Solution. Let $D=\{|z|<1 / 2\}$. The inverse image of $D$ under $T$ is

$$

\begin{aligned}

T^{-1}(D) &=\{z \in \mathbb{C}: T(z) \in D\}=\left\{|T(z)|<\frac{1}{2}\right\} \\

&=\left\{z:\left|\frac{z}{z+1}\right|<\frac{1}{2}\right\}=\{2|z|<|z+1|\}

\end{aligned}

$$

Let $z=x+y i$. Then

$$

\begin{aligned}

2|z|<|z+1| & \Leftrightarrow 4\left(x^{2}+y^{2}\right)<(x+1)^{2}+y^{2} \\

& \Leftrightarrow 3 x^{2}-2 x+3 y^{2}<1 \\

& \Leftrightarrow\left(x-\frac{1}{3}\right)^{2}+y^{2}<\frac{4}{9} \\

& \Leftrightarrow\left|z-\frac{1}{3}\right|<\frac{2}{3}

\end{aligned}

$$

So

$$

T^{-1}(D)=\left\{z:\left|z-\frac{1}{3}\right|<\frac{2}{3}\right\}

$$

is the disk with centre at $1 / 3$ and radius $2 / 3$.