Question

# The value of $\sqrt {2i}$:A) $1 + i$B) $- 1 - i$C) $- \sqrt {2i}$D) None of these

Hint: As the root of the complex number will also be a complex number, let the root of $2i$ as $x + iy$. Then, the equation will be $\sqrt {2i} = x + iy$. Square on both sides and compare the real and imaginary part of the equation and form equations. Solve the equations to find the value of $x$ and $y$. Then, substitute the values in $\sqrt {2i} = x + iy$ to find the required value.

Complete step by step solution: Since the given number is a complex number, therefore the square root of this number will also be a complex number.
So, let the square root of the given number $2i$ as $x + iy$ . Thus we can write
${\left( {x + iy} \right)^2} = 2i$
On simplifying the above equation we get,
${x^2} - {y^2} + 2ixy = 2i{\text{ }}\left( {\text{1}} \right)$
The part of the complex number without the $i$ as coefficient is called the real part of the complex number, and the part of the complex number with $i$ as its coefficient is called the imaginary part of the complex number.
On comparing the real and imaginary part of the equation $\left( 1 \right)$, we get
${x^2} - {y^2} = 0 \\ - 2xy = 2 \\$
On simplifying the equation we get
${x^2} - {y^2} = 0{\text{ }}\left( {\text{2}} \right) \\ xy = 1{\text{ }}\left( {\text{3}} \right) \\$
Find the value of ${x^2} + {y^2}$using the formula ${\left( {{m^2} + {n^2}} \right)^2} = {\left( {{m^2} - {n^2}} \right)^2} + 4{m^2}{n^2}$
${\left( {{x^2} + {y^2}} \right)^2} = {\left( {{x^2} - {y^2}} \right)^2} + {\left( {2xy} \right)^2}$
Substituting the values from equation $\left( 2 \right){\text{ and }}\left( 3 \right)$ we get
${\left( {{x^2} + {y^2}} \right)^2} = {\left( 0 \right)^2} + {\left( {2\left( { - 1} \right)} \right)^2} \\ = 0 + 4 \\ = 4 \\ {x^2} + {y^2} = 2{\text{ }}\left( {\text{4}} \right) \\$
Adding equation $2$ and $4$we get
${x^2} - {y^2} + {x^2} + {y^2} = 0 + 2 \\ {\text{2}}{x^2} = 2 \\ {x^2} = 1 \\ x = \pm 1 \\$
Substituting the $\pm 1$ for $x$in the equation $1$.
$xy = - 1$
When $x = 1$
Then,
$\left( 1 \right)y = 1 \\ \Rightarrow y = 1 \\$
Similarly, When $x = - 1$
Then,
$\left( { - 1} \right)y = 1 \\ \Rightarrow y = - 1 \\$

Substituting the values of $x,y$ in the required complex number
$x + iy = 1 + i$ and $x + iy = - 1 - i$

Hence, option A and B are correct.

Note: We can also do this question by adding and subtracting 1 to $2i$ and we will get $2i = 1 + 2i - 1$. The term can be rewritten as ${1^2} + 2\left( 1 \right)i - {1^2} \Rightarrow {1^2} + 2\left( 1 \right)i + {i^2}$. Hence, $2i = {\left( {1 + i} \right)^2}$. On taking square roots both sides, we will get, $\sqrt {2i} = \pm \left( {1 + i} \right)$. Also, the square root of the complex number will also be a complex number.