## Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $z = x$, $w = \dfrac{x + 2\text{i}}{\text{i}x}$ | M1A1 | M1: for replacing at least one $z$ with $x$ (showing appreciation that $y = 0$); A1: $w = \dfrac{x+2\text{i}}{\text{i}x}$ |
| $w = \dfrac{1}{\text{i}} + \dfrac{2\text{i}}{\text{i}x}$ | | |
| $u + \text{i}v = -\text{i} + \dfrac{2}{x}$ | | |
| $u = \dfrac{2}{x}$, $v = -1$ | M1 | for writing $w$ as $u + \text{i}v$ and equating real or imaginary parts to obtain either $u$ or $v$ in terms of $x$, or just a number |
| $\therefore w$ is on the line $v + 1 = 0$ | A1 | for giving the equation of the line $v + 1 = 0$; oe — must be in terms of $v$ |
| **Total** | **4 marks** | |

# Question 1 (Alternative 1):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $w = \frac{x + iy + 2i}{i(x+iy)}$ | | Replacing $z$ with $x+iy$ |
| $w = \frac{x+iy+2i}{-y+ix} \times \frac{-y-ix}{-y-ix}$ | | |
| $w = \frac{(x+i(y+2))(-y-ix)}{y^2+x^2}$ | | |
| $w = \frac{2x - i(x^2+y^2+2y)}{y^2+x^2}$ | | |
| $w = \frac{2x-ix^2}{x^2} = \frac{2}{x} - i$ using $y=0$ | M1A1 | First M1 awarded here; A1 if correct even if unsimplified **but denominator must be real** |
| $v = -1$ | M1A1 | M1, A1 as in main scheme |

# Question 1 (Alternative 2):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $z = \frac{2i}{iw-1}$ | | Writing the transformation as a function of $w$ |
| $z = \frac{2i}{i(u+iv)-1}$ | | |
| $z = \frac{2i}{(-v-1)+iu} \times \frac{(-v-1)-iu}{(-v-1)-iu}$ | | |
| $z = \frac{2u}{(-v-1)^2+u^2} + i\left(\frac{2(-v-1)}{(-v-1)^2+u^2}\right)$ | | |
| $\frac{2(-v-1)}{(v-1)^2+u^2} = 0$ or $-2(v+1)=0$ using $y=0$ | M1A1 | First M1 awarded here; A1 if correct even if unsimplified **but denominator must be real** |
| $v = -1$ | M1A1 | M1, A1 as in main scheme |

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