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 $x+iy=\sqrt{i}$
$\Rightarrow (x+iy{)}^2=i$
$\Rightarrow x^2-y^2+2ixy=i$
$\Rightarrow x^2-y^2=0$....(i)
$\text{ and }2xy=1$
${\left(x^2+y^2\right)}^2={\left(x^2-y^2\right)}^2+(2xy{)}^2$ $\left(x^2+y^2\right)=1$....(ii)
Solving Eqs. (i) and (ii), we get
$x=\pm \frac{1}{\sqrt{2}},y=\pm \frac{1}{\sqrt{2}}$
$\because$ Product of $xy$ 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 $\sqrt{1+i}=x+iy$
$\Rightarrow x^2-y^2+2ixy=1+i$
$\Rightarrow x^2-y^2=1$....(i)
$\Rightarrow 2xy=1$
$\Rightarrow {\left(x^2+y^2\right)}^2=\left(x^2-y^2\right)+(2xy{)}^2$
$\Rightarrow x^2+y^2=\sqrt{2}$.....(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
$\left(\pm \sqrt{\frac{\sqrt{2}+1}{2}}\pm \sqrt{\frac{\sqrt{2}-1}{2}}i\right).$