How do you solve the power of an imaginary number?

Answer 1

It helps to think about imaginary and complex numbers graphically.

Euler's Formula (see related link): eiΘ = cos(Θ) + i sin(Θ) {Θ is in radians}. Note that both eiΘ and [cos(Θ) + i sin(Θ)] have a magnitude of 1, so multiply by the magnitude: AeiΘ = Acos(Θ) + Ai sin(Θ). You now have a graphical representation of complex numbers, with real numbers on the horizontal axis, pure imaginaries on the vertical axis, and all other complex numbers placed on the 'complex plane'. The angle is a direction, from the origin, and the magnitude A tells how far away from the origin that the position is.

With pure imaginary numbers you can have Θ = pi/2 radians (90°, vertical), and let A be either positive or negative (up or down). From the rules for exponents and powers, you now have the imaginary number z = ei*pi/2, and (ex)n = ex*n, so zn = (ei*pi/2)n = ei*n*pi/2 , so switching to degrees for simplicity:

n Θ

0 0° (Points to the right: positive real)

1 90° (Pointing straight up: imaginary positive number)

2 180° (Points to the left: real negative number)

3 270° (Points straight down: imaginary negative)

4 360° (Points to the right: real positive ), same as 0°

Note it goes in a circle and repeats. Odd integer values of n will be pure imaginary and even integers will be real numbers. Non-integers will put the angle so it is a complex number. Negative exponents cause it to move in a clockwise direction on the circle, rather than counterclockwise (for positive exponents).

Now that you know the direction, you only need to take An, as a power, and then point it in the proper direction. So if the power of A yields a positive number, the answer will be in the direction, but if it yields a negative number (odd integer powers of a negative A), then it's in the opposite direction (add 180° to the angle).