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Question

If $\sqrt{a+ib}=x+iy,$ then possible value of $\sqrt{a-ib}$ is (a) ${x}^{2}+{y}^{2}$ (b) $\sqrt{{x}^{2}+{y}^{2}}$ (c) x + iy (d) x − iy (e) $\sqrt{{x}^{2}-{y}^{2}}$

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Solution

(d) x $-$ iy $\sqrt{a+ib}=x+iy\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get},\phantom{\rule{0ex}{0ex}}a+ib={x}^{2}+\left(iy{\right)}^{2}+2ixy\phantom{\rule{0ex}{0ex}}⇒a+ib=\left({x}^{2}-{y}^{2}\right)+2ixy\phantom{\rule{0ex}{0ex}}\therefore a=\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{and}b=2xy\phantom{\rule{0ex}{0ex}}\therefore a-ib=\left({x}^{2}-{y}^{2}\right)-2ixy\phantom{\rule{0ex}{0ex}}⇒a-ib={x}^{2}+{i}^{2}{y}^{2}-2ixy\left[\because {i}^{2}=-1\right]\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\sqrt{a-ib}=x-iy$

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