How to graph the inequality $1<|2z-6|<2$ in the complex plane?
My work so far is below:
I assume $z$ is complex so it can be represented as $z=a+bi$
\begin{align}
1<|2z-6|<2 \\
1<2|(a+bi)-3|<2 \\
\frac{1}{2}<|a-3+bi|<1\\
\frac{1}{2}<|a-3+bi|<1 \\
\frac{1}{2}<\sqrt{a^2-6a+9+b^2}<1 \\
\frac{-35}{4}
Graph the inequality $1<|2z-6|<2$.
2
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complex-analysis
2 Answers
2
hint
$$1<|2z-6|<2 \Leftrightarrow\frac{1}{2}<|z-3|<1 \Leftrightarrow \frac{1}{4}<|z-3|^2<1$$
$$\frac{1}{4}<(a-3)^2+b^2<1$$
What is:
$$(a-3)^2+b^2<1?$$
and
$$(a-3)^2+b^2>1/4?$$
what about the intersection?
4
Hint: think geometrically.
Divide the inequality through by $2$. Where will you find the points in the plane whose distance from $3+0i$ is between $1/2$ and $1$?