Given $Z=\frac{4-z}{4+z}$, find the locus of $Z$ if $|z|=4$ Given $Z=\frac{4-z}{4+z}$, find the locus of $Z$ if $|z|=4$
I tried letting $z=x+iy$ and subbing into $Z=\frac{4-z}{4+z}$, rationalising the denominator but I always end up with $\frac{3-2yi}{5+2x}$ and I don't know how to find the locus from that. Am I doing something wrong?
Thanks so much in advance
 A: Write $Z=-1+\frac{8}{4+z}$. The points satisfying $|z|=4$ is a circle with center at the origin and radius $4$. That means that those points applying $z+4$ is a circle of the same radius and center at $4$. This is a circle that passes through the origin. If you apply to them the function $1/z$ they become a perpendicular line passing through $1/8$. If you multiply the result by $8$ it becomes a perpendicular line passing through $1$. Finally, if you subtract $1$ it becomes the $Y$ axis.
A: You can rewrite $w=Z$, then $$z={4(1-w)\over 1+w}\;\;\; \Longrightarrow \;\;\; |z|=4\Big|{1-w\over 1+w}\Big|$$ 
so $|w-(-1)|=|w-1|$ so $w$ is equally distance from $-1$ and $1$, thus $w$ describes $Y$-axis.
A: Let $z =4 e^{i \theta}$, $ \quad \theta \in (-\pi,\pi)$ (as $\theta =\pm \pi$ will make denominator zer0)
$$Z=\frac{1-e^{i \theta}}{1+e^{i \theta}} \\= \frac{e^{-\frac{i \theta}{2}}-e^{\frac{i \theta}{2}}}{e^{-\frac{i \theta}{2}}+e^{i \frac{\theta}{2}}}
$$
Use $\cos \phi=\frac{e^{i \phi}+e^{-i \phi}}{2}, \quad \sin \phi=\frac{e^{i \phi}-e^{-i \phi}}{2 i}$
To get, $$ Z=-i \tan\left(\frac{\theta}{2}\right)$$
$\tan\left(\frac{\theta}{2}\right)$ stretches over all real numbers, hence locus of $Z$ is the whole imaginary axis.
