# How to plot $\{z \in \mathbb{C} : |z-i|>|z+i|\}$

How would I draw the set $\{z \in \mathbb{C} : |z-i|>|z+i|\}$ and $\{z \in \mathbb{C} : |z-i|\not=|z+i|\}$?

Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no where.

$|z-i|^2>|z+i|^2\\\\(z-i)(\bar z+i)>(z+i)(\bar z-i)\\ z\bar z+1+i(z -\bar z)>z \bar z+1 +i(\bar z -z)\\i(z-\bar z)>i(\bar z-z)$

What would the 'general' method/approach be for drawing the sets?

Edit: How would I draw $\{z \in \mathbb{C} : |z-i|\not=|z+i|\}$?

After a similar calculation using Zev Chonoles' post, I got that $-b\not=b$, hence $z=a+ib$ satisfies $|z-i|\not=|z+i|$ if and only if $-b\not=b$.

For $\{z \in \mathbb{C} : |z-i|>|z+i|\}$

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Write out a complex number $z$ with real and imaginary components, i.e. as $z=a+bi$. Then $$|z-i|=|a+(b-1)i|=\sqrt{a^2+(b-1)^2}$$ $$|z+i|=|a+(b+1)i|=\sqrt{a^2+(b+1)^2}$$ so \begin{align*}|z-i|>|z+i|&\iff\sqrt{a^2+(b-1)^2}>\sqrt{a^2+(b+1)^2}\\&\iff a^2+(b-1)^2>a^2+(b+1)^2\\ &\iff (b-1)^2>(b+1)^2\\ &\iff -2b>2b\\ &\iff b<0\end{align*} Thus, the complex number $z=a+bi$ satisfies $|z-i|>|z+i|$ if and only if $b<0$, i.e. if and only if it lies below the real axis in the complex plane. Thinking geometrically (i.e. with complex numbers as points in the plane), it might also help to note that $$|z-i|=\sqrt{a^2+(b-1)^2}=\text{distance from }(a,b)\text{ to }(0,1)$$ $$|z+i|=\sqrt{a^2+(b+1)^2}=\text{distance from }(a,b)\text{ to }(0,-1)$$

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Thanks, @ZevChonoles . Just to make sure, is my diagram correct for the first graph? Also, how would I draw the 2nd graph? – Derrick May 23 '12 at 23:11
@Derrick: Yup, it's correct - you have the dotted line along the real axis indicating that it is not included in the set, and you've shaded in the right region. Now, note that $-b\neq b$ if and only if $b\neq 0$. Does that help you see the graph for the second question? – Zev Chonoles May 23 '12 at 23:12
Thanks @ZevChonoles, so I take it, it would be shading the entire region with a dotted line along the real axis (see edited post with new picture)? – Derrick May 23 '12 at 23:19
@Derrick: Exactly! – Zev Chonoles May 23 '12 at 23:32
Okay, thanks again for all your help! – Derrick May 23 '12 at 23:33

Try writing $z=x+iy$ with $x,y\in\mathbb{R}$.

Then for the first inequality you get (just try it for yourself):

$$|z-i|>|z+i| \Leftrightarrow \mbox{Im}(z) < 0$$

so the solutions is the whole lower half-plane (without the real axis).

For the second one, you get $\mathbb{C}\setminus\mathbb{R}$, because of a similar condition for the imaginary part ($\mbox{Im}(z)\not=0$).

Also, your last line $i(z-\bar{z})>i(\bar{z}-z)$ translates for $z=x+iy$ into $0>y$, which gives the same result.

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Thanks, most appreciated :) – Derrick May 23 '12 at 23:34

One could simply look at the absolute value of a complex number as it's distance from origin, or in other words, $|a-b|$ is the distance of $a$ from $b$.

Now, in your problem you want find all the points such that their distance from $i$ is more than their distance from $-i$. Well, find all the points who have the same distance from $i$ and $-i$. That's a line (namely the line that is perpendicular to the line segment $\overline{i,-i}$).

So far you've got the solution to the second part of the problem. Now, choose the half-plane which includes $-i$. That is that answer for the first part.

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