Equidistant Points and the Locus of a Complex Number

When dealing with complex numbers, understanding the geometric relationships can provide a deeper insight into the underlying mathematical structures. One such interesting relationship is given by the equation:

Introduction to the Equation

The equation z1 - z-1 0 involves complex numbers and their distances in the complex plane. To solve this equation, we start by rearranging it as:

z1 z-1

This implies that the distance from the complex number z to -1 is equal to the distance from z to 1. Geometrically, this means that the point z is equidistant from the points -1 and 1 on the complex plane.

Geometric Interpretation

The points -1 and 1 lie on the real axis, and the locus of points that are equidistant from these two points is the perpendicular bisector of the segment connecting -1 and 1. The midpoint of the segment from -1 to 1 is:

( frac{-1 1}{2} 0 )

The perpendicular bisector of the segment is the vertical line that passes through this midpoint 0 on the real axis. Thus, the equation of this line is:

Re{z} 0

This means that z must lie on the imaginary axis.

Conclusion of the Solution

In conclusion, the solution to the equation z1 - z-1 0 is:

z bi, where b is any real number representing points along the imaginary axis.

Further Insights

The equation z1 - z-1 0 represents the locus of points in the complex plane that form a hyperbola. This hyperbola is a curve where the difference of the distances from any point on the curve to two fixed points (in this case, -1 and 1) is a constant. This geometric property provides a visual and intuitive understanding of complex numbers and their relationships.

From a geometric perspective, the points -1 and 1 on the real axis define a line segment, and the locus of points equidistant from these two points is the vertical line through their midpoint. This line is the perpendicular bisector, and it includes all points on the imaginary axis.

Counterexample and Verification

Initially, z 0 might seem like a solution, but it is not the most general solution. Let's verify this step by step:

Consider z abi, where a and b are real numbers. Then we have:

z1 abi

z-1 ( frac{1}{abi} - frac{i}{a} )

z1 - z-1 ( abi - left(- frac{i}{a} right) abi frac{i}{a} 0 )

Simplifying, we get:

abi frac{i}{a} 0 Rightarrow 4a 0 Rightarrow a 0 )

Since a 0, we see that our solution is indeed independent of b, and thus, any imaginary number bi works. This confirms that the solution set is all points on the imaginary axis.

The geometric picture of the triangle with sides 1 and b having the same hypotenuse as one with sides -1 and b further illustrates the symmetry and the balance of distances in the complex plane.