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 i^{2}=-1.

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

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

**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=±x^{1/2}

∴ $\sqrt{i}= \pm e^{\pi/4}$

= ± [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.

**ALSO READ:**

**How to multiply complex numbers**

**FAQs**

### Q1: What is square root of i?

Answer: The square root of i is ±(1+i)/√2.