if `(1+i)z=(1-i)barz` then `z` is

Options

A. `t ( 1- i)t, in R `
B. `t ( 1+ i)r, in R `
C. `t/(1+i),tin R ^(+) `
D. None of these

Answer

To solve the equation \((1+i)z = (1-i)\bar{z}\), we will follow these steps:

### Step 1: Express \(z\) in terms of its real and imaginary parts
Let \(z = a + bi\), where \(a\) and \(b\) are real numbers. The conjugate of \(z\) is given by \(\bar{z} = a - bi\).

### Step 2: Substitute \(z\) and \(\bar{z}\) into the equation
Substituting \(z\) and \(\bar{z}\) into the equation, we have:
\[ (1+i)(a+bi) = (1-i)(a-bi) \]

### Step 3: Expand both sides
Now, we will expand both sides of the equation:
- Left-hand side:
\[ (1+i)(a+bi) = a + ai + bi + b(i^2) = a + ai + bi - b = (a-b) + (a+b)i \]
- Right-hand side:
\[ (1-i)(a-bi) = a - ai - bi + b(i^2) = a - ai - bi - b = (a-b) - (a+b)i \]

### Step 4: Set the real and imaginary parts equal
Now we equate the real and imaginary parts from both sides:
1. Real part: \(a - b = a - b\) (This is always true)
2. Imaginary part: \(a + b = -(a + b)\)

### Step 5: Solve the imaginary part equation
From the imaginary part equation:
\[ a + b = -(a + b) \]
This simplifies to:
\[ a + b + a + b = 0 \implies 2(a + b) = 0 \implies a + b = 0 \]
Thus, we can express \(b\) in terms of \(a\):
\[ b = -a \]

### Step 6: Write \(z\) in terms of \(a\)
Substituting \(b = -a\) back into the expression for \(z\):
\[ z = a + bi = a - ai = a(1 - i) \]

### Step 7: General solution
Since \(a\) can be any real number, we can express \(z\) as:
\[ z = t(1 - i) \quad \text{where } t \in \mathbb{R} \]

### Conclusion
Thus, the solution to the equation \((1+i)z = (1-i)\bar{z}\) is:
\[ z = t(1 - i) \quad \text{for any real number } t \]