I was able to solve this problem and get that Re(z)=-3/2. I think the locus of points then is the vertical line in the Argand plane at x=-3/2. My question is how do we know that the imaginary part can be any real value as opposed to just the point 0? Or maybe my answer of the locus being the vertical line at x=-3/2 is incorrect and the answer is just the single point x=-3/2. Please let me know which is correct? Thanks. ### Description of the Mathematical Problem**Problem Statement:**Describe the locus of points $ z $ satisfying the equation:\[ |z - 1|^2 = |z + 1|^2 + 6. \]### Explanation and SolutionThe given equation involves complex numbers and is written in terms of modulus and squares. Here's a step-by-step approach to solve the problem:1. **Rewrite $ z $:** Let $ z = x + yi $, where $ x $ and $ y $ are real numbers, and $ i $ is the imaginary unit.2. **Expand the Modulus Squares:** - Calculate $ |z - 1|^2 $: \[ |z - 1|^2 = (x - 1)^2 + y^2 \] - Calculate $ |z + 1|^2 $: \[ |z + 1|^2 = (x + 1)^2 + y^2 \]3. **Substitute and Simplify the Equation:** Substitute the expressions for the modulus squares into the equation: \[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 + 6 \] Simplify by canceling out $ y^2 $ on both sides: \[ (x - 1)^2 = (x + 1)^2 + 6 \]4. **Expand and Solve for $ x $:** Expand the equation: \[ x^2 - 2x + 1 = x^2 + 2x + 1 + 6 \] Cancel $ x^2 + 1 $ on both sides: \[ -2x = 2x + 6 \] Combine like terms and solve for $ x $: \[ -4x = 6 \quad \Rightarrow \quad x = -\frac{3}{2} \]5. **Describe the Locus:** Since $ x = -\frac{3}{2} $ and there is no condition on $ y $, the locus of points $ z $ is a vertical line in the complex plane at \( x = -\frac{

Find the locus of point z satisfying the equation:z-12=z+12+6.

Answered: Describe the locus of points z…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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I was able to solve this problem and get that Re(z)=-3/2. I think the locus of points then is the vertical line in the Argand plane at x=-3/2. My question is how do we know that the imaginary part can be any real value as opposed to just the point 0? Or maybe my answer of the locus being the vertical line at x=-3/2 is incorrect and the answer is just the single point x=-3/2. Please let me know which is correct? Thanks.

### Description of the Mathematical Problem

**Problem Statement:**

Describe the locus of points $ z $ satisfying the equation:

\[ |z - 1|^2 = |z + 1|^2 + 6. \]

### Explanation and Solution

The given equation involves complex numbers and is written in terms of modulus and squares. Here's a step-by-step approach to solve the problem:

1. **Rewrite $ z $:**

Let $ z = x + yi $, where $ x $ and $ y $ are real numbers, and $ i $ is the imaginary unit.

2. **Expand the Modulus Squares:**

- Calculate $ |z - 1|^2 $:

\[
|z - 1|^2 = (x - 1)^2 + y^2
\]

- Calculate $ |z + 1|^2 $:

\[
|z + 1|^2 = (x + 1)^2 + y^2
\]

3. **Substitute and Simplify the Equation:**

Substitute the expressions for the modulus squares into the equation:

\[
(x - 1)^2 + y^2 = (x + 1)^2 + y^2 + 6
\]

Simplify by canceling out $ y^2 $ on both sides:

\[
(x - 1)^2 = (x + 1)^2 + 6
\]

4. **Expand and Solve for $ x $:**

Expand the equation:

\[
x^2 - 2x + 1 = x^2 + 2x + 1 + 6
\]

Cancel $ x^2 + 1 $ on both sides:

\[
-2x = 2x + 6
\]

Combine like terms and solve for $ x $:

\[
-4x = 6 \quad \Rightarrow \quad x = -\frac{3}{2}
\]

5. **Describe the Locus:**

Since $ x = -\frac{3}{2} $ and there is no condition on $ y $, the locus of points $ z $ is a vertical line in the complex plane at \( x = -\frac{

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