Question

How do you find `z, z^2, z^3, z^4` given `z=1/2(1+sqrt3i)`?

Answer 1

`z = cos(pi/3) + isin(pi/3)`

`z^2 = cos(2pi/3) + isin(2pi/3) = 1/2(-1 + sqrt(3)i)`

`z^3 = cos(3pi/3) + isin(3pi/3) = -1`

`z^4 = cos(4pi/3) + isin(4pi/3) = -1/2(1+sqrt(3)i)`

Explanation:

The easiest method is to use De Moivre's theorem. For complex number `z`

`z= r(costheta + isintheta)`

`z^n = r^n(cosntheta + isinntheta)`

So we want to convert our complex number to polar form. The modulus `r` of a complex number `a+bi` is given by

`r = sqrt(a^2+b^2)`

`r = sqrt((1/2)^2 + (sqrt(3)/2)^2) = sqrt(1/4 + 3/4) = 1`

The complex number will be in the first quadrant of an Argand diagram so the argument is given by:

`theta = tan^(-1)(b/a)`

`theta = tan^(-1)((sqrt(3)/2)/(1/2)) = tan^(-1)(sqrt(3)) = pi/3`

`z = cos(pi/3) + isin(pi/3)`

`z^2 = cos(2pi/3) + isin(2pi/3) = 1/2(-1 + sqrt(3)i)`

`z^3 = cos(3pi/3) + isin(3pi/3) = -1`

`z^4 = cos(4pi/3) + isin(4pi/3) = -1/2(1+sqrt(3)i)`