Hello, anthmoo!

I assume you're not allowed to use DeMoivre's Theorem.

. . It would be too easy!

Let: .z .= .a + bFind the square root ofiwherei= √(-1)i, whereaandbare real numbers and z² =i.

Then we have: .(a + bi)² .= .i

Expand: .a² + 2abi+ b²i² .= .i

We have: .(a² - b²) + 2abi.= .i

Equate real and imaginary components:

. . a² - b² .= .0 .[1]

- - - .2ab . = .1 .[2]

[1] gives us: .b² = a² . → . b = ±a

Solve [2] forb: .b = 1/(2a)

Substitute into [1]: .a² - (1/2a)² .= .0 . → . a² - 1/(4a²) .= .0

Multiply by 4a²: .4a^4 - 1 .= .0 . → . a^4 .= .1/4

. . . . - . - . - . ._ . . . . . . . - . . ._

Hence: .a = ±√2/2 .and .b = ±√2/2

. . . . . . . . . _ . . . . . _

Therefore: .√i .= .±(√2/2)(1 +i)