# Sq root of complex numbers

1. Oct 14, 2005

### UnD

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)
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

2. Oct 14, 2005

### 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.

3. Oct 14, 2005

### HallsofIvy

Staff Emeritus
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.

Last edited: Oct 14, 2005
4. Oct 14, 2005

### 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