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Let
$$T(z)=\frac{z}{z+1}$$
Find the inverse image of the disk $|z|<1 / 2$ under $T$ and sketch it.
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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 $2 / 3$.
by Gold Status (10,261 points)

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