### The filled Julia set of *x*^{2} + 0

^{2}+ 0

To work out the filled Julia set of *x ^{2} + 0 = x^{2}* we need to look at all the points on the complex plane and decide whether or not their orbits escape to infinity. Those whose orbit doesn't escape are part of the filled Julia set.

Let's start with a complex number *a + ib* and let's assume that its distance from the complex number 0 is greater than 1. On the plane, the number *a + ib* is represented by the point with co-ordinates *(a,b)* and 0 is the point with co-ordinates *(0,0)*.

This means that *a + ib* has distance $\sqrt{a^2+ b^2}$ from 0, and so, by our assumption, we have $$\sqrt{a^2+ b^2} > 1.$$

Now using *a + ib* as the seed *x _{0}* we get $$x_1 = (a + ib)^2 = a^2- b^2 + i2ab,$$ So the distance of

*x*to 0 is $$ \sqrt{(a^2- b^2)^2 + 4a^2b^2}.$$ So is

_{1}*x*closer to 0 than

_{1}*x*or further away? Well, we have $$ (a^2 - b^2)^2 + 4a^2b^2 \, \, = \,\, a^4 - 2a^2 b^2 + b^4 + 4a^2b^2\, \, = \,\, a^4 + 2a^2b^2 + b^4\, \, = \,\,(a^2 + b^2)^2.$$ Hence, the distance of

_{0}*x*to 0 is $a^2 + b^2.$ But since $\sqrt{a^2 + b^2} > 1$ we know that $\sqrt{a^2 + b^2} < a^2 + b^2,$ and so

_{1}*x*is further away from 0 than

_{1}*x*. Repeating this argument shows that

_{0}*x*is further away from 0 than

_{2}*x*,

_{1}*x*is further away from 0 than

_{3}*x*, and so forth. In other words, points in the orbit of

_{2}*x*move further and further out — the orbit tends to infinity. This means that

_{0}*x*does not lie in the filled Julia set. And since

_{0}*x*represented

_{0}*any*point with distance greater than 1 from 0, we know that no such point can lie in the filled Julia set.

A very similar calculation shows that if the distance between *x _{0} = a + ib* and 0 is less than 1, then

*x*is closer to 0 than

_{1}*x*, and this means that the orbit cannot possibly tend to infinity. So points whose distance is to 0 is less than 1 lie in the filled Julia set.

_{0}And what if the distance between *x _{0} = a + ib* and 0 is equal to 1? A calculation shows that the distance between

*x*and 0 is also equal to 1. Thus, points in the orbit of

_{1}*x*always remain at distance 1 from 0 — the orbit does not tend to infinity and therefore

_{0}*x*lies in the filled Julia set.

_{0}We've now accounted for all the points on the plane and seen that only those whose distance to 0 is less than or equal to 1 belong to the filled Julia set. The filled Julia set, then, is the disc in the plane with centre 0 and radius 1.

If you are familiar with complex numbers the calculation is actually easier. In this case you will know that a complex number *x _{0} = x + iy* can be written as

*x*, where

_{0}= re^{iθ}*r*is its distance to the point 0 and

*θ*is the angle that the line from 0 to

*x*makes with the

_{0}*x*-axis. Now

*x*so

_{1}= r^{2}e^{2iθ},*x*has distance

_{1}*r*from 0. This means that

^{2}*x*is further a way from 0 than

_{1}*x*precisely when

_{0}*r > 1*.