All the roots of `(z + 1)^(4) = z^(4)` lie on

Options

A. a straight line parallel to x-axis
B. a straight line parallel to y-axis
C. a circle with centre at -1 + 0i
D. a circle with centre at 1 + i

Answer

To solve the equation \((z + 1)^4 = z^4\) and determine where all the roots lie, we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ (z + 1)^4 - z^4 = 0 \]
This can be rewritten as:
\[ (z + 1)^4 = z^4 \]

### Step 2: Factoring the Equation
We can factor the left-hand side using the difference of squares:
\[ (z + 1)^4 - z^4 = ((z + 1)^2 + z^2)((z + 1)^2 - z^2) = 0 \]
This gives us two equations to solve:
1. \((z + 1)^2 + z^2 = 0\)
2. \((z + 1)^2 - z^2 = 0\)

### Step 3: Solving the First Equation
For the first equation:
\[ (z + 1)^2 + z^2 = 0 \]
Expanding \((z + 1)^2\):
\[ z^2 + 2z + 1 + z^2 = 0 \implies 2z^2 + 2z + 1 = 0 \]
Dividing by 2:
\[ z^2 + z + \frac{1}{2} = 0 \]
Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ z = \frac{-1 \pm \sqrt{1 - 2}}{2} = \frac{-1 \pm i}{2} \]
This gives us two roots:
\[ z_1 = \frac{-1 + i}{2}, \quad z_2 = \frac{-1 - i}{2} \]

### Step 4: Solving the Second Equation
For the second equation:
\[ (z + 1)^2 - z^2 = 0 \]
Expanding:
\[ z^2 + 2z + 1 - z^2 = 0 \implies 2z + 1 = 0 \]
Solving for \(z\):
\[ z = -\frac{1}{2} \]

### Step 5: Analyzing the Roots
Now we have three roots:
1. \(z_1 = \frac{-1 + i}{2}\)
2. \(z_2 = \frac{-1 - i}{2}\)
3. \(z_3 = -\frac{1}{2}\)

### Step 6: Finding the Geometric Representation
The roots \(z_1\) and \(z_2\) lie on the line \(x = -\frac{1}{2}\) (the vertical line in the complex plane), and the root \(z_3\) also lies on this line.

### Conclusion
All the roots of the equation \((z + 1)^4 = z^4\) lie on the vertical line given by:
\[ \text{Re}(z) = -\frac{1}{2} \]