Understanding the Complex Number Locus Problem

The question asks us to find the geometric locus of a complex number \(z = x + iy\) that satisfies a given equation involving \(z\) and its conjugate \(z̅\). The equation is \(z z̅ = |z + z̅ |\), where \(i = \sqrt{-1}\).

To find the locus, we need to express the given equation in terms of the real part (\(x\)) and the imaginary part (\(y\)) of the complex number \(z\).

Analyzing the Complex Number Equation

Let \(z = x + iy\). Then the complex conjugate is \(z̅ = x - iy\).

Now, let's evaluate the terms in the equation \(z z̅ = |z + z̅ |\):

• The product \(z z̅\) is given by: \(z z̅ = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 - i^2 y^2 = x^2 - (-1)y^2 = x^2 + y^2\).
• The sum \(z + z̅\) is given by: \(z + z̅ = (x + iy) + (x - iy) = x + iy + x - iy = 2x\).
• The magnitude \(|z + z̅|\) is the magnitude of the real number \(2x\). The magnitude of a real number is its absolute value: \(|z + z̅ | = |2x|\).

Substituting these back into the original equation \(z z̅ = |z + z̅ |\), we get:

\(x^2 + y^2 = |2x|\)

The absolute value \(|2x|\) means that \(|2x|\) can be \(2x\) if \(2x \geq 0\) (i.e., \(x \geq 0\)) or \(-2x\) if \(2x < 0\) (i.e., \(x < 0\)). We must consider these two cases separately.

Case 1: \(x \geq 0\)

If \(x \geq 0\), then \(|2x| = 2x\). The equation becomes:

\(x^2 + y^2 = 2x\)

To identify the locus, we rearrange this equation to the standard form of a geometric shape:

\(x^2 - 2x + y^2 = 0\)

We can complete the square for the \(x\) terms:

\((x^2 - 2x + 1) - 1 + y^2 = 0\)

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

This is the equation of a circle with center \((1, 0)\) and radius \(r = \sqrt{1} = 1\). This part of the locus is valid for all points on this circle where the x-coordinate is greater than or equal to zero.

Case 2: \(x < 0\)

If \(x < 0\), then \(|2x| = -2x\). The equation becomes:

\(x^2 + y^2 = -2x\)

Rearranging this equation:

\(x^2 + 2x + y^2 = 0\)

Complete the square for the \(x\) terms:

\((x^2 + 2x + 1) - 1 + y^2 = 0\)

\((x + 1)^2 + y^2 = 1\)

This is the equation of a circle with center \((-1, 0)\) and radius \(r = \sqrt{1} = 1\). This part of the locus is valid for all points on this circle where the x-coordinate is less than zero.

Combining the Cases to Find the Locus

The locus of \(z\) is the union of the solutions from Case 1 and Case 2. The first case \((x - 1)^2 + y^2 = 1\) for \(x \geq 0\) represents the portion of the circle centered at \((1,0)\) with radius 1 that lies in the region \(x \geq 0\). The second case \((x + 1)^2 + y^2 = 1\) for \(x < 0\) represents the portion of the circle centered at \((-1,0)\) with radius 1 that lies in the region \(x < 0\).

Let's check the point \(x = 0\). For \(x=0\), the equation \(x^2 + y^2 = |2x|\) becomes \(0^2 + y^2 = |0|\), which simplifies to \(y^2 = 0\), so \(y = 0\). The point \((0,0)\) is on the locus. Let's see if \((0,0)\) is on both circles:

• For \((x - 1)^2 + y^2 = 1\): \((0 - 1)^2 + 0^2 = (-1)^2 + 0 = 1\). Yes, \((0,0)\) is on this circle.
• For \((x + 1)^2 + y^2 = 1\): \((0 + 1)^2 + 0^2 = 1^2 + 0 = 1\). Yes, \((0,0)\) is on this circle.

Since both circles pass through the origin \((0,0)\) (which is the point where \(x=0\)), the two portions connect at the origin. The first circle \((x-1)^2 + y^2 = 1\) covers the right half, including the origin and extending up to \(x=2\). The second circle \((x+1)^2 + y^2 = 1\) covers the left half, including the origin and extending down to \(x=-2\). Together, they form two complete circles that touch at the origin.

Therefore, the locus of \(z\) is a pair of circles.

Summary of the Locus

The combination of these two parts gives a pair of circles.

Conclusion on the Locus of z

Based on the analysis, the equation \(z z̅ = |z + z̅ |\) translates into two separate equations in terms of \(x\) and \(y\), each representing a circle, depending on the sign of \(x\). The two circles are \((x - 1)^2 + y^2 = 1\) and \((x + 1)^2 + y^2 = 1\).

These equations describe a pair of circles in the complex plane (or the Cartesian plane representing the complex plane).

Revision Table: Key Concepts in Complex Number Locus

Additional Information on Locus of Complex Numbers

Finding the locus of a complex number \(z\) often involves converting the given equation in terms of \(z\) and \(z̅\) into an equation involving \(x\) and \(y\), where \(z = x + iy\). The resulting equation in \(x\) and \(y\) represents the geometric shape of the locus in the Cartesian plane.

Common types of loci encountered in complex numbers include:

• Straight Lines: Equations often linear in \(x\) and \(y\), e.g., \(ax + by + c = 0\), or involving conditions like \(\text{Re}(az+b)=0\) or \(\text{Im}(az+b)=0\).
• Circles: Equations typically of the form \((x-h)^2 + (y-k)^2 = r^2\), or \(x^2 + y^2 + Dx + Ey + F = 0\). This can arise from conditions like \(|z-a| = r\) (distance from a fixed point \(a\) is constant) or \(|z-a| = |z-b|\) (equidistant from two fixed points \(a\) and \(b\), which gives a perpendicular bisector, a line).
• Parabolas: Arise from conditions related to distance from a point and a line, though less common directly from basic \(z, z̅\) equations unless involving specific forms or manipulations.
• Hyperbolas/Ellipses: Can arise from conditions like \(|z-a| + |z-b| = k\) (ellipse) or \(||z-a| - |z-b|| = k\) (hyperbola), or from specific quadratic forms in \(x\) and \(y\). Rectangular hyperbolas are a special case.

In this specific problem, the presence of \(|2x|\) leads to the splitting into two cases based on the sign of \(x\), resulting in two distinct equations, each corresponding to a circle. The absolute value function is key to understanding why the locus is split into parts.