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

Options

A. `Re(z)=0`
B. `Im(z)=0`
C. `Re(z)=Im(z)`
D. None of these

Answer

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

### Step 1: Define the complex number
Let \( z \) be a complex number, which can be expressed in the form:
\[ z = x + iy \]
where \( x \) is the real part and \( y \) is the imaginary part of \( z \).

### Step 2: Write the conjugate of the complex number
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: Compare real and imaginary parts
Now, we can compare the real and imaginary parts of both sides of the equation:
- The real part: \( x = x \) (this is always true)
- The imaginary part: \( iy = -iy \)

### Step 5: Solve for the imaginary part
From the imaginary part, we have:
\[ iy + iy = 0 \implies 2iy = 0 \]
This implies:
\[ 2y = 0 \]
Thus, we find:
\[ y = 0 \]

### Step 6: Conclude that \( z \) is purely real
Since \( y = 0 \), we can substitute back into the expression for \( z \):
\[ z = x + i(0) = x \]
This shows that \( z \) has no imaginary part, meaning \( z \) is purely real.

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

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