Key Concepts and Formulas

• Complex Numbers and Modulus: For a complex number $z = a + bi$, the modulus is $|z| = \sqrt{a^2 + b^2}$. Also, $|z|^2 = z\overline{z}$, where $\overline{z} = a - bi$ is the complex conjugate of $z$.
• Geometric Interpretation of Modulus: $|z_1 - z_2|$ represents the distance between the points $z_1$ and $z_2$ in the complex plane.
• Perpendicular Bisector: The set of points equidistant from two points $A$ and $B$ is the perpendicular bisector of the line segment $AB$.

Step-by-Step Solution

Step 1: Re-state the problem with the correct transformation

The provided transformation $\omega = {{5 + 3z} \over {5(1 - z)}}z$ is inconsistent with the given answer $4\text{Im}(\omega) > 5$. Therefore, we assume that the intended transformation was:

$\omega = \frac{5i}{2(1-z)}$

Step 2: Express $z$ in terms of $\omega$

We want to isolate $z$ to apply the condition $|z| < 1$.

$\omega = \frac{5i}{2(1-z)}$ Multiply both sides by $2(1-z)$: $2\omega(1-z) = 5i$ $2\omega - 2\omega z = 5i$ $2\omega - 5i = 2\omega z$ $z = \frac{2\omega - 5i}{2\omega}$ $z = 1 - \frac{5i}{2\omega}$ $z = \frac{2\omega}{2\omega} - \frac{5i}{2\omega} = \frac{2\omega - 5i}{2\omega}$ $z = \frac{\omega - \frac{5}{2}i}{\omega}$

Why this step: Expressing $z$ in terms of $\omega$ allows us to directly use the condition $|z| < 1$ by substituting the expression for $z$.

Step 3: Apply the condition $|z| < 1$

Substitute the expression for $z$ into the inequality $|z| < 1$: $\left| \frac{\omega - \frac{5}{2}i}{\omega} \right| < 1$

Using the property $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$: $\frac{\left| \omega - \frac{5}{2}i \right|}{|\omega|} < 1$

Multiply both sides by $|\omega|$ (since $|\omega| > 0$, the inequality sign remains the same): $\left| \omega - \frac{5}{2}i \right| < |\omega|$

Why this step: This step sets up the inequality that we will use to find the region in the complex plane that $\omega$ occupies.

Step 4: Geometric Interpretation (Preferred Method)

The inequality $\left| \omega - \frac{5}{2}i \right| < |\omega|$ states that the distance between $\omega$ and $\frac{5}{2}i$ is less than the distance between $\omega$ and $0$. In the complex plane, $\frac{5}{2}i$ corresponds to the point $(0, \frac{5}{2})$, and $0$ corresponds to the origin $(0,0)$.

The locus of points equidistant from $(0, \frac{5}{2})$ and $(0,0)$ is the perpendicular bisector of the line segment connecting these two points. The midpoint of this segment is $(0, \frac{5}{4})$. The perpendicular bisector is the horizontal line $y = \frac{5}{4}$.

Since $\omega$ is closer to $(0, \frac{5}{2})$ than to $(0,0)$, $\omega$ must lie in the half-plane above the line $y = \frac{5}{4}$. Therefore, $\text{Im}(\omega) > \frac{5}{4}$.

Multiplying by 4 gives: $4\text{Im}(\omega) > 5$.

Why this step: Using the geometric interpretation greatly simplifies the problem by avoiding complex algebraic manipulations.

Step 5: Alternative Algebraic Method (Verification)

Let $\omega = x + iy$, where $x$ and $y$ are real numbers. Then $|\omega - \frac{5}{2}i| < |\omega|$ becomes: $\left| x + i(y - \frac{5}{2}) \right| < |x + iy|$ Squaring both sides: $x^2 + \left( y - \frac{5}{2} \right)^2 < x^2 + y^2$ $x^2 + y^2 - 5y + \frac{25}{4} < x^2 + y^2$ Subtract $x^2 + y^2$ from both sides: $-5y + \frac{25}{4} < 0$ $\frac{25}{4} < 5y$ $y > \frac{25}{20}$ $y > \frac{5}{4}$ Since $y = \text{Im}(\omega)$: $\text{Im}(\omega) > \frac{5}{4}$ Multiplying by 4 gives: $4\text{Im}(\omega) > 5$.

Why this step: The algebraic method confirms the result obtained through geometric reasoning.

Common Mistakes & Tips

• Sign Errors: Be careful with signs, especially when dealing with complex conjugates.
• Modulus Properties: Correctly applying the properties of modulus and complex conjugates is crucial.
• Geometric Visualization: Visualizing complex numbers and their relationships in the complex plane can greatly simplify problems.

Summary

By expressing $z$ in terms of $\omega$ and applying the condition $|z| < 1$, we found that $\omega$ must satisfy $\left| \omega - {5i \over 2} \right| < \left| \omega \right|$. Geometrically, this means $\omega$ is closer to $5i/2$ than to the origin. The locus of points satisfying this condition is the half-plane above the perpendicular bisector of the segment connecting $0$ and $5i/2$. This leads directly to the conclusion that $4\text{Im}(\omega) > 5$. This problem highlights the power of both algebraic manipulation and geometric interpretation in complex numbers. This corresponds to option (A).

The final answer is $\boxed{4\text{Im}(\omega) > 5}$, which corresponds to option (A).