Circle in Complex Form

Let $$z = x + iy$$ where $$x, y \in \mathbb{R}.$$

From $$|z - 1| = 1$$: $$|x - 1 + iy| = 1$$

$$(x-1)^2 + y^2 = 1$$

This is a circle centered at $$(1, 0)$$ with radius 1.

From $$|z - i| = 1$$: $$|x + i(y-1)| = 1$$

$$x^2 + (y-1)^2 = 1$$

This is a circle centered at $$(0, 1)$$ with radius 1.

Finding the intersection of the two circles:

From equation 1: $$x^2 - 2x + 1 + y^2 = 1 \Rightarrow x^2 + y^2 = 2x$$

From equation 2: $$x^2 + y^2 - 2y + 1 = 1 \Rightarrow x^2 + y^2 = 2y$$

Therefore: $$2x = 2y \Rightarrow x = y$$

Substituting into $$x^2 + y^2 = 2x$$:
$$x^2 + x^2 = 2x$$
$$2x^2 = 2x$$
$$x^2 = x$$
$$x(x-1) = 0$$

So $$x = 0$$ or $$x = 1$$

If $$x = 0$$: $$y = 0$$, so $$z = 0$$

Check: $$|0-1| = 1$$  and $$|0-i| = 1$$ 

If $$x = 1$$: $$y = 1$$, so $$z = 1 + i$$

Check: $$|1+i-1| = |i| = 1$$  and $$|1+i-i| = |1| = 1$$ 

For $$z = 0$$: $$|z| = 0$$
For $$z = 1 + i$$: $$|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$$

since  $$\sqrt{2}$$ is only there among 2 solutions, $$\sqrt{2}$$ is the answer

$$\text{Re}\left(\frac{z_1 + z_2}{z_3}\right) = -1$$
Let $$z_1 + z_2 = z_s$$ (the resultant vector of $$z_1$$ and $$z_2$$). The equation becomes:
$$\text{Re}\left(\frac{z_s}{z_3}\right) = -1 \implies \frac{|z_s|}{|z_3|} \cos(\theta) = -1$$
Where $\theta$ is the angle between the vector $$(z_1 + z_2)$$ and $$z_3$$. Since $$|z_3| = 2$$:

$$|z_1 + z_2| \cos(\theta) = -2$$

For any triangle inscribed in a circle, the product of its sides $$|z_1 - z_2| \cdot |z_2 - z_3| \cdot |z_3 - z_1|$$ is maximized when the triangle is equilateral.

In an equilateral triangle, the angle between any two position vectors from the origin is $$120^\circ$$.
The resultant $$|z_1 + z_2|$$ of two vectors of length 2 at $$120^\circ$$ is exactly $$2$$
The vector $$(z_1 + z_2)$$ points directly opposite to the third vertex $$z_3$$, meaning the angle $$\theta$$ between them is $$180^\circ$$.

$$|z_1 + z_2| \cos(\theta) = 2 \cdot \cos(180^\circ) = 2(-1) = -2$$
For an equilateral triangle inscribed in a circle of radius R = 2, the side length s is:
$$s = R\sqrt{3} = 2\sqrt{3}$$The required product is simply $$s^3$$
$$\text{Maximum Value} = (2\sqrt{3})^3 = 8 \cdot 3\sqrt{3} = 24\sqrt{3}$$

The given equation is:

$$|z - 4| = |z + 4|$$

Geometrically, this represents the set of all points $$z$$ in the complex plane that are equidistant from the fixed points $$(4, 0)$$ and $$(-4, 0)$$. The locus of such points is the perpendicular bisector of the line segment joining $$(4, 0)$$ and $$(-4, 0)$$, which is the y-axis ($$x = 0$$).

Alternatively, we can substitute $$z = x + iy$$ to solve algebraically:

$$|x + iy - 4| = |x + iy + 4|$$ $$\sqrt{(x - 4)^2 + y^2} = \sqrt{(x + 4)^2 + y^2}$$

Square both sides to eliminate the square roots:

$$(x - 4)^2 + y^2 = (x + 4)^2 + y^2$$ $$x^2 - 8x + 16 = x^2 + 8x + 16$$ $$-8x = 8x \implies 16x = 0 \implies x = 0$$

Thus, the locus line is the line $$x = 0$$.

We need to find the points on this line that lie strictly inside the circle $$|z| = 6$$.

The equation of the circle centered at the origin with radius $$R = 6$$ is:

$$x^2 + y^2 < 6^2 \implies x^2 + y^2 < 36$$

Since the points must lie on our locus line ($$x = 0$$), substitute $$x = 0$$ into the inequality:

$$0^2 + y^2 < 36$$  $$y^2 < 36 \implies -6 < y < 6$$

A lattice point must have integer coordinates. Since $$x = 0$$, we just need to count the integer values of $$y$$ that satisfy the inequality $$-6 < y < 6$$:

The valid integer values for $$y$$ are:

$$\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$$

Counting these values, we find there are exactly 11 valid integer values for $$y$$.

Therefore, there are $$11$$ lattice points on the line inside the circle, matching Option A.