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Describe all points in the complex plane that verify these inequalities:

$| z + 3 i | \geq 3 | z − i |$ & $| z - 3/2 i | \geq 1$

The answer is supposed to be the region between two disks of center $(0, 3/2)$ and of radius $1$ and $3/2$ respectively.

I tried squaring both sides of each inequalities but in vain. I have no idea how to get there.

Kevin Long
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jen mueller
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  • \begin{align} \lvert z + 3i \rvert \ge 3 \lvert z - i \rvert &\implies \lvert x + i(y + 3) \rvert \ge 3 \lvert x + i(y-1) \rvert \ &\implies \sqrt{x^{2} + (y + 3)^{2}} \ge 3 \sqrt{x^{2} + (y-1)^{2}} \end{align} – Matthew Cassell Mar 23 '18 at 02:50
  • Square both sides then substitute $z=x+iy$. – David Mar 23 '18 at 02:50
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    The equalities give two circles, first one being an Apollonian circle. Inequalities give the intersection between the regions interior/exterior to those circles. – dxiv Mar 23 '18 at 02:55

1 Answers1

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Let $z=x+yi$

Hence $$x^2+(y+3)^2\ge 9x^2+9(y-1)^2$$ $$\Rightarrow 0\ge 8x^2+8y^2-24y=x^2+y^2-3y$$ Which is a disc with centre $(0,3/2)$ and radius $3/2$

And $$x^2+(y-3/2)^2\ge 1\Rightarrow x^2+y^2-3y+5/4\ge 0$$

Hence the answer follows

Rohan Shinde
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