To prove that if [tex]\(\sqrt{x - i y} = a - i b\)[/tex], then [tex]\(\sqrt{x + i y} = a + i b\)[/tex], we can follow a step-by-step logical approach using properties of complex numbers:
1. Given Equation:
[tex]\[
\sqrt{x - i y} = a - i b \quad \text{(1)}
\][/tex]
2. Conjugate Both Sides:
The conjugate of a complex number [tex]\(z = u + iv\)[/tex] is [tex]\(\overline{z} = u - iv\)[/tex].
Applying the conjugate to both sides of equation (1):
[tex]\[
\overline{\sqrt{x - i y}} = \overline{a - i b}
\][/tex]
3. Properties of Conjugates:
Using the property of complex conjugates, the conjugate of the square root of a complex number is the same as the square root of the conjugate of that number. Thus,
[tex]\[
\sqrt{\overline{x - i y}} = a + i b \quad \text{(since} \ \overline{a - i b} = a + i b\text{)}
\][/tex]
4. Simplify the Conjugate:
The conjugate of [tex]\(x - i y\)[/tex] is [tex]\(x + i y\)[/tex]. Therefore,
[tex]\[
\sqrt{x + i y} = a + i b
\][/tex]
Thus, we have shown that if [tex]\(\sqrt{x - i y} = a - i b\)[/tex], then it follows that [tex]\(\sqrt{x + i y} = a + i b\)[/tex].
This completes the proof.
