Key Concepts and Formulas

• Modulus of a Complex Number: For a complex number $z = x + iy$, where $x$ and $y$ are real numbers, the modulus is defined as $|z| = \sqrt{x^2 + y^2}$. Geometrically, $|z|$ represents the distance of the point $(x, y)$ from the origin in the complex plane.
• Geometric Interpretation of Modulus Difference: $|z_1 - z_2|$ represents the distance between the points corresponding to the complex numbers $z_1$ and $z_2$ in the Argand plane.
• Inequality Properties: Squaring both sides of an inequality $a < b$ is valid (i.e., $a^2 < b^2$) if both $a$ and $b$ are non-negative.

Step-by-Step Solution

1. Represent the Complex Number in Cartesian Form To solve the inequality, we express the complex number $z$ in its Cartesian form: $z = x + iy$ where $x$ is the real part of $z$ (${\mathop{\rm Re}\nolimits} (z)$) and $y$ is the imaginary part of $z$ (${\mathop{\rm Im}\nolimits} (z)$), with $x, y \in \mathbb{R}$.

Why this step? Converting to Cartesian form allows us to transform the complex number inequality into an algebraic inequality involving real variables $x$ and $y$, which can be solved using standard algebraic techniques.

2. Substitute into the Inequality and Group Terms Substitute $z = x + iy$ into the given inequality $|z - 4| < |z - 2|$: $|(x + iy) - 4| < |(x + iy) - 2|$ Group the real and imaginary parts: $|(x - 4) + iy| < |(x - 2) + iy|$

Why this step? Grouping the real and imaginary parts makes it clear how to apply the modulus formula in the next step.

3. Apply the Modulus Formula Apply the definition of the modulus, $|a + ib| = \sqrt{a^2 + b^2}$, to both sides of the inequality: $\sqrt{(x - 4)^2 + y^2} < \sqrt{(x - 2)^2 + y^2}$

Why this step? This step eliminates the complex number notation and introduces real algebraic expressions under square roots, moving us closer to a solvable algebraic inequality.

4. Eliminate Square Roots by Squaring Both Sides Since both sides of the inequality represent distances, they are non-negative. Therefore, we can square both sides without changing the inequality: $(x - 4)^2 + y^2 < (x - 2)^2 + y^2$

Why this step? Squaring both sides removes the square roots, simplifying the expression for further algebraic manipulation.

5. Expand and Simplify the Inequality Expand the squared binomials: $x^2 - 8x + 16 + y^2 < x^2 - 4x + 4 + y^2$ Subtract $x^2$ and $y^2$ from both sides: $-8x + 16 < -4x + 4$

Why this step? Expanding and simplifying by canceling common terms helps reduce the complexity of the inequality, making it easier to isolate the variable $x$.

6. Isolate the Variable $x$ Gather terms containing $x$ on one side and constant terms on the other: Add $8x$ to both sides: $16 < 4x + 4$ Subtract $4$ from both sides: $12 < 4x$

Why this step? These are standard algebraic procedures applied to inequalities. The goal is to isolate the variable $x$ to determine the range of values it can take.

7. Solve for $x$ Divide both sides by $4$: $\frac{12}{4} < x$ $3 < x$ Or: $x > 3$

Why this step? This is the final algebraic step that provides the specific condition on the real variable $x$ that satisfies the initial complex inequality.

8. Relate Back to the Real Part of $z$ Since $z = x + iy$, the real part of $z$ is $x$ (i.e., ${\mathop{\rm Re}\nolimits} (z) = x$). Therefore, the solution to the inequality is: ${\mathop{\rm Re}\nolimits} (z) > 3$

Common Mistakes & Tips:

• Geometric Interpretation: Always remember the geometric meaning. The inequality $|z-a| < |z-b|$ means $z$ is closer to $a$ than to $b$.
• Squaring Inequalities: Only square both sides of an inequality if both sides are known to be non-negative. Moduli are always non-negative.
• Algebraic Precision: Pay close attention to signs when moving terms across the inequality.

Summary

The problem involves solving an inequality with complex numbers using the modulus. By substituting $z = x + iy$ and applying the modulus definition, we transform the complex inequality into a real algebraic inequality. Solving for $x$ gives us the condition on the real part of $z$. The solution ${\mathop{\rm Re}\nolimits} (z) > 3$ means the real part of $z$ must be greater than $3$.

The final answer is \boxed{{\mathop{\rm Re}\nolimits} (z) > 3}, which corresponds to option (C).