The following list is what I have noticed from the given GeoGebra interactivity:
1) When the radius and z2 are fixed, the distance between z1 and z3 and the angle created by the moduli of the two points are always consistent. For example, if I keep the radius to be 2 and z2 on the imaginary axis on 0+2i. The angle created by z1 and z3 will always be 90 degrees and the distance between the two points is always be 2 root5 units.
Proof that angle remains constant: Keep z1 on 2+0i, this way, the angle created by z1 and z3 will simply be the argument of z3 (since that is the angle made by the real axis and the line and z1 will be on the real axis).
Considering the position of z1 and z2, z3 will be on:
(2+0i)(0+2i) = 0+4i
Since z1 is on the real axis and z3 is on the imaginary one, and we know that the axes are perpendicular to each other, we know that z1 and z3 create a 90-degree angle.
Now, if we move z1 to 0+2i, z3 moves to -4+0i. This time z1 is on the imaginary axis and z3 is on the real one, due to the same reasons, we know that z1 and z3 are still at a 90-degree angle.
Regardless of where we place z1, as long as z2 stays fixed, the angle will remain constant.
Proof that the length remains constant: when z1 was on 2+0i and z3 was on 0+4i, the distance between the two points was:
root(2^2 + 4^2) = root(4+16) = root20 = 2 root5.
When z1 is moved to 0+2i, z3 moves to -4+0i, the distance remains constant:
root(2^2 + (-4)^2) = root(4+16) = root20 = 2 root5
Again, regardless of where z1 is, the distance between the two points remains constant.
2) When z2 is fixed and z1 is kept in one position on the circle (i.e: z1 only moves in a straight line as the radius of the circle of the radius increases or decreases), the radius simply acts as the multiplier to the product of z2 and the original z1.
Let z1 initially be on 0-1i (which means that the circle radius is 1) and z2 be on -2+2i, z3 will be on
(0-1i)(-2+2i) = 2+2i
Increasing the circle radius to 2 means that z1 is now at 0-2i (which is basically 2(0-i) )
The new position of z3 is
(0-2i)(-2+2i) = 4+4i
Notice that 4+4i = 2(2+2i) thus showing that the radius acted as a multiplier to the product of z2 and the original z1.
Therefore, z3 = z2 * original z1 * (new radius/original radius).
3) When the radius and z1 are fixed, z3 moves away from z2 at a constant rate. The distance between the two points increases by the same value when either the real or the imaginary value increases by 1. The said value is the distance between the two points when either the real or the imaginary value is 1.
Proof: Let radius = 1.5, and z1 = 0+1.5i
When z2 = -1+i, z3 = -1.5-1.5i, distance is (root26)/2
z2 = -2+21, z3 = -3-3i, distance is root26 (previous distance + (root26)/2)
z2 = -3+3i, z3 = -4.5-4.5i, distance is 3(root26)/2 (again, previous distance + (root26)/2)
Now let's change the radius to 2, so z1 moves to 0+2i
z2 = -1+i, z3 = -2-2i, distance = root10
z2 = -2+2i, z3 = -4-4i, distance = 2 root10 (previous distance + root10)
z2 = -3+3i, z3 = -6-6i, distance = 3 root 10 (previous distance + root10)
Finally, let radius=1.5, and z1 is 1.5+0i (There is no reason for which I have moved z1 from the imaginary axis to the real one)
z2 = -3+i (The purpose of this section of the proof is to show that the statement above stands even if the imaginary value is not equal to the real value), z3 = -4.5+1.5i, distance = (root10)/2
z2 = -6+2i, z3 = -9+3i, distance = root 10 (previous distance + (root10)/2)
z2 = -9+3i, z3 = -13.5+4.5i, distance = (3 root10)/2 (previous distance + (root10)/2)
4) The line going through z1 and z3 passes through the origin when z2 is on the real axis (when the imaginary value = 0). This is because when a real value (z2) is multiplied by the complex number (z1), the real component of the complex number remains real and the imaginary component remains imaginary. Thus keeping the gradient of the lines z1 and z3 equal.
Eg: z2 = -4+0i, z1 = -1+1.1i, therefore z3 = 4+4.4i (The real and the complex component is multiplied by a real number, thus not changing it).
If z2 = 0-4i, and z1 = -1+1.1i, z3 = 4i+4.4. Rather than z1 being merely elongated by the "multiplier" that z2 would be if real, z1's complex component has been changed to real (i*i = -1) and the real component has been changed to imaginary, thus making z3 perpendicular to z1.
Whether the real component of z2 is positive or negative determines whether z3 and z1 are in the same quadrants or in diagonally opposite ones, if positive, both points are in the same quadrant. If negative, they are in diagonally opposite quadrants.
5) z2 and z3 are equidistant from the origin when the radius of the circle is 1 (the modulus of z1 is 1). This is the case because the product of the moduli of z1 and z2 gives the modulus of z3:
Eg: z1 = 0+2i (modulus = 2), z2 = 2+2i (modulus = 2 root 2), z3 = -4+4i (modulus = 4 root2 = 2 root2 * 2)
Because of this, when the modulus of z1 is 1, the modulus of z2 and z3 are equal, which means that they are equidistant to the origin.