If `z` is a complex number and `z=bar(z)`, then prove that `z` is a purely real number.

Explanation

To prove that if \( z \) is a complex number and \( z = \bar{z} \), then \( z \) is a purely real number, we can follow these steps:

### Step 1: Define the complex number
Let \( z \) be a complex number defined as:
\[ z = x + iy \]
where \( x \) is the real part and \( y \) is the imaginary part of the complex number.

### Step 2: Write the conjugate of \( z \)
The conjugate of \( z \), denoted as \( \bar{z} \), is given by:
\[ \bar{z} = x - iy \]

### Step 3: Set \( z \) equal to \( \bar{z} \)
According to the problem statement, we have:
\[ z = \bar{z} \]
Substituting the expressions for \( z \) and \( \bar{z} \), we get:
\[ x + iy = x - iy \]

### Step 4: Equate the real and imaginary parts
From the equation \( x + iy = x - iy \), we can separate the real and imaginary parts. The real parts are equal:
\[ x = x \]
And the imaginary parts give us:
\[ iy = -iy \]

### Step 5: Solve for \( y \)
From the equation \( iy = -iy \), we can simplify it to:
\[ 2iy = 0 \]
This implies:
\[ y = 0 \]

### Step 6: Substitute \( y \) back into \( z \)
Now that we have found \( y = 0 \), we can substitute this back into the expression for \( z \):
\[ z = x + i(0) = x \]

### Conclusion
Since \( z = x \) and \( y = 0 \), we conclude that \( z \) is a purely real number. Hence, we have proved that if \( z = \bar{z} \), then \( z \) is a purely real number.