Let z be a complex number (not lying on x-axis) of maximum modulus such that `|z+1/z|=1`. Then,

Options

A. Im(z)=0
B. Re(z)=0
C. amp(z)=`pi`
D. Re(z)=1

Answer

To solve the problem, we need to find the complex number \( z \) such that \( |z + \frac{1}{z}| = 1 \) and \( z \) does not lie on the x-axis.

Let's denote \( z \) as \( z = x + iy \), where \( x \) and \( y \) are real numbers, and \( y \neq 0 \) since \( z \) does not lie on the x-axis.

### Step 1: Rewrite the condition
We start with the condition:
\[ |z + \frac{1}{z}| = 1 \]
Substituting \( z = x + iy \):
\[ |x + iy + \frac{1}{x + iy}| = 1 \]

### Step 2: Find \( \frac{1}{z} \)
To find \( \frac{1}{z} \):
\[ \frac{1}{z} = \frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} \]
Thus,
\[ z + \frac{1}{z} = (x + iy) + \left(\frac{x}{x^2 + y^2} - \frac{iy}{x^2 + y^2}\right) \]
Combining the real and imaginary parts:
\[ z + \frac{1}{z} = \left(x + \frac{x}{x^2 + y^2}\right) + i\left(y - \frac{y}{x^2 + y^2}\right) \]

### Step 3: Set up the modulus condition
The modulus is given by:
\[ \sqrt{\left(x + \frac{x}{x^2 + y^2}\right)^2 + \left(y - \frac{y}{x^2 + y^2}\right)^2} = 1 \]
Squaring both sides gives:
\[ \left(x + \frac{x}{x^2 + y^2}\right)^2 + \left(y - \frac{y}{x^2 + y^2}\right)^2 = 1 \]

### Step 4: Simplify the equation
Let’s denote:
\[ A = x + \frac{x}{x^2 + y^2}, \quad B = y - \frac{y}{x^2 + y^2} \]
Then, we have:
\[ A^2 + B^2 = 1 \]

### Step 5: Analyze the imaginary part
The imaginary part \( B \) simplifies to:
\[ B = y\left(1 - \frac{1}{x^2 + y^2}\right) = y\frac{x^2 + y^2 - 1}{x^2 + y^2} \]
For \( B \) to be zero (since \( |z + \frac{1}{z}| = 1 \) implies the imaginary part must equal zero), we have:
\[ y\frac{x^2 + y^2 - 1}{x^2 + y^2} = 0 \]
Since \( y \neq 0 \), we must have:
\[ x^2 + y^2 - 1 = 0 \quad \Rightarrow \quad x^2 + y^2 = 1 \]

### Step 6: Conclude the result
Since \( x^2 + y^2 = 1 \) represents a circle of radius 1 in the complex plane, and since \( y \neq 0 \), we conclude that the real part of \( z \) must be zero for maximum modulus. Thus, \( z \) lies on the imaginary axis.

### Final Answer
The real part of \( z \) is:
\[ \text{Re}(z) = 0 \]