If sqrt{a + ib} = x + iy,then the possible va | vedclass

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(D) Given $\sqrt{a + ib} = x + iy$. Squaring both sides,we get $a + ib = (x + iy)^2 = x^2 - y^2 + 2xyi$. Comparing real and imaginary parts,we have $a

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$x - iy$

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(D) Given $\sqrt{a + ib} = x + iy$. Squaring both sides,we get $a + ib = (x + iy)^2 = x^2 - y^2 + 2xyi$. Comparing real and imaginary parts,we have $a = x^2 - y^2$ and $b = 2xy$. Now,consider $\sqrt{a - ib} = \sqrt{(x^2 - y^2) - 2xyi}$. This can be written as $\sqrt{x^2 + (iy)^2 - 2x(iy)} = \sqrt{(x - iy)^2} = x - iy$. Thus,the correct option is $D$.

If $\sqrt{a + ib} = x + iy$,then the possible value of $\sqrt{a - ib}$ is

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