2 edits
Originally posted by Mexicowe seek a complex number a+bi, whose square: (a^2 - b^2) + (2ab)i = -i
Don't suppose you want to show me how to get that?
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.
Originally posted by Mexico-i = e^(3pi/2 * i)
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.
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)).