If `((1+i) z= (1-i))bar(z),` then

Options

A. z lies on a straight line
B. for all such `z ne 0, "arg" (z) = 3pi//4`
C. all such z are given by `z=t (1-i), t in R`
D. z must be purely real

Answer

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

### Step 1: Substitute \( z \) and \( \overline{z} \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The conjugate \( \overline{z} \) will be \( x - iy \).

### Step 2: Rewrite the equation
Substituting \( z \) and \( \overline{z} \) into the equation gives us:
\[ (1+i)(x+iy) = (1-i)(x-iy) \]

### Step 3: Expand both sides
Now, we expand both sides:
- Left side:
\[ (1+i)(x+iy) = x + ix + iy - y = (x - y) + i(x + y) \]
- Right side:
\[ (1-i)(x-iy) = x - ix - iy + y = (x + y) + i(-x + y) \]

### Step 4: Set real and imaginary parts equal
Now we equate the real and imaginary parts from both sides:
1. Real part: \( x - y = x + y \)
2. Imaginary part: \( x + y = -x + y \)

### Step 5: Solve the equations
From the first equation:
\[ x - y = x + y \implies -y = y \implies 2y = 0 \implies y = 0 \]

From the second equation:
\[ x + y = -x + y \implies x = -x \implies 2x = 0 \implies x = 0 \]

### Step 6: Conclusion
Thus, we find that \( z = x + iy = 0 + 0i = 0 \).