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I have this complex equation $|z+2i|=| z-2 |$. How can i resolve it? Please help me

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3 Answers

up vote 2 down vote accepted

We have $|z+2i|^2=| z-2 |^2$, which implies that $(z+2i)\overline{(z+2i)}=(z-2)\overline{(z-2)}$, that is $(z+2i)(\overline{z}-2i)=(z-2)(\overline{z}-2)$. This implies that $$z\overline{z}-2iz+2i\overline{z}+4=z\overline{z}-2z-2\overline{z}+4,$$ that is $-iz+i\overline{z}=-z-\overline{z}$, or $$i(z-\overline{z})=z+\overline{z}.$$ Now if we write $z=a+bi$, we get $i(2bi)=2a$, or $b=-a$.

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Let $z=a+bi$ be a complex number that satisfies $$|z+2i|=|z-2|.$$ Remember that for any complex number $c+di$, the definition of $|c+di|$ is $$|c+di|=\sqrt{c^2+d^2}$$ So, we have that $$|z+2i|=|(a+bi)+2i|=|a+(b+2)i|=\sqrt{a^2+(b+2)^2}$$ and $$|z-2|=|(a+bi)-2|=|(a-2)+bi|=\sqrt{(a-2)^2+b^2}.$$ Are you able to solve for the values of $a$ and $b$ such that the two expressions above are equal?

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See also: en.wikipedia.org/wiki/Apollonian_circles –  t.b. Nov 23 '11 at 10:13
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The geometric way

The points $z$ that satisfy the equation are at the same distance of the points $2$ and $-2\,i$, that is, they are on the perpendicular bisector of the segment joining $2$ and $-2\,i$. This is a line, whose equation you should be able to find.

The algebraic way

When dealing vith equations with $|w|$, it is usually convenient to consider $|w|^2=w\,\bar w$. In your equation, if $z=x+y\,i$, this leads to $x^2+(y+2)^2=(x-2)^2+y^2$, whose solution I'll leave to you.

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I really like solving these kind of equations geometrically. Keeping in mind the geometrical interpretation of complex numbers can often help when solving exercises. –  Beni Bogosel Nov 23 '11 at 10:36
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