Question

What is the square root of `2i`?

Answer 1

`sqrt{2i}={1+i, -1-i}`

Let us look at some details.

Let `z=sqrt{2i}`.
(Note that `z` are complex numbers.)

by squaring,

`Rightarrow z^2=2i`

by using the exponential form `z=re^{i theta}`,

`Rightarrow r^2e^{i(2theta)}=2i=2e^{i(pi/2+2npi)}`

`Rightarrow {(r^2=2 Rightarrow r=sqrt{2}), (2theta=pi/2+2npi Rightarrow theta=pi/4+npi):}`

So, `z=sqrt{2}e^{i(pi/4+npi)}`

by Eular's Formula: `e^{i theta}=cos theta +isin theta`

`Rightarrow z=sqrt{2}[cos(pi/4+npi)+isin(pi/4+npi)]`

`=sqrt{2}(pm1/sqrt{2}pm1/sqrt{2}i)=pm1pmi`

I kept the following original post just in case someone needs it.
`(2i)^(1/2)` = `(2)^(1/2)` `(i)^(1/2)`,

`(i)^(1/2)` = -1

`(2i)^(1/2)` = `(2)^(1/2)` x -1

`(2)^(1/2)` = 1.41

`(2i)^(1/2)` = 1.41 x -1 = -1.41