If n is a positive integer greater than unity z is a complex number satisfying the equation `z^n=(z+1)^n,` then

Options

A. `"Re"(z) lt 0`
B. `"Re"(z) gt 0`
C. Re(z) =0
D. none of these

Answer

To solve the equation \( z^n = (z + 1)^n \) for a complex number \( z \) where \( n \) is a positive integer greater than unity, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ z^n = (z + 1)^n \]
Dividing both sides by \( (z + 1)^n \) gives:
\[ \frac{z^n}{(z + 1)^n} = 1 \]

### Step 2: Take the nth root
Taking the nth root of both sides, we have:
\[ \frac{z}{z + 1} = \omega \]
where \( \omega \) is an nth root of unity. This implies that:
\[ \left| \frac{z}{z + 1} \right| = 1 \]

### Step 3: Modulus equality
From the modulus condition, we can deduce:
\[ |z| = |z + 1| \]
This means that the distance from \( z \) to the origin is equal to the distance from \( z \) to the point \( -1 \) on the complex plane.

### Step 4: Square both sides
Squaring both sides gives:
\[ |z|^2 = |z + 1|^2 \]
Using the property of modulus, we can express this as:
\[ z \overline{z} = (z + 1)(\overline{z} + 1) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side:
\[ z \overline{z} = z \overline{z} + z + \overline{z} + 1 \]
Subtracting \( z \overline{z} \) from both sides results in:
\[ 0 = z + \overline{z} + 1 \]

### Step 6: Express in terms of real and imaginary parts
Let \( z = x + i y \), where \( x \) and \( y \) are real numbers. Then:
\[ 0 = (x + i y) + (x - i y) + 1 \]
This simplifies to:
\[ 0 = 2x + 1 \]

### Step 7: Solve for \( x \)
Solving for \( x \):
\[ 2x = -1 \implies x = -\frac{1}{2} \]

### Conclusion
The real part of the complex number \( z \) is:
\[ \text{Re}(z) = -\frac{1}{2} \]