Question

Find the square root of $2i$.

Answer 1

In this particular question add and subtract 1 from $2i$ and use ${i^2} = - 1$ to make in form of ${a^2} + 2ab + {b^2}$ and after that use the formula ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$ and then take square root to reach the solution of the question

** Complete step-by-step answer :**
There are many different ways to find the square root of a complex number.
So, one of the ways is that we have to make that complex number such that it becomes the square of any other number then we can easily find the square root of the given complex number.
So, now let us add and subtract 1 from $2i$ to make it a perfect square and as we know that adding and subtracting a constant number to any equation does not affect the solution.
So, $2i = 2i + 1 - 1$ (1)
Now as we know that if $i$ is a complex then the value of $i$ is equal to $\sqrt { - 1}$ .
So, $i = \sqrt { - 1}$
And squaring both the sides of the above equation.
$\Rightarrow {i^2} = {\left( {\sqrt { - 1} } \right)^2} = - 1$ (2)
So, now putting the value of –1 form equation 2 to equation 1.
$2i = 2i + {\left( 1 \right)^2} + {\left( i \right)^2} = {\left( 1 \right)^2} + 2i + {\left( i \right)^2}$
Now as we know that ${a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}$. So, in the above equation a will be 1 and b will be $i$.
So, $2i = {\left( {1 + i} \right)^2}$ (3)
Now taking the square root on both sides of the above equation.
$\Rightarrow \sqrt {2i} = \sqrt {{{\left( {1 + i} \right)}^2}} = \pm \left( {1 + i} \right)$
Hence, the square root of $2i$ will be equal to $\left( {1 + i} \right)$ or $- \left( {1 + i} \right)$

Note :Whenever we face such type of questions the key concept, we had to remember is that always change the given equation to a square or cube of any other number as per requirement like if square root is asked then change it into a square of another number or if cube root is asked then change it into a cube root of any other number. And for that we have to add and subtract some constant terms and recall the complex identities such that $i = \sqrt { - 1}$, ${i^2} = - 1$, ${i^3} = - i$ and ${i^4} = 1$. And then take the square root on both sides of the equation and not neglect the negative sign because when we take the square root of a number then there are two possible answers positive and negative.