Can anyone help me visualize and understand how this sort of inequality works? $$\Big\{z:\left|z-1\right|<2\Big\}$$ It's a set of complex numbers, so I know this corresponds to a certain area in the complex plane, but I need some help to draw the corresponding graphs or get a sense of what this area looks like.
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3Remember that $|z|=\sqrt{a^2+b^2}$ so when we talk about $|z|=1$ we essentially have the unit circle. If $z$ is replaced by $z-1$, the circle is shifted $1$ unit to the right. In your case, the radius is $2$ and you can figure out now what the inequality sign does? – imranfat Aug 25 '16 at 17:14
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To put it slightly differently from @imranfat’s explanation, $|z-2|$ is just the distance from $z$ to $2$ in the Gaussian plane, just as $|z-2|$ is the distance from $z$ to $2$ on the real line. – Lubin Aug 25 '16 at 17:42
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The distance between two complex numbers $z_1$ and $z_2$ is $|z_1-z_2|$.
So your set is the set of all complex numbers which are less than $2$ units away from $1$. That is, the open disk of radius $2$ centered at $1$.
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