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This question requires finding the Cartesian equation for the locus:

$|z-1| = 2|z+1|$

that is, where the modulus of $z -1$ is twice the modulus of $z+1$

I've solved this problem algebraically (by letting $z=x+iy$) as follows:

$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$

$(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$

$x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$

$3x^2 + 10x + 3y^2 = -3$

$x^2 + \frac{10}{3}x + y^2 = -1$

$(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$

therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle.

However, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?

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  • $\begingroup$ I always liked to link complex number problems to geometry. Not only it simplifies problems sometimes, but gives you an insight to what's really going on. $\endgroup$ – Beni Bogosel Jun 13 '11 at 16:09
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View the point $z$ as a vector in $\mathbb R^2$. Without doing any calculations, we know that $\lVert z - (1,0) \rVert^2$ and $\lVert z - (-1,0) \rVert^2$ are of the form $z^Tz + b^T z + c$ for some $b$ and $c$. So the equation of the locus is $$(z^Tz + b_1^Tz + c_1) = 4(z^Tz + b_2^Tz + c_2),$$ or $$3z^Tz + b_3^Tz + c_3 = 0,$$ which is the equation of a circle.

In general, the form $x^TAx + b^Tx + c$ defines a quadric, and if you know something about the properties of the matrix $A$, it can tell you what the shape of the quadric is. In this case, $A$ is a multiple of the identity, so it's a circle/sphere/hypersphere.

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A perhaps easier way to see that it is a circle is to rewrite as

$\displaystyle |z'| = 2|z' + 2|, \text{where } z' = z -1$

And so

$\displaystyle 1/4 = |1/2 + 1/z'|$

Thus $\displaystyle \frac{1}{z'}$ gives a circle centred at $(-1/2,0)$ and radius $1/4$.

As we can see from the answers and comments to this recent question: If $0$, $z_1$, $z_2$ and $z_3$ are concyclic, then $\frac{1}{z_1}$,$\frac{1}{z_2}$,$\frac{1}{z_3}$ are collinear that, the map $\displaystyle z \to \frac{1}{z}$ maps circles to circles, if the circle does not pass through the origin.

In this case, it does not pass through the origin, and so the original curve you seek would be a circle too.

Of course the fact that $\displaystyle z \to \frac{1}{\bar{z}}$ is inversion with respect to the unit circle probably uses the Appollonius theorem, but it looks like a useful fact to keep in mind.

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Just to add on to Aryabhata's comment above. The map $f(z) = \frac{1}{z}$ for $ z \in \mathbb{C} -\{0\}$, $f(0) = \infty$ and $f(\infty) = 0$ is a circle preserving homeomorphism of $\bar{\mathbb{C}}$. To see this, one needs to prove that it is continuous on $\bar{\mathbb{C}}$, and since $f(z)$ is an involution proving this would mean that its inverse is continuous as well. It is also not hard to show that $f(z)$ is bijective.

Lastly use the general equation of a circle in $\bar{\mathbb{C}}$ to see that circles in $\bar{\mathbb{C}}$ are preserved under this map.

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