The transformation at $T$ given by $w=kz/(i+z)$ where $z\neq -i$, $k$ is a real number, maps the complex number $2+i$ in the $z$-plane to its image $1/2(3-i)$ in the $w$-plane.
a) Show that $k=2$
Point $P$ represents the complex number $z$ where $|z|=\sqrt{3}$. $T$ maps the point $P$ to point $Q$ in the w-plane.
b) Show that the locus of $Q$ is a circle with the cartesian equation given by: $(u-3)^2+v^2=3$ for $u, v\in\mathbb{R}$.
$T$ maps the point $z_0$ in the locus of $P$ to the point $w_0$ in the locus of $Q$, where the acute angle $\arg w_0$ is as large as possible.
c) Find the exact value of $|i+z_0|$
I did part (a) easily, but I couldn't do part (b). So far I did this: $$ iw=z(2-w)\Rightarrow z=iw/(2-w)\Rightarrow |z|=\left|\frac{iw}{2-w}\right|\Rightarrow \sqrt{3}=\frac{|iw|}{|(2-w)|}$$
for part (b), assuming $|z| = \sqrt 3, k = 2$, $$w = {kz \over i + z} = {kz(-i+\bar z) \over (i + z)(-i +\bar z)} = {k|z|^2-ikx + ky \over 1 + |z|^2 +ix + y -ix + y} ={k|z|^2+ ky -ikx \over 1 + |z|^2 + 2y} = {6 + 2y - 2ix \over 4 + 2y}$$
if $w = u + iv,$ then $$u = {3 + y \over 2 + y}, v = {-x \over 2 + y}$$
now we need to solve this for $x, y.$ we find $$y = {2u-3 \over 1 - u}, x = {v \over 1 - u}$$
the fact $x^2 + y^2 = 3$ translates to $$(2u-3)^2 + v^2 = 3(1-u)^2$$ which simplifies to $$(u-3)^2 + v^2 = 3$$
i don't know if there are still some mistakes. again, please check it out.