# Question 8:

## Part (a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\omega = a + \mathrm{i}b$ | **M1** | Writing $\omega$ in a form which allows 2 equations to be found |
| $a + 2 = 3a$; $b + 7 = -3b - 1$ | **M1** | Equating real and imaginary parts |
| $a = 1$ | **A1** | |
| $\omega = 1 - 2\mathrm{i}$ | **A1** | Need to see answer as a complex number. Ignore presence of $\omega^* = 1+2\mathrm{i}$ |

## Part (b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| If $z$ is purely imaginary, so $z = k\mathrm{i}$ for some real $k$, then $z^* = -k\mathrm{i} = -z$ as required | **B1** | $\Leftarrow$. This could, with care, be included in the $\Rightarrow$ proof |
| If $z = r + s\mathrm{i}$ ($r$, $s$ real) then $z^* = r - s\mathrm{i}$ so $z = -z^* \Rightarrow r + s\mathrm{i} = -(r-s\mathrm{i}) = -r + s\mathrm{i} \Rightarrow r = 0$ (so $s \neq 0$ since $z$ is non-zero) so $z$ is purely imaginary | **B1** | $\Rightarrow$. $z$ being non-zero does not have to be rigorously dealt with |

## Part (c)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Reflection in the real axis | **B1** | Must be real, rather than $x$, axis |

## Part (c)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $z = z^*$ means $A$ and $B$ are coincident so $A$ is invariant, so $A$ must lie on the mirror line (real axis), so $A$ represents a purely real number so $z$ is purely real | **B1** | $\Rightarrow$. Could, with care, be included in $\Leftarrow$ proof. Needs to be a geometric explanation |
| If $z$ is purely real then $A$ lies on real axis so invariant under reflection in real axis so $z^*$ is represented by same point so $z = z^*$ | **B1** | $\Leftarrow$ |