What is the square root of complex numbers?

[UnD]

Please help with these simple questions just not understanding it properly.

Find square root, of -6i
let sqroot of -6i= x+ yi
then -6i=x^2 - y^2 +2xyi
x^2 - y^2 = 0 and 2xy=-6
then xy=-3
x=-3/y
and then solve simu..
i got y= 3 and x=-1 y=-3 x=1
so the anser is +_(-1+3i)
BUt that isn't the answer, please help
one more
square root of
i,
sqroot of i= x+yi
i=x^2-y^2+2xyi
then x^2-y^2= 0 and 2xy=sqroot -1
I am a bit lost around here. Please help.

 

[Tide]

Write the complex numbers in polar form then extract the square root!

If you're not familiar with that then try something like this:

If a + i b = \sqrt {x + i y} then (a+ib)^2 = (x + iy). Expand the square, equate real and imaginary parts of the two sides respectively, then solve the resulting system of equations for a and b.

 

[HallsofIvy]

Using polar form is the standard way of extracting roots, especially of higher order roots, but I think UnD is right to try setting it up in an elementary form just to see how it works!
Yes, it is true that (x+ iy)²= x²- y²+ (2xy)i= -6i so we must have both x²- y²= 0 and 2xy= -6.
Dividing both sides of the second equation by 2y gives x= -3/y.
You then say "and then solve simu.. i got y= 3". It is the ".." that you should have shown us because that's where the problem is! Putting x= -2/y into the first equation we get 9/y²- y²= 0 or
y²= 9/y². Multiplying both sides by y²,
y⁴= 9. Then y²= +/- 3 and so y appears to have 4 possible values:
y= \sqrt{3}, y= -\sqrt{3}, y= i\sqrt{3}, y= -i\sqrt{3}.
But since "y" is a real number (in x+ iy, both x and y are real), only the first two are plausible solutions.
If y= \sqrt{3}, the first equation becomes x²- 3= 0 so either x= \sqrt{3} or x= -\sqrt{3}.
If y= -\sqrt{3}, we get the same equation and the same solutions for x.

That is, we have as possible solutions:
\sqrt{3}+ i\sqrt{3}, -\sqrt{3}+i\sqrt{3}, \sqrt{3}-i\sqrt{3}, and -\sqrt{3}-i\sqrt{3}.

It is easy to check, by direct multiplication, that only
\sqrt{3}-i\sqrt{3} and -\sqrt{3}+ i\sqrt{3}
satisfy the equations (the other two "extraneous" roots were introduced when we multiplied by sides of the equation by y²).

We can use the "polar" form, geometrically, to check that. The point -6i is on the negative "y" (imaginary) axis in the complex plane so it's angle, with the positive (real) axis, is either 270 degrees or -90 degrees. Taking the square root halves that (square root is 1/2 power) giving either 135 or -45 degrees, the line y= -x, so the real and imaginary parts must be negatives of one another. |-6i|= 6 so the absolute value of the square root is the \sqrt{6} which is true for the solutions above.

 

[UnD]

Thanks very much. You are very helpfull.
I just made a silly mistake. That i didn't pick up even after i did it again.
Thanks very much