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I need to sketch this region $\left \{ z\in\mathbb{C}| |z-i|\leq |z-1| \right \}$. I'd like some assistance with solving this inequality because I think that's where I'm going wrong.

To solve the inequality I'm squaring both sides and trying to solve for that. Similar to this post. $$(z-i)^2 \leq (z-1)^2$$ $$0\leq (z-1)^2 - (z-i)^2$$ $$0\leq ((z-1) - (z-i)) ((z-1) + (z-i))$$ $$0\leq (-1+i)(2z-1-i)$$ $$0\leq -2z+2zi+-i^2+1$$

Here is the point where I get stuck. I'm not quite sure how to progress from here.

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  • $\begingroup$ Your first inequality's already wrong. We're dealing here with complex numbers, so their module is not to be taken as in the real case. It could help you to remeber this to take into account that in the complex plane the is no linear order: no complex number is greater than or smaller than any other. $\endgroup$
    – Timbuc
    Jun 17, 2015 at 7:50
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    $\begingroup$ Note that $$i^2=-1.$$ But $$|i|^2=1.$$ $\endgroup$
    – Bumblebee
    Jun 17, 2015 at 7:50

2 Answers 2

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note that : $z=x+iy \rightarrow \space |z|=\sqrt{x^2+y^2}$ $$|z-i|<|z-1|\\$$put $z=x+iy$ $$ |x+iy-i|<|x+iy-1|\\|x+i(y-1)|<|(x-1)+iy|\\\sqrt{x^2+(y-1)^2}<\sqrt{(x-1)^2+y^2}$$ now go on when you simplify $$x^2+(y-1)^2 <(x-1)^2+y^2\\-2y<-2x\\2x<2y\\x<y$$enter image description here

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  • $\begingroup$ You should allow for equality as well. $\endgroup$
    – Macavity
    Jun 17, 2015 at 15:41
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The geometric interpretation of your inequality is:

The distance from $z$ to $i$ is less than or equal to the distance from $z$ to $1$.

Draw the perpendicular bisector to the segment from $1$ to $i$ and think about what this means.

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