If sqrt{x + iy} = pm(a + ib),then sqrt{-x - i | vedclass

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(C) Given $\sqrt{x + iy} = \pm(a + ib)$.Squaring both sides,we get $x + iy = (a + ib)^2 = a^2 - b^2 + 2iab$.Comparing real and imaginary parts,$x = a^

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$\pm(b - ia)$

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(C) Given $\sqrt{x + iy} = \pm(a + ib)$.Squaring both sides,we get $x + iy = (a + ib)^2 = a^2 - b^2 + 2iab$.Comparing real and imaginary parts,$x = a^2 - b^2$ and $y = 2ab$.Now,we need to find $\sqrt{-x - iy} = \sqrt{-(x + iy)}$.Substituting the values,$\sqrt{-x - iy} = \sqrt{-(a^2 - b^2 + 2iab)} = \sqrt{b^2 - a^2 - 2iab}$.We can rewrite the expression inside the square root as $(b - ia)^2 = b^2 + (ia)^2 - 2iab = b^2 - a^2 - 2iab$.Therefore,$\sqrt{-x - iy} = \sqrt{(b - ia)^2} = \pm(b - ia)$.

If $\sqrt{x + iy} = \pm(a + ib)$,then $\sqrt{-x - iy}$ is equal to

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