# What is the Square Root of i [Polar Form]

Note that i=√-1 is an imaginary complex number. The square root of i is equal to ±(1+i)/√2. In this post, we will learn how to find the square root of i.

## What is the Square Root of i

We have:

$i =\dfrac{1}{2} \cdot 2i$

= $\dfrac{1}{2} (1+2i-1)$, adding 1 and subtracting 1 will change nothing.

= $\dfrac{1}{2}(1+2i+i^2)$ as we know that i2=-1.

= $\dfrac{1}{2}(1+i)^2$ using the identity a2+2ab+b2=(a+b)2

So we have obtained that

$i =\dfrac{1}{2}(1+i)^2$

Now, taking the square root on both sides will give us the square root of i. That is,

$\sqrt{i} = \pm \dfrac{1}{\sqrt{2}}(1+i)$

So the value of the square root of i, that is, the square root of iota is equal to ±(1+i)/√2. We write it as follows:

## Square Root of i in Polar Form

We know that the polar form i is equal to

$i=e^{\pi/2}$

We take square root on both sides. Thus,

$\sqrt{i}= \pm e^{\pi/4}$ as √x=±x1/2

∴ $\sqrt{i}= \pm e^{\pi/4}$

= ± [cos(π/4) + i sin(π/4)]

= ± [1/√2 + i/√2]

= ±(1+i)/√2.

Therefore, the value of square root of i is equal to ±(1+i)/√2 and this is obtained by using the polar form of a complex number.