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) $\sqrt{x^2-y^2}$

Answer 1

(d) x $-$ iy

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