Why is it impossible to find distinct $z_1,z_2,z_3, z_4\in \mathbb C$ such that $|z_1- z_2|=|z_1-z_3|=|z_2-z_3|=|z_1-z_4|=|z_2-z_4|=|z_3-z_4|$? A. It is possible to find distinct $z_1,z_2,z_3\in \mathbb C$  such that $|z_1-z_2|=|z_1-z_3|=|z_2-z_3|$.
Answer: True
B. It is possible to find distinct $z_1,z_2,z_3, z_4\in \mathbb C$ such that $|z_1-   z_2|=|z_1-z_3|=|z_2-z_3|=|z_1-z_4|=|z_2-z_4|=|z_3-z_4|$.
Answer: False
I don't understand why the second one is false. Can anyone help me out?
 A: In the first case, take three points in the complex plane forming an equilateral triangle.
For the second case, assume that $|z_j-z_k| = R$ for all $j\neq k$. Looking at $z_1$ first, it follows that $z_2$, $z_3$ and $z_4$ all lie on a circle centered at $z_1$ with radius $R$. Then look at the circle of radius $R$ centered at $z_2$. This circle intersects the original one in just two points, which gives you your contradiction.
A: To make three points equally spaced, the the third must be at the intersection of the circles whose radius is the given distance centered at the other two points. This forms an equilateral triangle, as shown below. Note that the blue and green circles intersect at two points, so we could have chosen either for the red point.
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However, to make four points equally spaced, the fourth point must be at the intersection of the three circles since those are the loci of points at the requisite distance from the other three points. Unfortunately, there is no such point. For example, the point at the intersection of the green and red circles would be at the wrong distance from the blue point.
A: You can think about $\mathbb{C}$ as a two dimensional plane (that is, like $\mathbb{R}^2$). Each complex number defines a point in the plane, and given two complex numbers $z$ and $w$, $|z - w|$ is the distance between them, or the length of the line segment which joins them.
For three complex numbers $z_1$, $z_2$, $z_3$, the condition $|z_1 - z_2| = |z_1 - z_3| = |z_2 - z_3|$ implies that the three points are the vertices of an equilateral triangle. So if you draw an equilateral triangle in the complex plane, and label the vertices $z_1$, $z_2$ and $z_3$, you will have three complex numbers which satisfy the required condition. In your comment you asked whether $z_1 = z_2 = z_3$. This is not the case as the question asked for the numbers to be distinct.
For four complex numbers $z_1$, $z_2$, $z_3$, $z_4$, the condition $$|z_1 -  z_2| = |z_2 - z_3| = |z_3 - z_4| = |z_1 - z_4|$$ implies that the four points are the vertices of a rhombus (labelled cyclically). However, the question asks about four complex numbers which satisfy the stronger condition $$|z_1 -  z_2| = |z_2 - z_3| = |z_3 - z_4| = |z_1 - z_4| = |z_1 - z_3| = |z_2 - z_4|.$$ If four such complex numbers did exist, they would have to be the vertices of a rhombus labelled cyclically, but note that $|z_1 - z_3|$ and $|z_2 - z_4|$ are the lengths of the diagonals of the rhombus. So $z_1$, $z_2$, $z_3$, $z_4$ would have to be the vertices of a rhombus which had equal side and diagonal lengths, but no such rhombus exists (the diagonal length is always greater than the side length).
A: Complex number is a point on the complex plane, and absolute value of the difference between two complex numbers is essentially the euclidean distance between the points representing them. It is possible to find three points on a  plane such that the distance between them is the same (they will form equilateral triangle), but it is impossible to place four points on a plane in this way.
