Solving Complex Number Equations to Find Modulus

This problem requires us to find the modulus \(|z|\) of a complex number \(z\) that satisfies two given equations involving its modulus. Let's analyze each equation step by step.

Analyzing the First Complex Number Equation

The first equation is: \[ \left| \frac{z - 4}{z - 8} \right| = 1 \] Using the property \( \left| \frac{w_1}{w_2} \right| = \frac{|w_1|}{|w_2|} \), we can rewrite this as: \[ \frac{|z - 4|}{|z - 8|} = 1 \] This implies: \[ |z - 4| = |z - 8| \] Geometrically, this equation represents the locus of points \(z\) in the complex plane that are equidistant from the points 4 (which corresponds to the complex number \(4 + 0i\)) and 8 (which corresponds to the complex number \(8 + 0i\)). The locus of such points is the perpendicular bisector of the line segment joining 4 and 8. The midpoint of this segment is \( (4+8)/2 = 6 \), and the line segment lies on the real axis. Therefore, the perpendicular bisector is a vertical line with equation \(x = 6\).

Algebraically, let \(z = x + iy\), where \(x\) and \(y\) are real numbers. \[ |(x + iy) - 4| = |(x + iy) - 8| \] \[ |(x - 4) + iy| = |(x - 8) + iy| \] Using the definition of modulus \(|a + bi| = \sqrt{a^2 + b^2}\): \[ \sqrt{(x - 4)^2 + y^2} = \sqrt{(x - 8)^2 + y^2} \] Squaring both sides: \[ (x - 4)^2 + y^2 = (x - 8)^2 + y^2 \] \[ x^2 - 8x + 16 + y^2 = x^2 - 16x + 64 + y^2 \] Subtract \(x^2 + y^2\) from both sides: \[ -8x + 16 = -16x + 64 \] Rearranging terms to solve for \(x\): \[ 16x - 8x = 64 - 16 \] \[ 8x = 48 \] \[ x = \frac{48}{8} \] \[ x = 6 \] So, the real part of the complex number \(z\) is 6. \(z\) must be of the form \(6 + iy\).

Analyzing the Second Complex Number Equation

The second equation is: \[ \left| \frac{z}{z - 2} \right| = \frac{3}{2} \] Using the property \( \left| \frac{w_1}{w_2} \right| = \frac{|w_1|}{|w_2|} \), we can rewrite this as: \[ \frac{|z|}{|z - 2|} = \frac{3}{2} \] Cross-multiplying gives: \[ 2|z| = 3|z - 2| \] We know from the first equation that \(z = 6 + iy\). Substitute this into the second equation: \[ 2|6 + iy| = 3|(6 + iy) - 2| \] \[ 2|6 + iy| = 3|4 + iy| \] Using the definition of modulus \(|a + bi| = \sqrt{a^2 + b^2}\): \[ 2\sqrt{6^2 + y^2} = 3\sqrt{4^2 + y^2} \] \[ 2\sqrt{36 + y^2} = 3\sqrt{16 + y^2} \] Squaring both sides: \[ (2\sqrt{36 + y^2})^2 = (3\sqrt{16 + y^2})^2 \] \[ 4(36 + y^2) = 9(16 + y^2) \] Distribute the numbers: \[ 144 + 4y^2 = 144 + 9y^2 \] Rearranging terms to solve for \(y\): \[ 144 - 144 = 9y^2 - 4y^2 \] \[ 0 = 5y^2 \] \[ y^2 = 0 \] \[ y = 0 \] So, the imaginary part of the complex number \(z\) is 0.

Determining the Complex Number z and its Modulus |z|

From the first equation, we found that the real part of \(z\) is \(x = 6\). From the second equation, we found that the imaginary part of \(z\) is \(y = 0\). Therefore, the complex number \(z\) is \(z = 6 + 0i = 6\).

Now we can find the modulus \(|z|\): \[ |z| = |6 + 0i| = \sqrt{6^2 + 0^2} = \sqrt{36 + 0} = \sqrt{36} = 6 \]

The value of \(|z|\) is 6.

Checking the Solution

Let's verify if \(z = 6\) satisfies the original equations.

First equation: \( \left| \frac{z - 4}{z - 8} \right| = \left| \frac{6 - 4}{6 - 8} \right| = \left| \frac{2}{-2} \right| = |-1| = 1 \). This matches the right side of the first equation.

Second equation: \( \left| \frac{z}{z - 2} \right| = \left| \frac{6}{6 - 2} \right| = \left| \frac{6}{4} \right| = \left| \frac{3}{2} \right| = \frac{3}{2} \). This matches the right side of the second equation.

Both equations are satisfied by \(z = 6\). The modulus \(|z|\) is \(|6| = 6\).

Revision Table: Complex Number Modulus

Additional Information on Complex Number Properties

Understanding the properties of complex numbers and their modulus is crucial for solving equations like these.

• Modulus as Distance: The expression \(|z - a|\) represents the distance between the complex number \(z\) and the complex number \(a\) in the complex plane. The equation \(|z - 4| = |z - 8|\) means the distance from \(z\) to 4 is equal to the distance from \(z\) to 8. As shown, this leads to \(x=6\).
• Equations involving Modulus: Equations involving moduli often simplify into algebraic equations involving the real and imaginary parts (\(x\) and \(y\)). Squaring both sides of an equation like \(|z - a| = k|z - b|\) is a common technique.
• Solving Simultaneous Equations: When you have two equations involving a complex number \(z\), you typically solve them simultaneously to find the values of \(x\) and \(y\) that satisfy both conditions. In this case, the first equation gave us the value of \(x\) directly, which simplified the second equation to find \(y\).

This problem effectively combines algebraic manipulation with the geometric interpretation of the modulus of complex numbers.