I encountered the following STEP III ($2011,Q8$) problem:
The complex numbers z and w are related by
$$w=\frac{1+iz}{i+z} \tag{1}$$
Let $z=x+iy$ and $w=u+iv$, where $x$, $y$, $u$ and $v$ are real.
I was first asked to express $u$ and $v$ in terms of $x$ and $y$:
$$u=\frac{2x}{x^2+(1+y)^2} \tag{2a}$$
$$v=\frac{x^2-(1-y^2)}{x^2+(1+y)^2} \tag{2b}$$
Then
Find the locus of w when the locus of z is the line $y=1$, $-\infty<x<\infty$
For which I plugged in $y=1 \to (2a),(2b)$, rearranging gives
$$u^2+\left(v-\frac{1}{2}\right)^2=\frac{1}{4} \tag{3}$$
which is a circle with the centre $(0,\frac{1}{2})$ of radius $\frac{1}{2}$
However, it seems that I missed a subtlety here as the mark scheme used a very different approach:
Let $$x=2 \tan\left(\frac{\theta}{2}\right), y=1 \tag{4}$$
so as $-\infty<x<\infty$, $-\pi<\theta<\pi$, then
$$u=\frac{1}{2}\sin(\theta) \tag{5a}$$
and
$$v=\frac{1}{2}(1-\cos(\theta)) \tag{5b}$$
so the locus is the circle $$u^2+\left(v-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2 \tag{6}$$
excluding the point $\theta=\pi$, which is $(0,1)$.
I understand that at $(0,1)$ and hence $\theta=\pi$, $x=2\tan\left(\dfrac{\theta}{2}\right)$ is undefined, but why should this additional constraint be considered? Why should we not use $(3)/(6)$ right away for the locus?
Best Answer
We get $$z=\frac{1-iw}{w-i}, w\ne i~~~~(1)$$ The locus of $z$ is $$y=1 \implies z-\bar z=2i~~~~(2)$$ $$\implies \frac{1-iw}{w-i}-\frac{1+i\bar w}{\bar w+i}=2i \implies2 w\bar w-i\bar w+i w =0 \implies u^2+v^2-v=0$$ $$\implies u^2+(v-1/2)^2=1/4~~~(2)$$ So the locus of $w$-points is a circle without the point $(0,1)$ (see Eq. (1) as it makes $w=i$) in $u-v$ plane. This confirms OP's point in his method.
The resolution of OPs question: But in the earlier part of the question $$u=\frac{2x}{x^2+(1+y)^2}~ \text{and} ~ v=\frac{x^2-(1-y^2)}{x^2+(1+y)^2}.$$ When $x=0, y=1$ we get $u=0, v-0$ a point which lies on the locus (2).
But it is the asymptotic value of $x=\infty$ that gives $u=0,v=1$, hence (0,1) cannot lie on (2).