Let `z = x + iy` be the given complex number and we have to obtain its square root.

Let `a+ib=(x+iy)^(1//2)=> a^2-b^2+2iab=x+iy`

`=> x=a^2-b^2` and `y=2ab`

`=> x^2=(a^2+b^2)^2-4a^2b^2=> x^2+y^2=(a^2+b^2)^2`

`=>a^2+b^2=|z|`......................(1)

`=> a^2-b^2=x`.................(2)

`=> a^2=(|z|+x)/2=> a= pm sqrt((|z|+x)/2)`

`=>b^2=(|z|-x)/2=> b= pm sqrt((|z|-x)/2)`

`:. sqrt(x+iy)=a+ib= pm (sqrt((|z|+Re(z))/2)+i sqrt((|z|-Re(z))/2))`

Replacing `i` by `- i`, we get

`sqrt(x+iy)= pm (sqrt((|z|+Re(z))/2)-i sqrt((|z|-Re(z))/2))`

`text(Aliter :)`

If `sqrt(x+iy)`, where `x, y in R` and `i = sqrt(-1),` then

1. If y is not even, then multiply and divide in y by 2, then `sqrt{x + iy)` convert in `sqrt(x+ysqrt(-1)) = sqrt()x + 2sqrt(-y^2/4)`

2. Factorise `: - y^2/4 ` say `alpha , beta (alpha < beta)`

Take that possible factor which satisfy

`x = (alpha i)^2 + beta^2 , ` if `x > 0`

or `x = alpha^2 + ( i beta)^2 ,` if `x < 0`

3. Finally, write `x + iy = (alphai)^2 + beta^2 + 2ialphabeta`

or `alpha^2 + ( i beta)^2 + 2 i alpha beta`

and take their square root.

4. `sqrt(x+ iy) =` `{ tt((,pm,alphai+beta),(or,pm,alpha + i beta))`

and `sqrt(x- iy) =` `{ tt((,pm,alphai-beta),(or,pm,alpha - i beta))`

`text(Note)`

`1.` The square root of i is `pm((1+i)/sqrt2),` where `i= sqrt(-1)`

`2.` The square root of `(-i)` is `((1-i)/sqrt2).`