If `sqrt(a+ib)=x+iy`, then value of `sqrt(a-ib)` is:

Options

A. `x+iy`
B. `x-iy`
C. `sqrt(x^(2)-y^(2))`
D. `sqrt(x^(2)+y^(2))`.

Answer

To solve the problem, we start with the equation given:

\[ \sqrt{a + ib} = x + iy \]

We need to find the value of \(\sqrt{a - ib}\).

### Step 1: Square both sides of the equation
Squaring both sides gives us:

\[ a + ib = (x + iy)^2 \]

Expanding the right-hand side:

\[ a + ib = x^2 + 2xyi - y^2 \]

### Step 2: Separate real and imaginary parts
From the equation \(a + ib = (x^2 - y^2) + (2xy)i\), we can equate the real and imaginary parts:

1. Real part: \(a = x^2 - y^2\)
2. Imaginary part: \(b = 2xy\)

### Step 3: Find \(\sqrt{a - ib}\)
Now, we want to find \(\sqrt{a - ib}\). We can express this as:

\[ \sqrt{a - ib} = z_2 \]

Assuming \(z_2 = u + iv\), we square both sides:

\[ a - ib = (u + iv)^2 \]

Expanding the right-hand side:

\[ a - ib = u^2 - v^2 + 2uvi \]

### Step 4: Equate real and imaginary parts again
From the equation \(a - ib = (u^2 - v^2) + (2uv)i\), we can equate the real and imaginary parts again:

1. Real part: \(a = u^2 - v^2\)
2. Imaginary part: \(-b = 2uv\) (or \(b = -2uv\))

### Step 5: Relate \(u\) and \(v\) to \(x\) and \(y\)
We already have from the first part:

1. \(a = x^2 - y^2\)
2. \(b = 2xy\)

From the imaginary part of \(\sqrt{a - ib}\):

\[ b = -2uv \]

### Step 6: Find the relationship between \(u\) and \(v\)
We can see that:

\[ uv = -\frac{b}{2} \]

### Step 7: Use the modulus property
The modulus of both complex numbers must be equal:

\[ \sqrt{a^2 + b^2} = \sqrt{x^2 + y^2} \]

Squaring both sides gives:

\[ a^2 + b^2 = x^2 + y^2 \]

### Step 8: Conclude the value of \(\sqrt{a - ib}\)
From the properties of complex numbers, we can conclude that:

\[ \sqrt{a - ib} = x - iy \]

### Final Answer
Thus, the value of \(\sqrt{a - ib}\) is:

\[ \sqrt{a - ib} = x - iy \]