \frac{u - u^2 + uvi - vi - uvi + i - ui - v^2 - v + iv + i}{(1-u)^2 + v^2}$, $\text{Re}(z) = 0 \Rightarrow u - u^2 - v^2 - v = 0$ | M1A1 | M1: Expands numerator, sets real part to zero. A1: Correct equation in any form. |
| $u - u^2 - v^2 - v = 0 \Rightarrow \left(u - \frac{1}{2}\right)^2 + \left(v + \frac{1}{2}\right)^2 = \frac{1}{2}$ | dM1 | Completes the square for $u$ and $v$. Must attempt completing the square on at least one of $u$ or $v$. |
| Centre: $\left(\frac{1}{2}, -\frac{1}{2}\right)$ Radius: $\frac{1}{\sqrt{2}}$ | A1A1 | A1: Correct centre (accept as coordinates or as $\frac{1}{2} - \frac{1}{2}i$). A1: Correct radius, must be exact. |

## Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{z-i}{z+1} \Rightarrow w = \frac{iy-i}{iy+1}$ | B1 | Replaces $z$ with $iy$. |
| $w = \frac{iy-i}{iy+1} \times \frac{1-iy}{1-iy}$ | M1 | Multiplies numerator and denominator by conjugate. |
| $u + iv = \frac{iy + y^2 - i - y}{1+y^2} = \frac{y^2-y}{1+y^2} + \frac{y-1}{1+y^2}i \Rightarrow \frac{u}{v} = y$ | M1A1 | M1: Rearranges into real and imaginary parts to obtain $y$ in terms of $u$ and $v$ and substitutes. A1: Correct equation. |
| $u - u^2 - v^2 - v = 0 \Rightarrow \left(u-\frac{1}{2}\right)^2 + \left(v+\frac{1}{2}\right)^2 = \frac{1}{2}$ | dM1 | Completes the square. |
| Centre: $\left(\frac{1}{2}, -\frac{1}{2}\right)$ Radius: $\frac{1}{\sqrt{2}}$ | A1A1 | As Way 1. |

## Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 \to -i, \ i \to 0, \ \infty \to 1$ | B1 | Identifies three correct image points on the circle. |
| E.g. Bisector of 0 and 1 is $u = \frac{1}{2}$ | M1 | Finds at least two image points and attempts perpendicular bisector of at least one pair. |
| Bisector of 0 and $-i$ is $v = -\frac{1}{2}$ | M1A1 | M1: Finds three image points and attempts two perpendicular bisectors. A1: Correct bisectors. |
| Bisectors meet at $u = \frac{1}{2}$ and $v = -\frac{1}{2}$ | dM1 | Finds intersection of bisectors. |
| Centre: $\left(\frac{1}{2}, -\frac{1}{2}\right)$ Radius: $\frac{1}{\sqrt{2}}$ | A1A1 | A1: Correct centre. A1: Correct radius. |

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