If `x, y, a, b in R, a ne 0 and (a + ib) (x + iy) = (a^(2) + b^(2))i`, then (x, y) equals

Options

A. (a, b)
B. (a, 0)
C. (0, b)
D. (b, a)

Answer

To solve the equation \((a + ib)(x + iy) = (a^2 + b^2)i\) where \(x, y, a, b \in \mathbb{R}\) and \(a \neq 0\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:
\[ (a + ib)(x + iy) = ax + iay + ibx + i^2by \]
Since \(i^2 = -1\), we can rewrite this as:
\[ ax + iay + ibx - by = (ax - by) + i(ay + bx) \]

### Step 2: Set equal to the right-hand side
Now we set the real and imaginary parts equal to the corresponding parts of the right-hand side:
\[ (ax - by) + i(ay + bx) = 0 + i(a^2 + b^2) \]
This gives us two equations:
1. \(ax - by = 0\) (Real part)
2. \(ay + bx = a^2 + b^2\) (Imaginary part)

### Step 3: Solve the first equation for \(x\)
From the first equation \(ax - by = 0\), we can express \(x\) in terms of \(y\):
\[ ax = by \implies x = \frac{by}{a} \]

### Step 4: Substitute \(x\) into the second equation
Now we substitute \(x = \frac{by}{a}\) into the second equation \(ay + bx = a^2 + b^2\):
\[ ay + b\left(\frac{by}{a}\right) = a^2 + b^2 \]
This simplifies to:
\[ ay + \frac{b^2y}{a} = a^2 + b^2 \]
Multiplying through by \(a\) to eliminate the fraction gives:
\[ a^2y + b^2y = a(a^2 + b^2) \]
Factoring out \(y\) from the left side:
\[ y(a^2 + b^2) = a(a^2 + b^2) \]

### Step 5: Solve for \(y\)
Assuming \(a^2 + b^2 \neq 0\) (which is true since \(a \neq 0\)), we can divide both sides by \(a^2 + b^2\):
\[ y = a \]

### Step 6: Substitute \(y\) back to find \(x\)
Now substitute \(y = a\) back into the expression for \(x\):
\[ x = \frac{b(a)}{a} = b \]

### Final Result
Thus, we find that:
\[ (x, y) = (b, a) \]