Really Complex

A= e^[i(pi/3)*n] using Euler theorem.

period of sin & cos is 2pi.

so pi/3 * n = 2pi this implies n = 6

You can start by putting the above $a + ib$ form into polar form $r e^{i\theta}$ by setting

$r = \sqrt{a^{2} + b^{2}}$

and

$\theta = tan^{-1}(b/a)$ .

You will find that $r = 1$ and that $a = 1/2$ . Knowing that $cos(\theta) = 1/2$ implies that $\theta = 60$ degrees or $\pi/3$ radians, which is an alternative way to find $\theta$ if you are unfamiliar with the tangent function. Since we have our complex number in polar form, we can rewrite the full set of $A$ as $A = e^{in(\pi/3)}$ . Now we see clearly that increasing n from 0 to 5 generates unique elements in A, though $A(n=6) = A(n=0)$ as $e^{i2\pi n \theta} = e^{i \theta}$ for all $n$ in the integers. Thus there are 6 unique elements in the complex set A.

Iam Pakistani

the argument of the given complex number is 60 degrees and we can have 6 points in the argand plane when we divide the whole 360* equally as 60*....Hence Six Elements in set A......The Thing to notice is that the nth roots of unity are divide the whole 360 degrees into n pieces using n points