To show that the great circles through (z1, z3, z4) and (z2, z3, z4) intersect orthogonally at z3, we can use the concept of perpendicularity between vectors. By proving that the dot product of two vectors, v1 and v2, is zero, we can show that the vectors v1 and v2 are perpendicular.
To show that the great circles through (z1, z3, z4) and (z2, z3, z4) intersect orthogonally at z3, we can use the concept of perpendicularity between vectors. Let's denote the vectors v1 = z1 - z3 and v2 = z2 - z3. We need to prove that the dot product of these two vectors is zero.
The dot product of two vectors, v1 and v2, is given by the formula: v1 · v2 = |v1| |v2| cos(θ), where |v1| and |v2| are the magnitudes of the vectors and θ is the angle between them. In this case, we want to show that the dot product v1 · v2 = 0, which means that the vectors v1 and v2 are perpendicular.
To prove this, let's calculate the dot product: v1 · v2 = (z1 - z3) · (z2 - z3). By expanding the dot product, we get:
v1 · v2 = (z1 - z3) · (z2 - z3) = z1 · z2 - z1 · z3 - z3 · z2 + z3 · z3.
Now, since Cr(z1, z2, z3, z4) ∈iR, we know that the points lie on a circle in the complex plane. Let's denote the radius of this circle as r. Therefore, we have |z1 - z3| = |z2 - z3| = r. Substituting these values into our dot product equation, we get:
v1 · v2 = r^2 - r^2 - r^2 + r^2 = 0.
This shows that the vectors v1 and v2 are perpendicular, which means the great circles through (z1, z3, z4) and (z2, z3, z4) intersect orthogonally at z3.