# Locus of Z as cartesian equation

Could you please help with this locus problem? I think I am aiming for a cartesian equation in terms of $x$ and $y$ that may look like a circle equation e.g. $(x+a)^2 + (y+b)^2$ but I'm not sure.

Given there is a locus of $z$ such that $$\frac{|z-12j|}{|z+36|}=3,$$ then $|z-12j| = 3|z+36|$.

Now I want to write the locus of $z$ as a cartesian equation in terms of $x$ and $y$. Let $z=x+yj$. $$\begin{align*} |x+yj - 12j| &= 3|x+yj+36|\\ |x+(y-12)j| &= 3|(x+36)+yj| \end{align*}$$ Where should I go from here?

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Square both sides and use the fact that $|a+bj|^2 = a^2 + b^2$, then simplify. – Eric Stucky May 28 '12 at 11:12
You can type equations directly, instead of creating images and then linking to them. In fact, it's better not to rely on images. – Arturo Magidin May 28 '12 at 11:46
Thanks @ArturoMagidin I'd like to learn how to type equations, is there some instructions for that somewhere on here? – NSDigital May 28 '12 at 11:50
– Arturo Magidin May 28 '12 at 11:53
Note: $(x+a)^2 + (y+b)^2$ is not an equation. A dead giveaway that it is not an equation is the fact that it does not have an equal sign in it. – Arturo Magidin May 29 '12 at 17:21

## 1 Answer

Just as with real numbers, where squaring removes absolute value bars, you can do something similar with complex numbers. Remember that if $a$ and $b$ are real numbers (as your $x$ and $y$ appear to be, though you never actually said so), we have $$|a+bj| = \sqrt{a^2+b^2},$$ so that $|a+bj|^2 = a^2+b^2$. So from $$|x + (y-12)j| = 3|(x+36)+yj|$$ we get $$|x+(y-12)j|^2 = 9|(x+36)+yj|^2$$ and whence to $$\begin{align*} x^2 + (y-12)^2 &= 9\Bigl((x+36)^2 + y^2\Bigr)\\ x^2 + y^2 - 24y + 144 &= 9\Bigl( x^2 + 72x + 1296 + y^2\Bigr)\\ 0 &= 8x^2 + 648x + 8y^2 + 24y + 11520. \end{align*}$$ Since both sides of your original equation necessarily have the same sign, there is no problem with the possible introduction of spurious solutions, so you'll get a specific conic out of this equation.

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