STEP 1988 Paper 3 Pure Maths Q3: Who is wrong? I have been doing some STEP papers for fun and I have noticed that for the problem in question there is disagreement on the answer in the student room discussion (as there is no official mark scheme for old past papers). The parts where people (including myself) disagree on are the following:
$(ii)$ If $w = \frac{z+i}{z-i}$ what is the locus of $w$ when $z$ is real?
My solution:
$z$ being real implies that $z=x$ for some $x \in \mathbb{R}$.  Thus we have
$$w=\frac{x+i}{x-i}=\frac{x^2-1}{x^2+1}+i \frac{2x}{x^2+1}$$
Suppose $w=a+bi$. Thus $a=\frac{x^2-1}{x^2+1}$ and $b= \frac{2x}{x^2+1}$.  We see from the parametric definition of these variables that $a$ is in the region $[-1,1)$ and similarly that $b$ is in the range $[-1,1]$.
Now I aim to find a relationship between the two.  From the first parametric derivation,  we have $x^2=\frac{a+1}{1-a}$ which implies that
$$b=\frac{2 \times \pm \sqrt{\frac{a+1}{1-a}}}{\frac{2}{1-a}}=\pm \frac{ \sqrt{\frac{a+1}{1-a}}}{\frac{1}{1-a}}.$$
And hence $b^2=(a+1)(1-a)=1-a^2$ which is the circle centered at the origin of radius 1 (with the point $(1,0)$ committed).
$(iii)$ If $w=\frac{z+i}{z-i}$ what is the locus of $w$ when $z$ is imaginary?
My solution:
If $z$ is imaginary then $z=iy$ for some $y \in \mathbb{R}$. And thus
$$w=\frac{yi+i}{yi-i}=\frac{y+1}{y-1}.$$
Therefore $w$ is real and its real coefficient $a$ is given parametrically by $a=\frac{y+1}{y-1}$ which therefore implies that $w$ is the real axis with the point $(1,0)$ ommited.
Supposed Answers provided by a user at the STR
Both of these answers are in disagreement with the pinned answer at the STEP paper discussion thread. Here is their answer

My question: Who is right and if I am wrong why am I wrong?
 A: Before we do anything, we can rule out some definitely wrong answers. The transformation $z \mapsto \frac{z+i}{z-i}$ is a Möbius transformation. So it takes lines in the complex plane to either lines or circles. Any locus that isn't a line or a circle is wrong: no "two axes" or "hyperbola" are allowed.
(Sometimes, we also get a missing point somewhere that corresponds to the "point at $\infty$" of a line, which is where the missing point in your work comes from.)
To be honest, your work is pretty solid on figuring out what the locus actually is. So I'm just going to say "yeah, that looks right" and say how to solve the problem without doing that algebra.
If we know the answer is a circle or line, we just have to figure out three points of the locus, and then we know what it is. So:

*

*$\frac{0+i}{0-i}=-1$, $\frac{1+i}{1-i}=i$, and $\frac{2+i}{2-i} = \frac35 + \frac45i$ all lie on the unit circle, therefore the locus in the first case is the unit circle. The missing point is $(1,0)$, the limit of $\frac{z+i}{z-i}$ as $z \to \infty$.

*$\frac{0+i}{0-i}=-1$, $\frac{2i+i}{2i-i}=3$, and $\frac{-i+i}{-i-i}=0$ all lie on the real axis, therefore the locus in the second case is the real axis. The missing point is again $(1,0)$ for the same reason.

