If `x` and `y` are complex numbers, then the system of equations `(1+i)x+(1-i)y=1,2i x+2y=1+i` has

Options

A. unique solution
B. no solution
C. infinte number of solutions
D. none of theses

Explanation

To solve the system of equations given by:

1. \((1+i)x + (1-i)y = 1\)
2. \(2ix + 2y = 1+i\)

we will follow these steps:

### Step 1: Rewrite the equations
Let's rewrite the equations for clarity:

1. \((1+i)x + (1-i)y = 1\) (Equation 1)
2. \(2ix + 2y = 1+i\) (Equation 2)

### Step 2: Simplify Equation 2
We can simplify Equation 2 by dividing all terms by 2:

\[ ix + y = \frac{1+i}{2} \]

### Step 3: Express \(y\) in terms of \(x\)
From the simplified Equation 2, we can express \(y\) in terms of \(x\):

\[ y = \frac{1+i}{2} - ix \]

### Step 4: Substitute \(y\) into Equation 1
Now, substitute the expression for \(y\) into Equation 1:

\[ (1+i)x + (1-i)\left(\frac{1+i}{2} - ix\right) = 1 \]

### Step 5: Expand and simplify
Expanding the equation gives:

\[ (1+i)x + (1-i)\left(\frac{1+i}{2}\right) - (1-i)ix = 1 \]

Now, let's simplify the terms:

1. The first term is \((1+i)x\).
2. The second term is \((1-i)\left(\frac{1+i}{2}\right) = \frac{(1-i)(1+i)}{2} = \frac{1^2 - i^2}{2} = \frac{1 - (-1)}{2} = \frac{2}{2} = 1\).
3. The third term is \(-(1-i)ix = -ix + i^2x = -ix - x = -(1+i)x\).

Putting it all together:

\[ (1+i)x + 1 - (1+i)x = 1 \]

### Step 6: Analyze the result
The equation simplifies to:

\[ 1 = 1 \]

This shows that the original equations are dependent, meaning they represent the same line in the complex plane.

### Conclusion
Since the two equations are dependent, there are infinitely many solutions for the system of equations.

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