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I am tasked with finding the region of the complex plane under condition: $$\left|\frac{z-2i}{z+2}\right|\ge 1$$ I can then calculate that $|z-2i|\ge|z+2|$. Thus, I can say I'm looking for the region where the distance from $z$ to $2i$ is greater than the distance from $z$ to $-2$. Imagining the plane, I feel as though that this is the region below and including the line $$z=a-ai, a\in\mathbb{R}$$ But I thought I would try it algebraically. So $$|z-2i|\ge|z+2|$$ $$\sqrt{a^2+[(b-2)i]^2}\ge\sqrt{(a+2)^2+(bi)^2}$$ $$\sqrt{a^2-(b^2-4b+4)}\ge\sqrt{a^2+4a+4-b^2}$$ $$\sqrt{a^2-b^2+4b-4}\ge\sqrt{a^2+4a+4-b^2}$$ From here, I can't assume that taking square roots will allow the inequality to hold, i think. If I didn't, then $$4b-4\ge 4a+4$$ Which doesn't match my intuition... any help?

EDIT: I think I found my error... i mistakenly put the $i$ in the definition of modulus. So... $$\sqrt{a^2+(b-2)^2}\ge\sqrt{(a+2)^2+b^2}$$ $$\sqrt{a^2+b^2-4b+4}\ge\sqrt{a^2+4a+4+b^2}$$ From here I can say that $a\le -b$ and the region can be described as $$\{z=a+bi|a\le -b, a,b\in \mathbb{R}\}$$ Is this correct?

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Given that only the absolute value is of interest

$$\color{red}{\left|\color{black}{\frac{z-2i}{z+2}}\right|}\ge 1$$

and it is of a complex number that is found by division of two other complex numbers

$$\left|\color{red}{\frac{\color{black}{z-2i}}{\color{black}{z+2}}}\right|\ge 1$$

it's the same as looking at the division of the individual absolute values of these two complex numbers.

$$\left|\frac{z-2i}{z+2}\right| = \frac{|z-2i|}{|z+2|}$$

Considering the result 1 as an edge case, this can only happen if both absolute values are the same. Geometrically speaking, this means they have the same distance from the origin.

Both complex numbers (numerator and denominator) have different but constant offsets from the number $z$, which positions them in a constant pattern relative to $z$

enter image description here

Dictated by the symmetry of this pattern, the two numbers $|z+2|$ and $|z-2i|$ can only have the same value if the origin lies on the symmetry line (the black line). Three possible positions of the origin have been added to the picture. Compare the red one with the other two green ones. Clearly, only on the symmetry line will both absolute values be equal and thus the result be 1.

enter image description here

In order to not only find the line that full fills the equality, but rather the region that full fills the inequality, one now only has to reason about what side of the line (left or right) is the one that should be added.

The value of a fraction increases above 1 if denominator < numerator. Or in this case

$$|z+2| < |z-2i|$$

As can be seen in the picture above, this is the case if $z$ moves to the left, which means that all points to the left (under) the symmetry line (which goes through the origin) are part of the solution of the inequality.

Your feeling and algebraic reasoning are both correct.

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  • $\begingroup$ That is nice. I drew a similar picture on graph paper when I was working pencil and paper on the problem. I've realized that, with these types of problems, sometimes the algebra is not as clear, and simply intuiting the problems with pictures really help develop the direction i go. To see you use the pictures was awesome. Thanks for taking the time to do that. $\endgroup$ – Lalaloopsy Sep 13 '15 at 4:34
  • $\begingroup$ @Lalaloopsy glad I could help. I agree that it's nice to find geometric solutions to problems like this. Be it my incapability or the complexity (pun intended) of other problems, but more often than not, I am unable to find one. Knowing a possible solution to work towards greatly improved my confidence in attempting this case. I'm not sure if what I have done would even hold as a proof or is merely showing the plausibility of yours. So don't feel down for not finding that solution geometrically, because you found one algebraically, which is a solution just the same. Good job! $\endgroup$ – null Sep 13 '15 at 9:14
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Your geometric interpretation and your corrected algebraic calculation are correct.

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  • $\begingroup$ Awesome. Thanks for the verification. $\endgroup$ – Lalaloopsy Sep 12 '15 at 17:43

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