# Question 8:

$$w=\frac{z+3\text{i}}{1+\text{i}z}$$

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z=\frac{w-3\text{i}}{1-\text{i}w}$ | M1A1 | M1: Attempt to make $z$ the subject; A1: Correct equation |
| $|z|=1\Rightarrow\left|\frac{w-3\text{i}}{1-\text{i}w}\right|=1\Rightarrow|w-3\text{i}|=|1-w\text{i}|$ | M1 | Uses $|z|=1$ and introduces $u+\text{i}v$ (or $x+\text{i}y$) for $w$ |
| $u^2+(v-3)^2=u^2+(v+1)^2$ | M1 | Correct use of Pythagoras on either side |
| $v=1$ | A1 | $v=1$ or $y=1$ |

**Alternative 1 for (a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. $w(1)=\frac{1+3\text{i}}{1+\text{i}}=2+\text{i}$ | M1A1 | M1: Maps one point on circle using given transformation; A1: Correct mapping |
| e.g. $w(-\text{i})=\frac{2\text{i}}{2}=\text{i}$ | M1 | Maps a second point on the circle |
| $v=1$ | M1A1 | M1: Forms Cartesian equation using their 2 points; A1: $v=1$ or $y=1$ |

**Alternative 2 for (a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z=\frac{w-3\text{i}}{1-\text{i}w}$ | M1A1 | M1: Attempt to make $z$ the subject; A1: Correct equation |
| $|z|=1\Rightarrow|w-3\text{i}|=|1-w\text{i}|=|w+\text{i}|=|w-(-\text{i})|$ | M1 | Uses $|z|=1$ and changes to form $|w-...|=|w-...|$ or draws diagram |
| Perpendicular bisector of points $(0,3)$ and $(0,-1)$ | M1 | Uses correct geometrical approach |
| $v=1$ | A1 | $v=1$ or $y=1$ |

## Alternative 3 for (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $z = x + iy$, $|z| = 1 \Rightarrow x^2 + y^2 = 1$ | | |
| $w = \frac{z+3i}{1+iz} = \frac{x+iy+3i}{1+i(x+iy)} = \frac{x+i(y+3)}{(1-y)+ix}$ | | |
| $w = \frac{x+i(y+3)}{(1-y)+ix} \times \frac{(1-y)-ix}{(1-y)-ix}$ | M1 | Substitute $z = x+iy$ and multiply numerator and denominator by complex conjugate of their denominator |
| $w = \frac{x(1-y)-ix^2+i(y+3)(1-y)-i^2x(y+3)}{(1-y)^2-ix(1-y)+ix(1-y)-i^2x^2}$ | | |
| $w = \frac{[x(1-y)+x(y+3)]+i[-x^2+(y+3)(1-y)]}{(1-y)^2+x^2}$ | M1 A1 | M1: Multiply out and collect real and imaginary parts in numerator. Denominator must be real. A1: all correct |
| $w = \frac{[x-xy+xy+3x]+i[-x^2+y-y^2+3-3y]}{1-2y+y^2+x^2}$ | | |
| $w = \frac{[4x]+i[-1+3-2y]}{2-2y}$ | M1 | Applies $x^2 + y^2 = 1$ |
| $w = \frac{4x+i[2-2y]}{2-2y} = \frac{4x}{2-2y}+i$ | | |
| $y = 1$ | A1 | $y=1$ or $v=1$ |

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## Question (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $|w|=5 \Rightarrow \left|\frac{z+3i}{1+iz}\right|=5 \Rightarrow |z+3i|=5|1+iz|$ $\therefore |x+iy+3i|=5|(x+iy)i+1|$ | M1 | Uses $|w|=5$ and introduce "$x+iy$" |
| $x^2+(y+3)^2=25\left(x^2+(1-y)^2\right)$ | M1A1 | M1: Correct use of Pythagoras, allow 25 or 5. A1: Correct equation |
| $x^2+y^2-\frac{7}{3}y+\frac{2}{3}=0$ | | |
| $x^2+\left(y-\frac{7}{6}\right)^2=\frac{25}{36}$ | M1 | Attempt circle form or attempt $r^2$ from the line above |
| $a=0,\ b=\frac{7}{6},\ c=\frac{5}{6}$ | A1, A1 | A1: 2 correct. A1: All correct |
| Or for last 3 marks: $\left|z-0-\frac{7}{6}i\right|=\frac{5}{6}$ | M1A1A1 | If 0 not shown score M1A1A0. No need to list $a$, $b$, $c$ separately if answer in this form. |

**Total: 11 marks (6 for part b)**