- 24 Feb '08 04:41 / 2 edits

we seek a complex number a+b*Originally posted by Mexico***Don't suppose you want to show me how to get that?****i**, 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+b**i**= [sqrt(2)/2]**i**- [sqrt(2)/2]**i^2**or finally:

a+b**i**= [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**. - 25 Feb '08 05:23

-i = e^(3pi/2 * i)*Originally posted by Mexico***Don't suppose you want to show me how to get that?**

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. - 29 Feb '08 21:12The 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)).