Find the locus of a complex number `z= x + yi`, satisfying the relation `|2z+ 3i| ge |2z + 5|`. Illustrate the locus in the Argand plane.

Answer

To find the locus of the complex number \( z = x + yi \) satisfying the relation \( |2z + 3i| \geq |2z + 5| \), we can follow these steps:

### Step 1: Substitute \( z \)
We start by substituting \( z \) into the given inequality:
\[ |2z + 3i| \geq |2z + 5| \]
Substituting \( z = x + yi \):
\[ |2(x + yi) + 3i| \geq |2(x + yi) + 5| \]
This simplifies to:
\[ |2x + 2yi + 3i| \geq |2x + 2yi + 5| \]
\[ |2x + (2y + 3)i| \geq |(2x + 5) + 2yi| \]

### Step 2: Express the moduli
Now, we can express the moduli:
\[ \sqrt{(2x)^2 + (2y + 3)^2} \geq \sqrt{(2x + 5)^2 + (2y)^2} \]

### Step 3: Square both sides
To eliminate the square roots, we square both sides:
\[ (2x)^2 + (2y + 3)^2 \geq (2x + 5)^2 + (2y)^2 \]
Expanding both sides:
\[ 4x^2 + (4y^2 + 12y + 9) \geq (4x^2 + 20x + 25) + 4y^2 \]

### Step 4: Simplify the inequality
Now, we simplify the inequality:
\[ 4x^2 + 4y^2 + 12y + 9 \geq 4x^2 + 20x + 25 + 4y^2 \]
Cancelling \( 4x^2 \) and \( 4y^2 \) from both sides:
\[ 12y + 9 \geq 20x + 25 \]
Rearranging gives:
\[ 20x - 12y + 16 \leq 0 \]
or
\[ 5x - 3y \leq -4 \]

### Step 5: Find the boundary line
The boundary line of the inequality \( 5x - 3y = -4 \) can be found by rewriting it in slope-intercept form:
\[ 3y = 5x + 4 \implies y = \frac{5}{3}x + \frac{4}{3} \]

### Step 6: Identify the region
To find the region satisfying the inequality \( 5x - 3y \leq -4 \), we can test a point not on the line, such as the origin \( (0, 0) \):
\[ 5(0) - 3(0) + 4 = 4 \quad \text{(which is greater than 0)} \]
Thus, the region satisfying the inequality is below the line \( 5x - 3y = -4 \).

### Step 7: Illustrate in the Argand plane
To illustrate the locus in the Argand plane, we can plot the line \( 5x - 3y = -4 \) and shade the region below it. The points satisfying the original inequality will lie in this shaded region.