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 \( z \).

### 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 up the equation
According to the problem, we have:
\[ z = \bar{z} \]
Substituting the expressions for \( z \) and \( \bar{z} \) gives:
\[ x + iy = x - iy \]

### Step 4: Equate the real and imaginary parts
From the equation \( x + iy = x - iy \), we can equate the real and imaginary parts:
- The real parts give us:
\[ x = x \]
This equation is always true.

- The imaginary parts give us:
\[ iy = -iy \]
This implies:
\[ 2iy = 0 \]

### Step 5: Solve for \( y \)
From \( 2iy = 0 \), we can divide both sides by \( 2i \) (assuming \( i \neq 0 \)):
\[ y = 0 \]

### Step 6: Conclusion
Since \( y = 0 \), we can substitute back into the expression for \( z \):
\[ z = x + i(0) = x \]
Thus, \( z \) is purely real, as it can be expressed as \( z = x \) where \( x \) is a real number.

### Final Statement
Therefore, we have proven that if \( z = \bar{z} \), then \( z \) is a purely real number.

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