If z satisfies `|z-1| lt |z +3| " then " omega = 2 z + 3 -i` satisfies

Options

A. `|omega - 5 -i|lt | omega + 3+i|`
B. `|omega - 5| lt | omega +3|`
C. `lm (iomega) lt 1`
D. `|arg (omega -1) | lt pi/2`

Answer

To solve the problem, we need to analyze the inequality given and how it relates to the expression for \( \omega \). Let's break it down step by step.

### Step 1: Understand the given inequality
The inequality we have is:
\[ |z - 1| < |z + 3| \]
This means that the distance from \( z \) to \( 1 \) is less than the distance from \( z \) to \( -3 \). Geometrically, this describes a region in the complex plane.

### Step 2: Rewrite \( z \) in terms of \( \omega \)
We have:
\[ \omega = 2z + 3 - i \]
To express \( z \) in terms of \( \omega \), we rearrange the equation:
\[ z = \frac{\omega - 3 + i}{2} \]

### Step 3: Substitute \( z \) into the inequality
Now, we substitute \( z \) back into the inequality:
\[ \left| \frac{\omega - 3 + i}{2} - 1 \right| < \left| \frac{\omega - 3 + i}{2} + 3 \right| \]
This simplifies to:
\[ \left| \frac{\omega - 5 + i}{2} \right| < \left| \frac{\omega + 3 + i}{2} \right| \]

### Step 4: Remove the factor of \( \frac{1}{2} \)
Since the factor of \( \frac{1}{2} \) is positive, we can multiply both sides by \( 2 \) without changing the inequality:
\[ |\omega - 5 + i| < |\omega + 3 + i| \]

### Step 5: Interpret the inequality geometrically
This inequality states that the distance from \( \omega \) to the point \( 5 - i \) is less than the distance from \( \omega \) to the point \( -3 - i \). This describes a region in the complex plane.

### Step 6: Find the boundary
To find the boundary, we set the two distances equal:
\[ |\omega - (5 - i)| = |\omega - (-3 - i)| \]
This represents the perpendicular bisector of the segment joining the points \( (5, -1) \) and \( (-3, -1) \) in the complex plane.

### Step 7: Determine the region
The region defined by the original inequality \( |z - 1| < |z + 3| \) translates to a specific area in the complex plane for \( \omega \).

### Conclusion
From the analysis, we can conclude that \( \omega \) must lie in a certain region defined by the inequality. Specifically, we can derive that:

- The argument of \( \omega - 1 \) is constrained, leading to the conclusion that:
\[ -\frac{\pi}{2} < \text{arg}(\omega - 1) < \frac{\pi}{2} \]
This means that \( \omega \) is in the right half-plane.