Anyone better at math than me....

Whats the square root of -i ??

-sqrt(2)/2+sqrt(2)i/2
and
sqrt(2)/2-sqrt(2)i/2
I think.

Originally posted by GregM
-sqrt(2)/2+sqrt(2)i/2
and
sqrt(2)/2-sqrt(2)i/2
I think.

Don't suppose you want to show me how to get that?

Originally posted by Mexico
Don't suppose you want to show me how to get that?

we seek a complex number a+bi, whose square: (a^2 - b^2) + (2ab)i = -i

so we seek a simultaneous solution to

(1) a^2 - b^2 = 0 and

(2) 2ab = -1.

clearly by equation (1), a=b will give us one solution. then by (2), a=b= [sqrt(2)/2]*i, and our complex number a+bi = [sqrt(2)/2]i - [sqrt(2)/2]i^2 or finally:
a+bi = [sqrt(2)/2]i - sqrt(2)/2

the other solution is given by a=-b. so then a = [sqrt(2)/2] and b = -[sqrt(2)/2].
so a + bi = [sqrt(2)/2] - [sqrt(2)/2]i.

Thank you....

you're welcome 🙂 it's easier than it looks with all the sqrt(---) notation. i wish equation editor worked outside of microsoft word haha

Another way is to use the identity:

r(cos x + i sin x) = e^(ix)

So -i = e^(-i PI/2)

=> sqrt(i) = +/- e^(-i PI/4)
= +/- [cos PI/4 - i sin PI/4]
= +/- [sqrt(2)/2 - i sqrt(2)/2

Originally posted by Mexico
Don't suppose you want to show me how to get that?

-i = e^(3pi/2 * i)
sqrt(-i) = (-i)^(1/2) = (e^(3pi/2 * i)) ^ (1/2) = e^(3pi/2 * i * 1/2) = e^(3pi/4 * i)

r*e^(theta * i) = r*(cos(theta) + i*sin(theta)), so that
e^(3pi/4 * i) = cos(3pi/4) + i*sin(3pi/4) = -sqrt(2)/2 + sqrt(2)i/2

But all numbers have two square roots (and, in general, n nth roots). The second square root is always the negative of the first, because if x = sqrt(y) then (-x)^2 = (-1 * x)^2 = (-1)^2 * (x)^2 = 1*x^2 = y, so that -x is another square root of y. So the other root is

-1*(-sqrt(2)/2 + sqrt(2)i/2) = sqrt(2)/2 - sqrt(2)i/2.

Originally posted by Mexico
Whats the square root of -i ??

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

Further to what GregM showed, use can use de Moivre's theorem to find the nth root of a complex number. Given a complex number c = a + bi = r*cis(x), where cis(x) = sin(x) + i*cos(x), the roots are:

c^(1/n) = r^(1/n) * (sin((x+2k*pi)/n) + i*cos((x+2k*pi)/n)) for k = 0,1,...(n-1)

simplest way:

transform < a , b > = a + bi

to < r , y > where r = radius and y = angle from positive real numbers.

then < r , y > ^n = < r^n , y*n >

to get all roots of a number

< r , y > ^(1/n) = < r^(1/n) , [y/n + (2k pi /n)] > k= (1, 2, ... , n)

The cool thing about the roots of an imaginary number is that when you plot them on the imaginary plane, they form a regular n-gon. If you can find one of the roots by inspection, this makes finding the other roots a snap.

For example, one of the cube roots of 1 is 1. To find the others, just rotate the vector (1,0) on the imaginary plane 120 degrees (360/3) until you're back where you started. The other two roots are then (-SQRT(2), SQRT(2)) and (-SQRT(2), -SQRT(2)).

this material is too complex for me.

and i've always loved that fact about the n-th roots of a number forming a regular n-gon. it's unreal.

Originally posted by Mexico
Whats the square root of -i ??

Do negative numbers have square roots? I don't think so, but my maths teacher says they do in a complicated way...

Originally posted by Mexico
Whats the square root of -i ??

You can check quaternions math. It's cool.