## Question 2:

$z = x+iy$ and $w = u+iv$ used.

$z = x \Rightarrow w = \frac{1-ix}{x}$ | M1 | Replaces at least one $z$ with $x$, ie indicate that $y=0$ (may be done later)

$w = \frac{1}{x} - i$ or $w = \frac{1-ix}{x}$ oe | A1 | Reach this statement somewhere

$u + iv = \frac{1}{x} - i$ | M1 | $w=u+iv$ and equating real or imaginary parts to obtain either $u$ or $v$ in terms of $x$ or just a (real) number

$v = -1$ oe $\left(u = \frac{1}{x}$ need not be shown$\right)$ | A1 | $v=-1$ or $v+1=0$ oe, ie equation of the line

**NB:** If $x+iy$ has been used for $z$ and then also for $w$ allow M1A1M1A0 max.

**ALT 1:**

$z = \frac{1}{w+i} = \frac{1}{u+iv+i} = \frac{u-i(v+1)}{u^2+(v+1)^2}$ | M1 | Multiplies numerator and denominator by complex conjugate

$\frac{u-i(v+1)}{u^2+(v+1)^2}$ | A1 |

$(y=0 \Rightarrow) \frac{(v+1)}{u^2+(v+1)^2} = 0 \Rightarrow v+1=0$ | M1 | Uses $y=0$ and equates real or imaginary parts to obtain either $u$ or $v$ in terms of $x$ or just a number

$v=-1$ or $v+1=0$ oe | A1 |

**NB1:** If $x+iy$ has been used for $z$ and then also for $w$ allow M1A1M1A0 max.

**NB2:** M1A0M1A1 is possible

**ALT 2:** $|z+i| = |z-i|$

$\left|\frac{1}{w+i}+i\right| = \left|\frac{1}{w+i}-i\right|$ | M1 A1 | M1: Use of real line and attempt to substitute. A1: Correct substitution

$\left|\frac{1+wi-1}{w+i}\right| = \left|\frac{1-wi+1}{w+i}\right|$

$|wi| = |2-wi|$ | M1 | Common denominator and equate numerators

$|w| = |w+2i|$ | A1 | Equation of the line – any form accepted

**ALT 3:** $z = \frac{1}{w+i}$

$z$ lies on real axis $\Rightarrow \frac{1}{w+i}$ is real | M1 | Re-arrange equation and state that $\frac{1}{w+i}$ is real

$\Rightarrow w+i$ is real | A1 | Deduce that $w+i$ is real

$w = u+iv$, $u+i(v+1)$ is real | M1 | Replace $w$ with $u+iv$ (any letters inc $x+iy$ allowed here)

$v+1=0$ | A1 | Deduce equation of the line

**ALT 4:** Choose any 2 points on the real axis in the $z$-plane:

$z=a$: $w_a = \frac{1-ia}{a}$ | M1 | Any one point

$z=b$: $w_b = \frac{1-ib}{b}$ | A1 | Any two points

$w_a = \frac{1}{a}-i \quad w_b = \frac{1}{b}-i$ | M1 | Simplify both

$v=-1$ oe | A1 | Any letter (inc $y$) allowed here

**NB:** The work can be done using arguments to find the equation. If seen, send to review.

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