Question

Statement I Square root of $i$ is $\left(\pm \frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} i\right)$.
Statement II Square root of $(1+i)$ is $\left(\pm \sqrt{\frac{\sqrt{2}+1}{3}} \pm \sqrt{\frac{\sqrt{2}-1}{3} i}\right)$.

• A. Statement I is correct
• B. Statement II is correct
• C. Both are correct
• D. Neither I nor II is correct

Answer 1

Answer: (A)

I. Let $x+i y=\sqrt{i}$
$\Rightarrow (x+i y)^2 =i $
$ \Rightarrow x^2-y^2+2 i x y =i $
$ \Rightarrow x^2-y^2 =0 $....(i)
$ \text { and } 2 x y =1 $
$ \left(x^2+y^2\right)^2 =\left(x^2-y^2\right)^2+(2 x y)^2 $ $\left(x^2+y^2\right) =1$....(ii)
Solving Eqs. (i) and (ii), we get
$x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{2}}$
$\because$ Product of $x y$ is positive.
$\therefore x=\frac{1}{\sqrt{2}}=y $
$\Rightarrow x=-\frac{1}{\sqrt{2}}=y$
Thus, square root of complex number $i$ is
$\left(\pm \frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} i\right)$
$\therefore$ Statement $I$ is correct.
II. Let $ \sqrt{1+i}=x+i y$
$\Rightarrow x^2-y^2+2 i x y=1+i$
$\Rightarrow x^2-y^2=1$....(i)
$\Rightarrow 2 x y=1$
$\Rightarrow \left(x^2+y^2\right)^2=\left(x^2-y^2\right)+(2 x y)^2$
$\Rightarrow x^2+y^2=\sqrt{2}$.....(ii)
Solving Eqs. (i) and (ii), we get
$x=\pm \sqrt{\frac{\sqrt{2}+1}{2}}, y=\pm \sqrt{\frac{\sqrt{2}-1}{2}}$
Thus, square root of complex number $1+i$ is
$\left(\pm \sqrt{\frac{\sqrt{2}+1}{2}} \pm \sqrt{\frac{\sqrt{2}-1}{2}} i\right) .$