If `z=ibarz` then

Options

A. z is purely real
B. z is purely imaginary
C. `z=x(1+i),x in R`
D. `z=0`

Answer

To solve the equation \( z = i \bar{z} \), we will follow these steps:

1. **Express \( z \) and \( \bar{z} \)**:
Let \( z = x + i y \), where \( x \) is the real part and \( y \) is the imaginary part. The conjugate of \( z \) is given by \( \bar{z} = x - i y \).

2. **Substitute into the equation**:
Substitute \( z \) and \( \bar{z} \) into the equation:
\[ z = i \bar{z} \implies x + i y = i (x - i y) \]

3. **Simplify the right-hand side**:
Calculate the right-hand side:
\[ i (x - i y) = i x + y \]
Therefore, the equation becomes:
\[ x + i y = y + i x \]

4. **Equate real and imaginary parts**:
From the equation \( x + i y = y + i x \), we can separate the real and imaginary parts:
- Real part: \( x = y \)
- Imaginary part: \( y = x \)

5. **Conclusion**:
Since both equations \( x = y \) and \( y = x \) are the same, we can conclude that:
\[ y = x \]
Thus, we can express \( z \) in terms of \( x \):
\[ z = x + i x = x(1 + i) \]

Hence, the solution is:
\[ z = x(1 + i) \]
where \( x \) is any real number.