Square Root of Negative i (Polar Form)

The square root of negative i is equal to ±(1-i)/√2 where i=√-1 is an imaginary complex number. That is, √-i= ±(1-i)/√2. In this post, we will find the square root of -i.

What is Square Root of Negative i

Note that we can write -i as follows.

$-i =\dfrac{1}{2} \cdot (-2i)$

Add 1 and subtract 1 from -2i. By doing so we get that

$-i = \dfrac{1}{2} (1-2i-1)$

= $\dfrac{1}{2}(1-2i+i^2)$. This follows because i²=-1.

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

Hence, we deduce from the above that

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

Now, we take the square root on both sides. This will give us the square root of negative i. In other words, we have

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

So the square root of negative i (iota) equals ±(1-i)/√2. That is,

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

Square Root of i in Polar Form

The polar form negative i is equal to

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

Taking square roots on both sides, we get that

$\sqrt{-i}= \pm e^{-\pi/4}$ as we know √a=±a^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 the square root of negative i is equal to ±(1-i)/√2 and this is obtained by using the polar form of a complex number.

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