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Squaring complex numbers is the same as squaring normal numbers. If you’re familiar with the F.O.I.L. method for squaring (first, outside, inside, last), (a+b)*(c+d) = first + outside + inside + last = (a*c) + (a*d) + (b*c) + (b*d). If (a+b)*(a+b) then you have (a*a) + (a*b) + (b*a) + (b*b) = a^2 + 2*a*b + b^2. (what ddun said)

So in your example: x=y^2+c, where y=z^2+c, z=0, and c is the complex number 0.5-0.75i
. . y=0^2+c=c
. . x=c^2+c
. . . = (0.5 – 0.75i) * (0.5 – 0.75i) + (0.5 – 0.75i)
. . . = (0.5*0.5 – 0.5*0.75i – 0.5*0.75i + 0.75i*0.75i) + (0.5 – 0.75i)
. . . = (0.25 – 0.375i – 0.375i + 0.5625*i^2) + (0.5 – 0.75i)
. . . = (0.25 – 0.375i – 0.375i + 0.5625*(-1)) + (0.5 – 0.75i)
. . . = (0.25 – 0.375i – 0.375i – 0.5625) + (0.5 – 0.75i)
. . . = (-0.3125-0.75i) + (0.5 – 0.75i)
. . . = 0.1875-1.5i (what D.W. and xaiym said)
So in the context of your Z series; Z={0,0.5-0.75i,0.1875-1.5i}.

I should probably point out that determing whether sets are "bounded" or not, is alot of work. I don’t think the mandelbrot set would have been invented without computers. It’s best to build the mandelbrot picture with a computer program.
References :

For example, (3 + 4i)^2 = 3^2 + 2(3)(4i) + (4i)^2
= 9 + 24i + 16(i^2)
= 9 + 24i -16
= -7 + 24i.
References :

So, if we have an imaginary number a + bi, and we want to find (a + bi)^2, all we do is treat it like a binomial, expand, and simplify.

(a + bi)^2 is no more difficult to expand than (x + y)^2. The process is exactly the same.

Since (x + y)^2 = x^2 + 2xy + y^2, we have

(a + bi)^2 = a^2 + 2abi + (bi)^2.

But we’re not done here.
Remember, i is a number (albeit imaginary), so we will perform the same algebraic functions on it like we would any real number.

a^2 is real, so we can leave that alone.
2abi is an imaginary component, but cannot be simplified further.

(bi)^2 is the same as (b^2)*(i^2).

And since i = (-1)^(1/2),

i^2 = -1.

Substituing this for i^2 yields
(bi)^2 = -bi. So our final equation will be

a^2 + 2abi – b^2.

The inital example, (-3 + 4i)^2, would give us:

(-3)^2 + 2(-3)(4)i – (4)^2 = 9-24i-16 = -7 -24i

The problem with x, y, z, and c would be solved as follows:

x = y^2 + c

= (z^2 + c)^2 + c

= (0 + c)^2 + c

= c^2 + c.

Using the formula we just derived, we can now find c^2:

(0.5)^2 + 2(0.5)(-0.75)i – (0.75)^2, which equals

0.25+-0.75i-0.5625.

Adding c and combining our real and imaginary parts gives us:

0.25 + 0.5 – 0.5625 – 0.75i – 0.75i =

0.1875 – 1.5i.
References :