If z satisfies `|z+1| lt |z-2|`, then `w=3z+2+i`

Options

A. `|w+1|lt|w-8|`
B. `|w+1|lt|w-7|`
C. `|w+barw|gt7`
D. `|w+5|ltw-4|`

Answer

To solve the problem where \( z \) satisfies \( |z + 1| < |z - 2| \) and we need to find \( w = 3z + 2 + i \), we can follow these steps:

### Step 1: Interpret the inequality
The inequality \( |z + 1| < |z - 2| \) means that the distance from the point \( z \) to the point \( -1 \) is less than the distance from \( z \) to the point \( 2 \) in the complex plane.

### Step 2: Geometric representation
This inequality describes a region in the complex plane. The points \( -1 \) and \( 2 \) can be represented as complex numbers \( -1 + 0i \) and \( 2 + 0i \). The line that is the perpendicular bisector of the segment joining these two points can be found.

### Step 3: Find the midpoint and the perpendicular bisector
The midpoint \( M \) of the segment joining \( -1 \) and \( 2 \) is:
\[ M = \left( \frac{-1 + 2}{2}, 0 \right) = \left( \frac{1}{2}, 0 \right) \]
The slope of the line connecting \( -1 \) and \( 2 \) is \( 0 \) (horizontal line), so the perpendicular bisector is a vertical line through \( x = \frac{1}{2} \).

### Step 4: Determine the region
The inequality \( |z + 1| < |z - 2| \) implies that \( z \) lies to the left of the line \( x = \frac{1}{2} \) in the complex plane.

### Step 5: Express \( w \) in terms of \( z \)
Given \( w = 3z + 2 + i \), we can express \( w \) in terms of the real and imaginary parts of \( z \):
Let \( z = x + yi \), then:
\[ w = 3(x + yi) + 2 + i = (3x + 2) + (3y + 1)i \]

### Step 6: Analyze the real and imaginary parts of \( w \)
The real part of \( w \) is \( 3x + 2 \) and the imaginary part is \( 3y + 1 \).

### Step 7: Determine the constraints on \( w \)
Since \( z \) must satisfy \( x < \frac{1}{2} \), we have:
\[ 3x + 2 < 3 \cdot \frac{1}{2} + 2 = \frac{3}{2} + 2 = \frac{7}{2} \]
Thus, the real part of \( w \) is constrained by:
\[ \text{Re}(w) < \frac{7}{2} \]

### Step 8: Find the range for the imaginary part
The imaginary part \( 3y + 1 \) can take any value depending on \( y \), as there are no restrictions on \( y \) from the original inequality.

### Conclusion
Thus, the final result is that \( w \) can take values such that:
\[ \text{Re}(w) < \frac{7}{2} \quad \text{and} \quad \text{Im}(w) \text{ can be any real number.} \]